Physics GRE Solutions Omnibus

Senior Editor: Taylor Faucett

Editor-in-Chief: Taylor Faucett

Associate Editor: Taylor Faucett

Editorial Assistant: Taylor Faucett

Art Studio: Taylor Faucett

Art Director: Taylor Faucett

Cover Design: Taylor Faucett

BiBTeX:

@BookFaucettOmnibus,

author = Taylor Faucett,

title = Physics GRE Solutions Omnibus,

pages = 1-560,

year = 2010,

edition = first,

c© 2010 by Taylor Faucett, all rights reserved.

The “Physics GRE Solutions Omnibus” is in no way affiliated with ETS or the GRE.All media and information contained within is to be used strictly for educational and non-profitpurposes. Any reproductions of this work, whether in part or in full, will be allowed provided

that proper attribution is given. Additionally, contacting me about its use would be a nice gesture.

Contents

1 Introduction 131.1 About the Omnibus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 PGRE8677 Solutions 152.1 PGRE8677 #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 PGRE8677 #2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 PGRE8677 #3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.4 PGRE8677 #4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.5 PGRE8677 #5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.6 PGRE8677 #6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.7 PGRE8677 #7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.8 PGRE8677 #8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.9 PGRE8677 #9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.10 PGRE8677 #10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.11 PGRE8677 #11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.12 PGRE8677 #12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.13 PGRE8677 #13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.14 PGRE8677 #14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.15 PGRE8677 #15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.16 PGRE8677 #16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.17 PGRE8677 #17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.18 PGRE8677 #18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.19 PGRE8677 #19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.20 PGRE8677 #20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.21 PGRE8677 #21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.22 PGRE8677 #22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.23 PGRE8677 #23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.24 PGRE8677 #24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.25 PGRE8677 #25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482.26 PGRE8677 #26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.27 PGRE8677 #27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.28 PGRE8677 #28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 532.29 PGRE8677 #29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542.30 PGRE8677 #30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552.31 PGRE8677 #31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562.32 PGRE8677 #32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582.33 PGRE8677 #33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

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2.34 PGRE8677 #34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 602.35 PGRE8677 #35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 622.36 PGRE8677 #36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 642.37 PGRE8677 #37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662.38 PGRE8677 #38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 682.39 PGRE8677 #39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 692.40 PGRE8677 #40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702.41 PGRE8677 #41 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 712.42 PGRE8677 #42 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 722.43 PGRE8677 #43 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 742.44 PGRE8677 #44 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 762.45 PGRE8677 #45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 782.46 PGRE8677 #46 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 792.47 PGRE8677 #47 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 802.48 PGRE8677 #48 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 812.49 PGRE8677 #49 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 822.50 PGRE8677 #50 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 832.51 PGRE8677 #51 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 842.52 PGRE8677 #52 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 852.53 PGRE8677 #53 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 862.54 PGRE8677 #54 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 872.55 PGRE8677 #55 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 882.56 PGRE8677 #56 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 892.57 PGRE8677 #57 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 902.58 PGRE8677 #58 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 912.59 PGRE8677 #59 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 922.60 PGRE8677 #60 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 932.61 PGRE8677 #61 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 952.62 PGRE8677 #62 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 962.63 PGRE8677 #63 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 982.64 PGRE8677 #64 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 992.65 PGRE8677 #65 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1002.66 PGRE8677 #66 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1012.67 PGRE8677 #67 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1032.68 PGRE8677 #68 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1042.69 PGRE8677 #69 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1052.70 PGRE8677 #70 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1072.71 PGRE8677 #71 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1092.72 PGRE8677 #72 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1112.73 PGRE8677 #73 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1122.74 PGRE8677 #74 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1142.75 PGRE8677 #75 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1152.76 PGRE8677 #76 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1172.77 PGRE8677 #77 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1192.78 PGRE8677 #78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1212.79 PGRE8677 #79 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1222.80 PGRE8677 #80 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1232.81 PGRE8677 #81 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

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2.82 PGRE8677 #82 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1262.83 PGRE8677 #83 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1272.84 PGRE8677 #84 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1282.85 PGRE8677 #85 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1292.86 PGRE8677 #86 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1302.87 PGRE8677 #87 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1312.88 PGRE8677 #88 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1322.89 PGRE8677 #89 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1332.90 PGRE8677 #90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1342.91 PGRE8677 #91 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1362.92 PGRE8677 #92 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1372.93 PGRE8677 #93 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1382.94 PGRE8677 #94 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1402.95 PGRE8677 #95 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1422.96 PGRE8677 #96 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1432.97 PGRE8677 #97 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1452.98 PGRE8677 #98 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1472.99 PGRE8677 #99 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1482.100PGRE8677 #100 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

3 PGRE9277 Solutions 1513.1 PGRE9277 #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1523.2 PGRE9277 #2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1533.3 PGRE9277 #3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1543.4 PGRE9277 #4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1553.5 PGRE9277 #5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1563.6 PGRE9277 #6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1583.7 PGRE9277 #7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1603.8 PGRE9277 #8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1613.9 PGRE9277 #9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1623.10 PGRE9277 #10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1643.11 PGRE9277 #11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1663.12 PGRE9277 #12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1683.13 PGRE9277 #13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1703.14 PGRE9277 #14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1713.15 PGRE9277 #15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1723.16 PGRE9277 #16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1733.17 PGRE9277 #17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1743.18 PGRE9277 #18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1763.19 PGRE9277 #19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1773.20 PGRE9277 #20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1783.21 PGRE9277 #21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1803.22 PGRE9277 #22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1813.23 PGRE9277 #23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1823.24 PGRE9277 #24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1833.25 PGRE9277 #25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1843.26 PGRE9277 #26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1853.27 PGRE9277 #27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

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3.76 PGRE9277 #76 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2543.77 PGRE9277 #77 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2553.78 PGRE9277 #78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2563.79 PGRE9277 #79 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2573.80 PGRE9277 #80 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2583.81 PGRE9277 #81 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2593.82 PGRE9277 #82 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2613.83 PGRE9277 #83 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2633.84 PGRE9277 #84 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2653.85 PGRE9277 #85 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2663.86 PGRE9277 #86 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2683.87 PGRE9277 #87 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2693.88 PGRE9277 #88 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2713.89 PGRE9277 #89 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2723.90 PGRE9277 #90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2733.91 PGRE9277 #91 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2743.92 PGRE9277 #92 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2753.93 PGRE9277 #93 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2763.94 PGRE9277 #94 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2773.95 PGRE9277 #95 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2783.96 PGRE9277 #96 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2793.97 PGRE9277 #97 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2813.98 PGRE9277 #98 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2823.99 PGRE9277 #99 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2833.100PGRE9277 #100 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

4 PGRE9677 Solutions 2854.1 PGRE9677 #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2864.2 PGRE9677 #2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2884.3 PGRE9677 #3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2904.4 PGRE9677 #4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2924.5 PGRE9677 #5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2954.6 PGRE9677 #6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2974.7 PGRE9677 #7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3004.8 PGRE9677 #8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3024.9 PGRE9677 #9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3034.10 PGRE9677 #10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3054.11 PGRE9677 #11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3064.12 PGRE9677 #12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3074.13 PGRE9677 #13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3094.14 PGRE9677 #14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3114.15 PGRE9677 #15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3134.16 PGRE9677 #16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3154.17 PGRE9677 #17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3164.18 PGRE9677 #18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3184.19 PGRE9677 #19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3204.20 PGRE9677 #20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3214.21 PGRE9677 #21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323

7

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8

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4.70 PGRE9677 #70 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3874.71 PGRE9677 #71 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3884.72 PGRE9677 #72 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3904.73 PGRE9677 #73 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3914.74 PGRE9677 #74 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3924.75 PGRE9677 #75 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3934.76 PGRE9677 #76 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3944.77 PGRE9677 #77 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3954.78 PGRE9677 #78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3974.79 PGRE9677 #79 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3984.80 PGRE9677 #80 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4004.81 PGRE9677 #81 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4024.82 PGRE9677 #82 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4034.83 PGRE9677 #83 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4054.84 PGRE9677 #84 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4064.85 PGRE9677 #85 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4074.86 PGRE9677 #86 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4094.87 PGRE9677 #87 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4114.88 PGRE9677 #88 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4124.89 PGRE9677 #89 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4144.90 PGRE9677 #90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4164.91 PGRE9677 #91 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4174.92 PGRE9677 #92 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4184.93 PGRE9677 #93 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4194.94 PGRE9677 #94 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4204.95 PGRE9677 #95 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4214.96 PGRE9677 #96 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4234.97 PGRE9677 #97 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4244.98 PGRE9677 #98 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4264.99 PGRE9677 #99 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4274.100PGRE9677 #100 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428

5 PGRE0177 Solutions 4295.1 PGRE0177 #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4305.2 PGRE0177 #2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4325.3 PGRE0177 #3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4345.4 PGRE0177 #4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4355.5 PGRE0177 #5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4365.6 PGRE0177 #6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4375.7 PGRE0177 #7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4385.8 PGRE0177 #8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4395.9 PGRE0177 #9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4405.10 PGRE0177 #10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4415.11 PGRE0177 #11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4425.12 PGRE0177 #12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4435.13 PGRE0177 #13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4455.14 PGRE0177 #14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4465.15 PGRE0177 #15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447

9

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10

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5.64 PGRE0177 #64 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5035.65 PGRE0177 #65 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5045.66 PGRE0177 #66 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5055.67 PGRE0177 #67 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5065.68 PGRE0177 #68 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5075.69 PGRE0177 #69 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5085.70 PGRE0177 #70 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5095.71 PGRE0177 #71 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5105.72 PGRE0177 #72 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5115.73 PGRE0177 #73 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5125.74 PGRE0177 #74 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5135.75 PGRE0177 #75 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5145.76 PGRE0177 #76 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5155.77 PGRE0177 #77 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5165.78 PGRE0177 #78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5175.79 PGRE0177 #79 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5185.80 PGRE0177 #80 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5195.81 PGRE0177 #81 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5205.82 PGRE0177 #82 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5215.83 PGRE0177 #83 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5225.84 PGRE0177 #84 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5245.85 PGRE0177 #85 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5255.86 PGRE0177 #86 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5265.87 PGRE0177 #87 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5285.88 PGRE0177 #88 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5295.89 PGRE0177 #89 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5305.90 PGRE0177 #90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5325.91 PGRE0177 #91 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5335.92 PGRE0177 #92 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5355.93 PGRE0177 #93 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5375.94 PGRE0177 #94 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5395.95 PGRE0177 #95 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5405.96 PGRE0177 #96 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5425.97 PGRE0177 #97 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5435.98 PGRE0177 #98 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5455.99 PGRE0177 #99 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5465.100PGRE0177 #100 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 547

6 Appendix 5486.1 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548

6.1.1 Classical Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5486.1.2 Electricity & Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5506.1.3 Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5526.1.4 Thermodynamics & Stat Mech . . . . . . . . . . . . . . . . . . . . . . . . . . 5536.1.5 Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5556.1.6 Special Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5566.1.7 Electronics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5576.1.8 Special Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558

11

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6.2 Units and Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5596.2.1 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5596.2.2 Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5606.2.3 SI Prefixes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560

12

Chapter 1

Introduction

13

1.1. ABOUT THE OMNIBUS CHAPTER 1. INTRODUCTION

1.1 About the Omnibus

The “Physics GRE Solutions Omnibus” is a compilation of 400 solutions to the 4 Physics GREpractice exams. In addition to the solutions, a section of useful equations, units and conversionsworth memorizing has been added.

A website has also been created, http://PhysicsGrad.com, as a companion to the book.PhysicsGrad.com is a community based site that differs from “The Omnibus” in that all 400problems feature both my solutions and solution contributions from registered members. Addition-ally, the website offers updates on University and departmental events and admissions, help withselecting universities and advice on test taking.

I gladly welcome any and all corrections and/or advice that any Omnibus reader happens tofind, whether it be typos, poor explanations or even outright falsehoods. The easiest way to contactme is via email through [email protected].

14

http://PhysicsGrad.com

mailto:[email protected]

Chapter 2

PGRE8677 Solutions

15

2.1. PGRE8677 #1 CHAPTER 2. PGRE8677 SOLUTIONS

2.1 PGRE8677 #1

Recommended Solution

(A) In a frictionless environment, this would be true. However, note that the problem tells us thatfriction is not negligible (in fact it is proportional to the rocks velocity −v) so it isn’t possiblefor the acceleration to always be g.

(B) Since the frictional force is always fighting the rocks motion proportional to its speed, the onlypoint at which frictional forces disappear will be when the rock stops moving (v = 0) at thetop of its flight. At this point, the only acceleration on the rock will then be the accelerationdue to gravity.

(C) As we just demonstrated in (B), the acceleration can be g at the top of the rocks flight

(D) In a frictionless environment, kinetic energy at the bottom of the rocks path will be convertedto potential at the peak of its flight. The rock will then head back to its initial point convertingall potential back to kinetic energy. However, in an environment with non-negligible friction,some of that initial kinetic energy is lost to the surroundings through heat, sound, etc and sothe initial energy can never be recovered at the end of the rocks flight.

(E) Even in an ideal (i.e. frictionless) environment, this can’t be true. This could only be true ifthere was some force applied to the rock on the way down which is not the case.

Correct Answer

(B)

Alternate Solution

Sum the forces in the vertical direction for the rocks motion,

16


Ftot = −Ffric − FG (2.1)

Ftot = −kv −mg (2.2)

(A) From the equation, we can see that it is not generally true that the total acceleration on therock is simply g

(B) Plug in v = 0 and g becomes the only acceleration and so this is correct

(C) As demonstrated in (B), acceleration is not less than g when v = 0

(D) This can’t be true because energy is dissipated from friction so the final velocity can’t be equalto its initial velocity

(E) Conservation of energy won’t allow you to have more velocity in the end than you startedout with and we would expect to have a lower final velocity than initial velocity based ondissipation of energy from friction

Correct Answer

(B)

17


2.2 PGRE8677 #2


We can first eliminate (D) because we would need a continuous force pushing the satellite awayfrom the Earth to get an orbit spiraling outward. We can then eliminate (E) because oscillationsarent going to occur unless the potential energy of the satellite remains the same after the force isapplied, which isnt the case. Next, eliminate (C) because, as a general rule, when we perturb anorbit with a brief bit of thrust, the satellite will return to the same point on the next pass but therest of its orbit will be altered, so it cant be circular again. Finally, when choosing between (A)and (B), a small energy change from a circular orbit is more likely to become an elliptical orbitand, in general, elliptical orbits are more common for lower energy orbits, like a satellite orbitingearth.

Correct Answer

(A)

18


2.3 PGRE8677 #3


Recall from quantum mechanics (specifically from the relativistic de Broglie relations) that wavespeed is

vp =E

p(2.3)

=γmc2

γmv(2.4)

=c2

v(2.5)

The velocity of light through a medium is related to its relative permittivity and relative per-meability by

v =1

εµ(2.6)

Plug this in and get

vp =c2

εµ=

c√2.1

(2.7)

Correct Answer

(D)

19


Alternate Solution

It is generally true that only particles with mass (i.e. not a photon) will have phase velocities fasterthan the speed of light. Checking the possibilities

(A)√

3.1c ≈ 1.7c

(B)√

2.1c ≈ 1.4c

(C) c/√

1.1 ≈ 0.95c

(D) c/√

2.1 ≈ 0.7c

(E) c/√

3.1 ≈ 0.56c

From this you can comfortably eliminate (A) and (B) and cautiously eliminate (C) because it isnearly the speed of light. With (D) and (E), ask yourself whether it makes more sense for thepermittivity and permeability to be additive (i.e. 1.0 + 2.1) or multiplicative (i.e. (1.0)(2.1)). Evenif you can’t recall that the velocity of an electromagnetic wave is the inverse product of the twovalues, you could at least make an educated guess by recalling that every time you have ever seenthe permittivity of free space (ε0), it has always acted as a scaling value. For example in theGravitational Potential equation

UG =1

4πε0

mM

r(2.8)

Correct Answer

(D)

20


2.4 PGRE8677 #4


The general equation of a traveling wave is

y(x, t) = Asin

(2π

λ

)(x± vt) (2.9)

Where A is the amplitude, T is the period, λ is wavelength, t is time and x is position.

(A) The amplitude in this instance is A. In order to get an amplitude of 2A, we would need ourequation to have the amplitdue, A, multiplied by 2

(B) From the general traveling wave equation, if the ± ends up being a negative then the wavepropagates in the positive direction and the converse is also true.

(C) The period is T = 1/f and propagation velocity is v = fλ so T = λ/v not T = T/λ. Also,T/λ would give the wrong units for the period.

(D) Propagation speed is v = fλ and the period T = 1/f so v = λ/T , not x/t.

(E) Propagation speed is v = fλ and the period T = 1/f so v = λ/T

Correct Answer

(E)

21


2.5 PGRE8677 #5


This problem asks us about an inelastic collision so we can’t assume that energy is conservedbetween the initial and final stages of the swinging spheres. Break the problem up into 3 separatetime frames with the first phase involving just the falling of sphere A. In this phase, the initialpotential energy is equal to the kinetic energy immediately before the two spheres collide

Mgh0 =1

2Mv2

0 (2.10)

so the velocity before impact is

v0 =√

2gh0 (2.11)

In the second phase, consider the collision of the ball, which should conserve momentum

Mvo = (M + 3M)vf = 4Mvf (2.12)

combine Equation 2.11 and Equation 2.12 to get

M(√

2gh0

)= 4Mvf (2.13)

so the final velocity of the spheres after they’ve combined is

22


vf =

√2gh0

16=

√gh0

8(2.14)

finally, in the third phase, we know that the kinetic energy of the combined spheres immediatelyafter collision will be equal to the potential energy at their peak

1

2(4M)v2

f = (4M)ghf (2.15)

Combine Equation 2.14 and 2.15 to get

1

2

√gh0

8

2

= ghf (2.16)

hf =1

16h0 (2.17)

Correct Answer

(A)

23


2.6 PGRE8677 #6


Note that as x → ∞, the track becomes infinitely steep and the particle should free fall at anacceleration of g. Only (D) approaches g in this limit.

Correct Answer

(D)

Alternate Solution

Consider the particle at some point on the curve and break it up into its components (Figure 2.1)The tangential acceleration is going to be the component of the acceleration due to gravity

multiplied by a unit vector in the same direction as the tangent

at = ~g · t (2.18)

24


O

dy

dx

at

θ

x

y

Figure 2.1: Velocity components on a particle as it moves through an arc

since we are talking about a simple unit vector, take the dot product of the two and get

at = gcos(θ) (2.19)

and with the angle θ, we get

at = gcos(θ) = gdy

at= g

dy√dx2 + dy2

(2.20)

recall that

dy

dx=

d

dx

(x2

4

)=x

2(2.21)

so Equation 2.20 becomes

at =gy′√

1 + (y′)2=

gx√4 + x2

(2.22)

Correct Answer

(D)

25


2.7 PGRE8677 #7


Draw out your force diagram with the applied force shown pulling with 10N to the right and atension force with an x component in the opposite direction of the applied force and a y componentalong the dotted line (pointing up). When the box is in equilibrium, acceleration will be 0 so weare concerned with the point at which the applied force of 10N is equal to the opposing tensionforce

FT−x = 10N (2.23)

Take tan(θ) to get

tan(θ) =FT−xFT−y

(2.24)

and re-arrange it to get

FT−x = FT−y tan(θ) (2.25)

The force in the vertical direction will only be due to gravity so we can use FT−y = 20N as anapproximation. Plug this into Equation 2.25 and then combine FT−x with Equation 2.23 to get

10N = (20N) tan(θ) (2.26)

θ = tan−1(

1

2

)(2.27)

26


Correct Answer

(A)

27


2.8 PGRE8677 #8


Using our kinematic equation

v2 = v20 + 2a∆x (2.28)

and assuming the initial velocity is zero, solve for the acceleration

(10 m/s)2 = 2a(0.025 m) (2.29)

a = 2000 m/s2 (2.30)

With a mass of 5 kg, use Newton’s second law to get

F = (5 kg)(2000 m/s2) = 10, 000 N (2.31)

Correct Answer

(D)

Alternate Solution

The kinetic energy of the stone right at impact will be

EK =1

2mv2 =

1

2(5 kg)(10 m/s)2 = 250 J (2.32)

This calculation doesn’t account for the small potential energy the stone will have once it hashit the nail and it also makes the assumption (which the problem doesn’t necessarily give us) thatenergy is being conserved. However, as an approximation, let’s assume that all energy is accountedfor and conserved so we can let all the kinetic energy translate into the work done when movingthe nail

W = F ·∆x = 250 J (2.33)

F =250 J

0.025 m= 10, 000 N (2.34)

28


Correct Answer

(D)

29


2.9 PGRE8677 #9


The drift velocity of a charged particle due to an electric field is

vd =i

nqA(2.35)

where i is the current, n is the number of particles, q is the charge of the particle and A is thecross sectional area. We could calculate the value exactly but it is generally quicker and sufficient,especially with the wide range of choices, to do a quick approximation.

vd =100 A

(1× 1028 e−/m3)(2× 10−19 C)(3× 10−3 m2)≈ 2× 10−5 (2.36)

which is closest to (D)

Correct Answer

(D)

30


2.10 PGRE8677 #10


From Gauss’s law, we have ∮SE · dA =

Q

ε0(2.37)

For a sphere, the surface area dA will be 4πr2 and the charge volume Q will be the volume ofa sphere, Q = 4/3 πr3 . Plug things in to get

E(4πr2

)=

(4

3ε0πr3

)(2.38)

E =r

3ε0(2.39)

so the curve must be linear, i.e. (C)

Correct Answer

(C)

31


2.11 PGRE8677 #11


You should recall from quantum mechanics (especially if you were forced to do the rigorous prooffrom Griffith’s canonical quantum mechanics textbook) that the divergence of the curl of any vectorfield is always 0. Take the divergence of each side to get

∇ ·(D + J

)= ∇ · (∇×H) (2.40)

∇ ·(D + J

)= 0 (2.41)

Correct Answer

(A)

32


2.12 PGRE8677 #12


Without knowing much about the physics involved, we can solve this problem with a bit of musicalknowledge. From the Doppler effect, we know that pitch (frequency) goes up as a source of soundmoves toward you, eliminating (A) and (B). Next, recall that the next octave of any tone is twicethat of the original, meaning that (C) and (D) would both be suggesting that even at near the speedof light the source hasn’t even exceeded an octave in sound. (E) is the most reasonable solution.

Correct Answer

(E)

Alternate Solution

If you recall the equation for the frequency of a moving sound wave, you can calculate the exactsolution

f =

[1

1± vsourcevwave

]f0 (2.42)

Plug everything into Equation 2.42 to get

f =

[1

1− 0.9

]f0 = 10f0 (2.43)

Correct Answer

(E)

33


2.13 PGRE8677 #13


We can eliminate (A) and (B) because we know that the interference pattern should be affected, atthe very least we know this is the case for the interference extrema as the relative phase changes.Eliminate (C) because the interference pattern certainly wont be destroyed for all but two points.Finally, eliminate (D) because the monochromacity is not destroyed and, even if it stopped beingmonochromatic, that wouldnt stop an interference pattern. Alternatively, instead of doing processof elimination you can directly figure that (E) is correct because the Flicker-Fusion Threshold (Thefrequency at which an objects movement starts to become imperceptible to the human eye) occursat roughly 60 Hz, which is a far cry from the 500 Hz given in the problem.

Correct Answer

(E)

34


2.14 PGRE8677 #14


When you hear ideal gas law, always go straight to the ideal gas law equation (the ‘Piv-nert’equation)

PV = nRT (2.44)

(A) Recall from the first law of thermodynamics that work done on the system is, W = PdV .Combining this with the ideal gas law shows that (A) is true.

(B) Recall that Cv =∣∣∣dQdT ∣∣∣v and Cp =

∣∣∣dQdT ∣∣∣p. If (B) was correct, then∣∣∣dQdT ∣∣∣v =

∣∣∣dQdT ∣∣∣p = Cp = Cv

which the initial question clearly tells us is not true.

(C) This isn’t necessarily true, especially with changes in volume.

(D) Recall that Cv =∣∣∣dUdT ∣∣∣ which tells us that the energy of an ideal gas is only dependent on

temperature.

(E) Presumably the correct answer will never be an ill-formed question (or it won’t be graded ifit is) so exclude (E)

Correct Answer

(E)

35


2.15 PGRE8677 #15


If we had no atoms in the box (N = 0) then the probability of not finding an atom would be 1.From this, eliminate (A) and (D). When we have only 1 atom in the box (N = 1) it would be veryunlikely that the single atom would occupy the specific volume of 1.0×10−6 cm3 so the probabilityof not finding it there should be close to 1 but not exactly 1. From this, eliminate (B) and (E).

Correct Answer

(C)

Alternate Solution

If we were to break the box up into small cubes, each with a volume of 1.0× 10−6 cm3, we wouldsee that the total numbers of cubes in the container is

cubes

container=

1.0× 10−6 cm3

1 cm3= 106 cubes (2.45)

Assuming that the probability of being in any given cube in the container is equally likely wewould expect the probability of finding 1 atom in a single cube to be

P1 atom =1

106(2.46)

Keep in mind, however, that this is the probability of finding the atom inside a single smallcube. What we want is the probability that we DON’T find the atom inside that cube. From this,we know that the probability of the atom being outside the cube will be

Pout = 1− Pin (2.47)

Pout = 1−(10−6

)(2.48)

Extrapolating this to multiple atoms, we use the multiplication rule to get

Pout =(1− 10−6

)n(2.49)

Correct Answer

(C)

36


2.16 PGRE8677 #16


The muon is a lepton/fermion (spin-1/2) and has charge of -1

(A) electron Electrons are leptons/fermions (spin-1/2) and have a charge of -1. They differ from

the muon in that their mass is about (1/200th) that of a muon. Muons are are frequentlyreferred to as heavy electrons because of their similar properties.

(B) graviton You could probably immediately throw this option out because we aren’t evenparticularly sure that these exist. However, if it exists, the graviton should be massless, haveno charge, have spin-2 and is a force particle (boson).

(C) photon Massless, no charge (at least for our purposes), spin-1 and a force particle (boson).

(D) pion Not an elementary particle, depending on which π −meson you get spin (+1, -1 or 0),depending on which π −meson you get charge (0, +e, -e).

(E) proton Not an elementary particle (hadron) and has a charge of +1.

Correct Answer

(A)

37


2.17 PGRE8677 #17


The problem tells us that we need to get from AZX to A−4

Z−1Y in two steps,

AZX →?

? X′ →?

? X′′ →A−4

Z−1 Y (2.50)

in β− decay, we have

AZX →A

Z+1 X′ + e− + ve (2.51)

for α decay,

AZX →A−4

Z−2 X′ + α (2.52)

and for γ decay,

AZX →A

Z+1 X′ + γ (2.53)

From these, it should be clear that we can get our solution by the combination

AZX → β− decay→ α decay→A−4

Z−1 Y

AZX →A

Z+1 X′ →A−4

Z−2 X′′ →A−4

Z−1 Y

From this, it should be clear that the only two step process which will give us

Correct Answer

(A)

38


2.18 PGRE8677 #18


Recall the time-independent Schrodinger equation (Equation 2.54)

Eψ(x) = − h2

2m∇2ψ(x) + V (x)ψ(x) (2.54)

take the second derivative of ψ(x) to get

ψ′(x) = −b2xψ(x) (2.55)

ψ′′(x) = b4x2ψ(x) (2.56)

plug everything in to get

Eψ(x) = − h2

2M

[b4x2ψ(x)

]+ V (x)ψ(x) (2.57)

and simplify Equation 2.57 to get

E = − h2b4x2

2M+ V (x) (2.58)

Plug x = 0 into Equation 2.58 to use the condition, V (0) = 0,

E = − h2b4(0)2

2M+ V (0) = 0 (2.59)

Which tells us that all of the energy we found previously is accounted for entirely by V (x),

39


V (x) =h2b4x2

2M(2.60)

Additionally, you can also get to this conclusion by realizing that E shouldn’t have any depen-dence on x but that V (x) should, so the only term we have must be accounted for by V (x).

Correct Answer

(B)

Alternate Solution

We can immediately eliminate (A) because it lacks the necessary dependence on x. Next, wecan eliminate (C) because the wavefunction (ψ) given makes our potential a quantum harmonicoscillator which must have a dependence on x2 rather than x4. Next, eliminate (D) because theenergy, E, must have some dependence on the mass of the particle. This will get you to the pointat which you can either guess between the two solutions or do enough of the time independentSchrdinger equation to see some dependence on x emerge.

Correct Answer

(B)

40


2.19 PGRE8677 #19


Recall from the Bohr Model of the Hydrogen atom that the total energy of the atom in relation ton is

Etot =−13.6 eV

n2(2.61)

which agrees with (E) when A = −13.6 eV.

Correct Answer

(E)

41


2.20 PGRE8677 #20


Recall Einstein’s famous equation for energy with a resting mass

E = mc2 (2.62)

and his lesser known equation of relativistic momentum

E = P0c4 +m0c

2 (2.63)

In the problem, the proton will need to use the resting mass energy equation and the kaon usesthe relativistic momentum equation, giving

mpc2 = Pkc

4 +mkc2 (2.64)

mp = Pkc2 +mk (2.65)

We could plug in the exact values given, however we can simplify the calculations by usingmasses K+ = 500MeV/c2 and P+ = 1000MeV/c2, giving

(1000 MeV/c2)2 = Pkc2 + (500 MeV/c2)2 (2.66)

Pk = 750, 000 MeV/c2 (2.67)

Recall that the relativistic momentum is

Pk =m0v√

1− v2/c2(2.68)

substituting that in to our previous equation gives

m0v√1− v2/c2

= 750, 000 MeV/c2 (2.69)

you can get the velocity by itself by moving the denominator on the LHS to the RHS, squaringboth sides and then grouping terms to get

42


V 2k =

3

4c2 (2.70)

Vk =

√3

4c (2.71)

Vk ≈ 0.86c (2.72)

Correct Answer

(E)

43


2.21 PGRE8677 #21


The equation for the space time interval is

∆s2 = ∆r2 − c2∆t2 (2.73)

but since the problem specifies that c = 1, Equation 2.73 becomes

∆s2 = ∆r2 −∆t2 (2.74)

Recall the fantastically cool fact that the Pythagorean theorem works the same in any numberof dimensions, so the length ∆r is

∆r2 = ∆x2 + ∆y2 + ∆z2 (2.75)

= 22 + 02 + 22 (2.76)

= 8 (2.77)

The time difference is ∆t = 2 so

∆s2 = 8− 4 (2.78)

∆s = 2 (2.79)

Correct Answer

(C)

44


2.22 PGRE8677 #22


Recall the electromagnetic field tensors

Fµν =

0 −Ex/c −Ey/c −Ez/c

Ex/c 0 −Bz ByEy/c Bz 0 −BxEz/c −By Bx 0

Which clearly demonstrates that both an electric field and/or a magnetic field may exist. This

also happens to contradict (A), (C), and (D).

Correct Answer

(B)

45


2.23 PGRE8677 #23


Most of these options can be analyzed by knowing that copper is a good conductor and silicon isgenerally used as a semi-conductor

(A) It is generally true that the conductivity is higher in magnitude for a conductor than a semi-conductor.

(B) Conductivity and temperature are inversely related for a conductor. In this specific option asthe temperature goes up we would expect the conductivity to go down.

(C) For semi-conductors, as temperature increases, conductivity increases and vice versa.

(D) It is generally true that impurities in a conductor will decrease conductivity.

(E) The addition of impurities to a semi-conductor, commonly called doping, is done to increaseconductivity rather than decrease conductivity. This is the only false statement of the bunch.

Correct Answer

(E)

46


2.24 PGRE8677 #24


Recall Kirchhoff’s voltage loop law which tells us that the sum of all voltages about a completecircuit must sum to 0 ∑

Vk = 0 (2.80)

Recalling that V = IR, sum the voltage of the battery, generator and resistor

G− V − VR = 0 (2.81)

However, since there is an internal resistance of 1 Ω, VR = (R+ 1 Ω)I

G− V − (R+ 1 Ω)I = 0 (2.82)

120 volts− 100 volts−R(10 amps)− 10 volts = 0 (2.83)

10 volts−R(10 amps) = 0 (2.84)

R = 1 Ω (2.85)

Correct Answer

(C)

47


2.25 PGRE8677 #25


From the lorentz force, we know that the magnetic field vector maximizes force when it is orthogonalto the electric field

F = q ~E +(qv × ~B

)(2.86)

= q ~E + qv ~Bsin(θ) (2.87)

Initially, the particle is at rest so the initial velocity is v = 0 and only the electric field willprovide any force. Once the particle is in motion, however, the velocity is non-zero and will bemoving in the direction of the electric field. Since this problem explicitly states that the electricfield and magnetic field are parallel, we know that v and B are also parallel and thus, our angle isθ = 0

F = q ~E + qv ~Bsin(0) = q ~E (2.88)

so we would only expect the electric field to influence the particle and for it to provide forcesolely in the direction of the field.

Correct Answer

(E)

48


2.26 PGRE8677 #26


Henry Moseley discovered, from the Bohr model, that when an external electron strikes and knocksloose one of the innermost electrons, a gap is created which allows an electron from the next Bohrlevel (n = 2) to fall into its place. This transition is known as a K series transition. In general, alltransitions from some Bohr level n ≥ 2 down to n = 1 is one of the K series transitions (Figure 2.2)

n=2

n=1

n=3

n=4n=5

Kα

Kβ

Kγ

Kδ

M - series

N - series

Lα

Lβ

Lγ

Mα

Mβ

L - series

K - series

Figure 2.2: K, L and M series electron transitions

The specific transition we are concerned with (i.e. the minimum energy transition) is that ofKα so we can use Moseley’s law

E = Re (Z − β)2

(1

n2f

− 1

n2i

)(2.89)

where Re is the Rydberg energy at Re = 13.6 eV and β corresponds to the transition type (K,L, M , N , . . .). To make things quicker, round all of your numbers off to get

49


E = (15 eV ) (30)2(

1

1− 1

4

)(2.90)

E = 10, 125 (2.91)

E ≈ 10, 000 (2.92)

Correct Answer

(D)

50


2.27 PGRE8677 #27


The solution to this answer should stick out like a sore thumb. Without knowing any informationabout any of the other possible answers, you should immediately know that any particle with acharge (like an electron) will experience deflection under a uniform magnetic field regardless ofwhether it has spin or not. The Lorentz force exhibits precisely this fact,

F = q ~E + q(~v × ~B

)(2.93)

where q is the particles charge. In fact, it wasn’t until we realized that neutral particles couldexperience deflection (I’m speaking of the Stern-Gerlach experiment)that we realized we had toincorporate particle spin into the mix.

Correct Answer

(D)

Alternate Solution

(A) Recall that electron spin, along with the Pauli exclusion principle, determined the arrangementof electrons in energy level diagrams and, from that, influences our understanding of thestructure of the periodic table.

(B) Einstein and Debye models of specific heat both eventually incorporated electron contributioninto their descriptions which necessitates a discussion on spin.

(C) The Zeeman effect is a method of distinguishing between electrons with the same energylevel by instituting a magnetic field to influence them to different energies. The “anomalousZeeman effect” appeared in all instances when the net spin of a grouping of electrons wasn’t0. This effect wasn’t understood until electron spin was understood, hence why it was called“anomalous”.

(D) Electron deflection could be explained using only the Lorentz force and its dependence on theelectrons charge.

(E) The “Gross” atomic structure describes the splitting of line spectra without factoring in theeffects of electron spin. “Fine” atomic structure was a correction to this model which factoredin spin and relativistic effects.

51


Correct Answer

(D)

52


2.28 PGRE8677 #28


The condition for normalizing a wavefunction is∫ ∣∣∣ψ2(x)∣∣∣ = 1 (2.94)

for this specific scenario, our wavefunction is for a rigid rotator with dependence on φ in whichthe dumbbell rotates from φ = −π to φ = π∫ π

−π

∣∣∣ψ2(φ)∣∣∣ =

∫ π

−πA2e2imφ = 1 (2.95)

Recall from doing Fourier series approximations that the RHS of Equation 2.95 is equivalent to2πA2, giving us

2πA2 = 1 (2.96)

A =1√2π

(2.97)

Correct Answer

(D)

53


2.29 PGRE8677 #29


Use Fleming’s left hand rule (Figure 2.3) to figure out the direction of force relative to a movingcharge

Figure 2.3: Fleming’s left hand rule

Using the wire in the image as the wire in the problem, you should see that we want to rotatethe disembodied left hand such that the force vector (thumb) is pointing in the same direction asthe wire. Doing so results in the current vector (middle finger) pointing towards the wire.

Correct Answer

(A)

54


2.30 PGRE8677 #30


(A) This could only be true if all subshells (S, P , D) were completely full, which isn’t the case.

(B) Since the valence shell is 4s1, it should be clear that this single electron doesn’t fill out thesubshell.

(C) We are only concerned with angular quantum number l = 2 which corresponds to s, not l = 4which corresponds to g. Why does ETS keep asking about the wrong quantum numbers?

(D) You can figure that this isn’t correct by having the periodic table memorized, in which caseyou would either know that Potassium has atomic number 19 or that the element with atomicnumber 17 is chlorine. Alternatively, because the electron configuration is described for itsground state, just add up all of the electrons (superscripts) to get a total of 19 electrons and,therefore, 19 protons.

(E) Atoms with a single electron in its outer shell can be thought of as analogous to a hydrogenatom. From this (i.e. without doing any painful calculations) you can convince yourself thatthe charge distribution of Potassium will be similar to that of the hydrogen atom, in factspherical.

Correct Answer

(E)

55


2.31 PGRE8677 #31


This problem is testing your knowledge of physics history. Phillip Lennard experimented withessentially the same setup (Figure 2.4) in the early stages of the development of the photoelectriceffect. Lennard fired different frequencies of light at a photocell and collected the photoelectronsemitted. attached to his photocell was a variable power supply, a voltmeter and a micro-ammeter

As the light knocked electrons from the photocell, it would gain a positive charge. The secondplate also gained a positive charge because it was connected in the same circuit as the photocell.The recently freed electrons would then be attracted to the now positively charged second plate andwould flow to it. Finally, the electrons would move through the circuit back to the photocell andstart the process over, generating a current. The variable power supply was placed in this circuitsuch that it would fight the flow of electrons from moving back to the photocell (which occursbecause the negative end of the potential is attached to the end of the circuit not receiving light).At low potentials, low energy electrons would be caught and pushed back while higher electrons

Figure 2.4: Phillip Lennard’s experimental photocell arrangment

56


could flow through, meaning the micro-ammeter would get a decreased reading. Lennard discoveredthat he could keep increasing the potential of the power supply until he reached a potential thatwould completely cease electron movement. At this point, that is when the micro-ammeter firstreads 0, we’ve found the potential V on the collector. From this, eliminate all choices but (A) and(C). Lastly, you can determine that the potential must be negative because the potential of thepower supply is fighting the electron movement so it must be negative valued.

Correct Answer

(A)

57


2.32 PGRE8677 #32


(A) The photoelectric equation was derived well before quantum mechanics came around.

(B) The important aspect of the photoelectric effect was the discrete quantization of energy. De-termination of an electrons wavelength won’t get you there.

(C) Eliminate this choice because, aside from not being relevant, it isn’t even true.

(D) This is the answer we’ve been looking for. The important distinction that the photoelectriceffect gave us was that Energy is discrete and that it is dependent on the frequency of a lightsource, not the intensity.

(E) The photoelectric effect necessitates that we think of light as particles rather than waves.

Correct Answer

(D)

58


2.33 PGRE8677 #33


I think it’s worth noting that this question does some strong hinting, whether intentional or un-intentional, that the previous 2 questions can be solved without knowing much if anything aboutthe equation they gave you. In the photoelectric equation, W represents the work function (in myexperience, φ is more commonly used) and it represents the minimum amount of work (energy)necessary to pop an electron from a material. Thus, you can read the photoelectric equation fromleft to right as ”The total kinetic energy of an ejected electron (|eV |) is equal to the energy of thephoton that hit it (hν) minus the energy it required to pop the electron out in the first place (W )”.Reading it like this should convince you that really the photoelectric equation is nothing more thana conservation of energy argument.

Correct Answer

(D)

59


2.34 PGRE8677 #34


Recall that Work is force integrated over the area in which the force is applied and that work hasthe opposite sign of the potential energy (−W = U),

−W = U = −∫F · ds (2.98)

60


Since we are given the energy and asked to find force, differentiate your work equation to getthe force equation

d

dx(U) =

d

dx

(−Kx4

)(2.99)

F = −4kx3 (2.100)

Correct Answer

(B)

61


2.35 PGRE8677 #35


You should remember that the Hamiltonian is the sum of kinetic energy and potential energy (asopposed to the Lagrangian which is the difference between the two). From this you should get

H = T + V (2.101)

H =p2

2m+ kx4 (2.102)

62


In case you immediately think of T = 12mv

2 for the kinetic energy, take a second to re-mind/convince yourself that

1

2mv2 =

p2

2m(2.103)

Correct Answer

(A)

63


2.36 PGRE8677 #36


You should recall from any course that was heavy into mechanics, that the equations of motion fora system come out as a result of applying a minimization on the actions of the system, which isknown as the principle of least action. From this, you should recall that our equations of motion aregenerally always time dependent, not position dependent and you can eliminate (D) and (E). Asfor the other three, you will just have to remember that the principle of least action is the integralwith respect to time of the Lagrangian,

64


∫ t2

t1

1

2mv2 − kx4 (2.104)

Correct Answer

(A)

65


2.37 PGRE8677 #37


Consider that when the radius, r, goes to 0, the tension force should be just the force as a resultof an object hanging from a string, i.e. FT = mg. In this limit, only (D) and (E) are remaining.Next, compare the units in both to see that in (D) they are asking you to add a term with units ofm2/s2 with another term with units of m2/s4 and that ain’t happening.

Correct Answer

(E)

Alternate Solution

Draw out your force diagram (Figure 2.5),It should be clear that the tension force is

F 2T = (FT−x)2 + (FT−y)

2 (2.105)

66


l

FT - x

FT - y

FG

FT

Figure 2.5: Component forces on a mass rotating on a cord

The vertical component of tension will be equal and opposite that of the force from gravitybecause it won’t be accelerating in that direction,

FT−y = FG = mg (2.106)

The tension in the horizontal will just be FT−x and it will be equal to the centripetal accelerationdue to its rotation,

FT−x = mac = mω2r (2.107)

Plug Equations 2.106 and 2.107 into Equation 2.105 to get

F 2T =

(mω2r

)2+ (mg)2 (2.108)

FT = m√ω4r2 + g2 (2.109)

Correct Answer

(E)

67


2.38 PGRE8677 #38


Construct a truth table based on the conditions of the problem

Input 1 Input 2 Output

0 0 01 0 10 1 11 1 1

which is the truth table of an OR gate.Recall that the truth table for an AND gate is

Input 1 Input 2 Output

0 0 01 0 00 1 01 1 1

To recall these tables, simply remember that for an OR gate, it must be true that either Input1 OR Input 2 must be true for the output to be true. For an AND gate, however, it must be truethat input 1 AND input 2 are true for the output to be true.

Correct Answer

(A)

68


2.39 PGRE8677 #39


In the area of ω = 3 × 105 second−1, the line is roughly linear and decreasing. From this, we caneliminate any solution which shows an increase in gain with an increase in angular frequency, i.e.eliminate (B) and (C). Next, consider that we are talking about a log-log graph and (A) won’t giveus a linearly decreasing function with this type of scaling. Finally, between (D) and (E), we need tolook at the slope near ω = 3× 105 second−1. The horizontal and vertical axes are scaled with thesame step size so we can approximate the slope as moving down 2 steps and to the right roughly4 steps, giving a slope of roughly 2/4 or 1/2. From this, we see that the gain is dependent on thesquared value of the angular frequency

Correct Answer

(E)

69


2.40 PGRE8677 #40


The standard deviation for radioactive emission is described by the Poisson noise,

σ =√λ (2.110)

where λ is the average number of radioactive samples. To simplify things, let λ ≈ 10, 000 so

σ =√

10, 000 = 100 (2.111)

Correct Answer

(A)

70


2.41 PGRE8677 #41


The binding energy for an atom is the amount of energy required to strip an electron from theatom. We can immediately eliminate (D) and (E) because the binding energy of these atoms is solow that it decomposes on its own (i.e. they are both radioactive). Next, eliminate (B) becauseit doesn’t have a full valence shell and will generally be more willing to give up an electron thanatoms with completed shells. Finally, choose (C) because Iron has a much stronger nucleus bindingelectrons in their shells and it is generally true that as you move down the periodic table, bindingenergy increases.

Correct Answer

(B)

71


2.42 PGRE8677 #42


If you happen to recall the mean free path equation for the probability of stopping a particle movingthrough a medium, then this problem is a relatively straightforward plug-n-chug problem

P (x) = nσdx (2.112)

Where P (x) is the probability of stopping the particle in the distance dx, n is the number ofnuclei, σ is the scattering cross section and dx is the thickness of the scatterer. Plug everything into get

σ =P

ndx(2.113)

=

(1× 10−6 nuclei

)(1020 nuclei

cm3

)(0.1cm)

(2.114)

= 1× 10−25cm2 (2.115)

Correct Answer

(C)

Alternate Solution

In case you don’t recall the relevant equation, examine the units of all the information given to youto see if you can derive it on the spot. You are given,

Scatterer Thickness 0.1 cm

Nuclei 1020 nuclei/cm3

Quantity making it a given distance 1 nuclei1×106

72


Since we ultimately want to get a result with units of cm2, we will have to combine our 3 knownvalues as

Quantity making it a given distance

(Nuclei)(Scatterer Thickness)=

nuclei(nucleicm3

)(cm)

(2.116)

Plug in your values to get (C).

Correct Answer

(C)

73


2.43 PGRE8677 #43


Ideally, you should recognize that the frequency given in the problem as being that of a SimpleHarmonic Oscilattor (SHO)

ωSHO =1

2π

√k

m(2.117)

The simple harmonic oscillator involves a mass oscillating about a fixed point. Based on thisfact, we should expect the frequency for this linearly oscillating system of springs and massesto exhibit the SHO frequency when they are also moving about a fixed point. Of the potentialsolutions, only (B) describes masses A and C as moving about a fixed point, specifically mass B.

Correct Answer

(B)

Alternate Solution

In case you don’t recall the frequency of a SHO, you can calculate it by first considering all of theforces on one of the masses,

Fnet = mx = −kx (2.118)

for a SHO, the position equation is

x(t) = Asin(2πωt+ φ) (2.119)

74


plug this into the force equation to get the differential equation

md

dt(Asin(2πωt+ φ)) = −k (Asin(2πωt+ φ)) (2.120)

and when you find the general solution to the diffeq, you should get

ω =1

2π

√k

m(2.121)

Correct Answer

(B)

75


2.44 PGRE8677 #44


The most difficult part of this problem isn’t performing the calculation but convincing yourself ofthe correct solution. When the particle strikes the stick, the particle stops which tells us that allof the momentum that the particle originally had has been transferred to the stick. Because theparticle doesn’t strike the stick at its center of mass, the stick will start to spin in addition tomoving to the right. This may convince you that if the momentum is conserved, that a fraction ofthat momentum results in a linear motion for the stick and the other fraction of that momentumgoes into the rotation. However, if we consider all of the momentum on the stick, we get thefollowing diagram.

Note that at any instant of time, the momentum of the ends of the stick (and every pointbetween the end and the COM) has a horizontal and vertical component of momentum exactlyopposite that of the opposite end of the stick (Figure 2.6). Since momentum must be conserved,we sum the momentum components to get

Pnet - x = PCOM + Px + (−Px) = PCOM (2.122)

Pnet - y = Py + (−Py) = 0 (2.123)

From this, we get that the net momentum transferred from the particle to the stick is allcontained within the center of mass. Finally, compare the initial and final momentum to get

76


Figure 2.6: Conservation of momentum in a rotating system

Pi = Pf (2.124)

mv = MV (2.125)

V =mv

M(2.126)

Correct Answer

(A)

77


2.45 PGRE8677 #45


This problem is making reference to Compton scattering. Compton scattering is the phenomenaby which photons collide with a particle, impart some of their kinetic energy to the particle andthen scatters off at a lower energy (Figure 2.7).

Figure 2.7: Photon wavelength change due to compton scattering

The Compton equation derived for these interactions is

λ′ − λ =h

mpc(1− cos(θ)) (2.127)

If we plug in all the values provided in the problem, we get

λ′ − λ =h

mpc

(1− cos

(π

2

))(2.128)

λ′ =h

mpc+ λ (2.129)

which tells us that the new wavelength, λ′ has an extra hmpc

added to it.

Correct Answer

(D)

78


2.46 PGRE8677 #46


According to Stefan’s Law (Stefan-Boltzmann’s Law), power radiation of a blackbody is only de-pendent on temperature according to

j∗ = σT 4 (2.130)

where σ is a constant equal to

σ =2π5k4

15c2h3= 5.67× 10−8 J

sm2K4 (2.131)

For our problem, we are doubling the temperature so our power is

j∗ = σ(2T )4 (2.132)

= 16σT 4 (2.133)

= 160 mW (2.134)

Correct Answer

(E)

79


2.47 PGRE8677 #47


Without knowing anything about the Frank-Hertz experiment, you can immediately eliminateoptions (A) and (B) from the list because it is neither true that electron collisions are alwayselastic, nor is it true that they are always inelastic. By a similar line of reasoning, since we’vealready argued that some electron scattering can be elastic, it wouldn’t make since to then say thatelectrons always lose some specific amount of energy, so we eliminate (D). Finally, when choosingbetween (C) and (E), you will need to know that the Nobel Prize winning Frank-Hertz experimentdemonstrated that at a specific energy range (4.9 volts to be specific) electrons begin to experienceinelastic collisions and that the energies lost in the collision came in discrete amounts.

Correct Answer

(C)

80


2.48 PGRE8677 #48


From our quantum mechanical selection rules, we know that any change of state must be accom-panied by a change in the quantum angular momentum number

∆l = ±1 (2.135)

This tells us that if the wave function is spherically symmetric, i.e. l = 0, then after a changeto a new wave function, l 6= 0 and so they can’t both be spherically symmetric.

Correct Answer

(D)

81


2.49 PGRE8677 #49


The classical Hamiltonian equation is

H = T + V (2.136)

=p2

2m+ V (2.137)

The quantum mechanical Hamiltonian is

H = − h2

2m∇2 + V (2.138)

examine both equations to see that we can move from the classical Hamiltonian to the quantummechanical one if we allow p = ih∇.

Correct Answer

(B)

82


2.50 PGRE8677 #50


If you place a flat conducting bar with a current passing through it into a magnetic field with fieldvectors pointing perpendicularly to the direction of current, the Lorentz force will cause the streamof electrons to curve to one face of the bar while the remaining positive charges are left on theopposite face of the bar (Figure 2.8). This results in a potential difference in the conductor knownas the ”Hall Voltage” which can be calculated using

VH = − IBdne

(2.139)

where I is the current, B is the magnetic flux density, d is the plate depth, e is the electroncharge and n is the charge carrier density. One of the benefits the Hall effect has to experimentalphysics relates to the direction in which a current will curve depending on the direction of themagnetic field, current and charge.

Effectively, the direction in which the electrons curve determines the sign of the Hall Voltageand provided the direction of the current and magnetic field are known, it is trivial to determinethe sign of the charge carriers.

Correct Answer

(C)

83


Figure 2.8: Moving charges tend to one side of a conducting bar from the Hall effect

2.51 PGRE8677 #51


The Debye and Einstein models of specific heat were essentially identical except for the way inwhich they were scaled. Debye scaled his model around a value he called the Debye temperature(TD) that was dependent on a number of properties of the material. Einstein, on the other hand,scaled his model based on a single frequency value (hν) and constant k. Although they wereidentical in terms of their predictions that vibrational energies were dependent on 3N independentharmonic oscillators, Debye’s scaling proved to be more accurate in the low temperature range andso it prevailed.

Correct Answer

(B)

84


2.52 PGRE8677 #52


From what you learned about electrostatics, you should recall that in a region where there are nocharges and with a uniform potential on the surface, electric potential is described by Laplace’sequation

∇2V = 0 (2.140)

It is also true that within the boundaries of any region which is satisfied by Laplace’s equations,there can be no local minima or maxima for V (courtesy of the Divergence Theorem). From thiscondition, we know that whatever value of potential we have on the surface, the value inside mustbe exactly the same, else we’ve found a maximum or minimum. Therefore, the potential at thecenter of the cube must be V , just as it is on the surface of the cube.

Correct Answer

(E)

85


2.53 PGRE8677 #53


When a charged particle oscillates in 1-dimension it will radiate electromagnetic radiation in adirection perpendicular to to the motion of the particle. For an imperfect but useful analogy,think along the lines of the Lorentz Force and the orthogonality of its electric and magnetic fields.Expanding on this Lorentz Force analogy a bit more, recall that the orthogonality condition comesas a sine function from a cross product

F = q(~E + ~v × ~B

)(2.141)

= qE + vB sin(θ) (2.142)

The dipole oscillation, much like the Lorentz Force, has this same sine dependence. Knowingonly these two facts we can conclude that the point P must be in the xy-plane, because the problemhas already specified that it lies in this plane, and that the amplitude is maximized when θ = 90

Correct Answer

(C)

86


2.54 PGRE8677 #54


This problem is extremely quick and easy if we can simply recall the definition of the dielectricconstant, K. The dielectric constant is a scaling factor which describes the concentration of elec-trostatic flux in any given material. It is calculated by taking the ratio between the amount ofelectrical energy stored by the material when some voltage is applied to it (absolute permittivityof the material, εs) relative to the permittivity of a vacuum, ε0,

K =εsε0

(2.143)

from the previous equation, it should be clear that K = 1 when the absolute permittivity ofthe material is equal to that of a vacuum. Since the polarization charge density, σp, is

σp = P · n (2.144)

= χe ~En (2.145)

where P is the electric polarization, χe is the electric susceptibility and ~E is the electric field.Since the electric susceptibility is

χe =P

ε0 ~E=εsε0− 1 (2.146)

We see that when εs = ε0 (i.e. when K = 1), the bound charge density should equal 0. Thislimit is only satisfied by (E)

Correct Answer

(E)

87


2.55 PGRE8677 #55


You should recall the thermal energy referenced in the problem from Boltzmann’s constant, whicheffectively tied together the microscopic affects of fermions to the macroscopic view of thermalenergy,

PV = NkT (2.147)

When dealing with individual fermions, in our case electrons, we are primarily concerned withthe Fermi Energy which describes the highest energy level occupied by an electron. The highestoccupied energy level can be found using energy level diagrams and continuously appending elec-trons to the lowest unoccupied levels. At the point that we’ve run out of electrons to assign toenergy levels, the highest level electron determines the Fermi Energy and, additionally, the FermiVelocity. If the Pauli exclusion principle weren’t true, then electrons would have no reason to buildinto higher energy levels (in fact they would likely all bunch into the ground state energy level)and so, without the Pauli exclusion principle, the mean kinetic energy of electrons wouldn’t bedependent on the thermal energy of the system.

Correct Answer

(B)

88


2.56 PGRE8677 #56


From quantum mechanics, we know that the expectation value or the mean value of an operator isfound by

〈ψ∗|Q|ψ〉 =

∫ ∞−∞

ψ∗Qψ dx (2.148)

where Q is some operator. In this problem they tell us that Q is the operator corresponding toobservable x and traditionally, the operator Q is used such that

(Qψ) (x) = (xψ) (x) (2.149)

so finding the expectation value of Q is effectively the same thing as finding the mean valueof x and so the correct answer is (C). Be aware, however, that an operator which “corresponds toa physical observable x” doesn’t necessarily imply that you are finding the mean value of x, eventhough it does mean you are finding the mean value of the observable itself.

Correct Answer

(C)

89


2.57 PGRE8677 #57


The eigenfunction of a linear operator is any non-zero function, f , that satisfies the condition

Af = λf (2.150)

where λ is the eigenvalue. From this, we will want to plug in all of the given functions, f , into

hkf = −ih∂f∂x

(2.151)

without doing any work, it should be obvious that both (A) and (B) won’t work because onlythe RHS of Equation 2.151 contains a derivative and so the trig functions will be different on eachside. However, for the purposes of rigor, we can compute each of the 5 possible solutions

(A) hk cos(kx) 6= ihk sin(kx)

(B) hk sin(kx) 6= −ihk cos(kx)

(C) hke−ikx = i2hke−ikx 6= −hkeikx

(D) hkeikx = −i2hkeikx = hkeikx

(E) hke−kx 6= ihke−kx

Correct Answer

(D)

90


2.58 PGRE8677 #58


Looking through the three options provided, we see that III, Wave-front angular frequency, i.e. thecolor of the object, is one of the properties which the problem claims a hologram might “record”. Ifyou’ve ever seen a real hologram (i.e. not one from star wars) then you know that color is somethingthat is not conserved (Figure 2.9). In fact, the question gives this away when it tells you that theresulting hologram is monochromatic.

Figure 2.9: Comparison of realistic holograms to cinematic holograms

From this, we can eliminate any solutions which claim to record Wave-front angular frequency,i.e. (C), (D) and (E). When choosing between (A) and (B), we simply need to determine whetherany phase information is saved by the hologram. Since a hologram is generated as the result ofinterference between two light beams and wave interference is dependent on wave phase, choose(B).

Correct Answer

(B)

91


2.59 PGRE8677 #59


The group velocity, vg, is related to the angular frequency, ω, by

vg =∂ω

∂k(2.152)

ω is given in the problem so we take the derivative with respect to k to get

∂ω

∂k=

∂

∂k

√c2k2 +m2 (2.153)

=c2k√

c2k2 +m2(2.154)

and plugging in the limit k → 0, you should get that vg = 0 which eliminates all possiblesolutions but (D) and (E). Next, letting k →∞, both terms with c and k will blow up and m willeffectively be 0, giving us

c2k√c2k2

=c2k

ck(2.155)

= c (2.156)

Correct Answer

(E)

92


2.60 PGRE8677 #60


For a simple harmonic oscillator the frequency is typically given as

f =1

2π

√k

m(2.157)

this gives you a mass dependence and allows you to eliminate (A) and (B). Additionally, it alsodemonstrates that there is no initial velocity dependence and we can eliminate (E). Finally, whenchoosing between (C) and (D), ask yourself where we get the potential energy equation in the firstplace. Recall that a mass which exhibits simple harmonic motion due to a spring has a spring forceof

Fs = −kx (2.158)

Additionally, recall that

V =

∫Fs dx (2.159)

which means we can solve for the potential energy of the SHO by integrating the spring forceequation,

V =

∫Fs dx (2.160)

= kx2 + C (2.161)

where C is just some constant. This expression is identical to the potential energy described inthe problem and so we conclude, by Hooke’s law, that the force and, by extension, the potentialenergy of the system is dependent on the constant of proportionality for position, b, but notdependent on the arbitrary constant, a.

Correct Answer

(C)

93


Alternate Solution

Recall that the potential energy of a system is related to the force on that system by

dV

dx= −F (x) (2.162)

so our potential energy, V , becomes

Fs = − d

dx

(a+ bx2

)(2.163)

= −2bx (2.164)

By Newton’s second law,

Fs = mx (2.165)

−2bx = mx (2.166)

but since x = ω2x, we get

2b = mω2 (2.167)

ω =

√2b

m(2.168)

finally, since the angular frequency, ω, is related to frequency, f , by ω = 2πf

f =1

2π

√2b

m(2.169)

Correct Answer

(C)

94


2.61 PGRE8677 #61


The problem specifies that the rocket has velocity v and that the motion of the rocket is determinedby the equation

mdv

dt+ u

dm

dt= 0 (2.170)

we will consider the rocket moving to the right and let this direction of motion be positive. Inorder for the net momentum of the system to be 0, there must be another velocity, in our case u,moving in a direction opposite to that of the rocket. For options (A), (B) and (C), each of thevelocities mentioned should be positive as they are all instances of the rocket in its motion in thepositive direction (to the right). Now, when choosing between (D) and (E), realize that our originalequation of motion describes a stationary frame viewing a moving frame. From this, we know thatu couldn’t be the exhaust speed in a stationary frame, else we wouldn’t be taking into account thevarying reference frames.

Correct Answer

(E)

95


2.62 PGRE8677 #62


The question gives you the initial conditions that when m = m0 we get v = 0. Substitute m = m0

into each to see which, if any, goes to 0.

(A) u(mm

)= u 6= 0

(B) uemm = ue 6= 0

(C) u sin(mm

)= u sin(1) 6= 0

(D) u tan(mm

)= u tan(1) 6= 0

(E) (A), (B), (C) and (D) were all wrong, therefore we choose (E)

Correct Answer

(E)

Alternate Solution

The quick method to solve this problem involves using some very irresponsible math that wephysicists (and to a larger extent, chemists) like to use to simplify differentials. multiply the dt outof our equation of motion to get

96


mdv + udm = 0 (2.171)

mdv = −udm (2.172)

get dv on its own and integrate both sides∫dv = −u

∫dm

m(2.173)

and we will just generalize the results as

∆v = −u [ln(m)]mm0(2.174)

vf − v0 = −u ln

(m

m0

)(2.175)

since the question tells us that v0 = 0, we make the substitutions to get,

vf = −u ln

(m

m0

)(2.176)

which doesn’t match any of the options given as possible solutions.

Correct Answer

(E)

97


2.63 PGRE8677 #63


Anytime you see different arrangements of variables in which all or most of the options will havedifferent units than one another, check the units. Recall that we are looking for units of chargerper area

(A) q4πD ≡

Cm

(B) qD2

2π ≡ C ·m2

(C) qd2πD2 ≡ C

m

(D) qd2πD3 ≡ C

m2

(E) qd4πε0D2 ≡ C

m

Correct Answer

(D)

98


2.64 PGRE8677 #64


In electronics, when we want to maximize power transfer we need to do some impedance matchingin which the load impedance and complex source impedance is

ZS = Z∗L (2.177)

plugging in given impedance values,

ZS = Z∗L (2.178)

Rg + jXg = Rg + jXl (2.179)

In order for the RHS to be equal to the LHS,

Xl = −Xg (2.180)

Correct Answer

(C)

99


2.65 PGRE8677 #65


A magnetic dipole occurs in any instance that you have a closed circulation of electrical current.In the case of our problem, we have an enclosed loop of wire with a current i. Any given dipolecan be described by its dipole moment and in the case of a magnetic dipole, we get the magneticdipole moment

µ =

∫i dA (2.181)

where dA is a differential piece of the area about which the current circulates. In our case, ourarea is simply that of a circle with radius b and so we get a magnetic moment of

µ = iπb2 (2.182)

∝ ib2 (2.183)

Correct Answer

(B)

100


2.66 PGRE8677 #66


Without knowing much about thermodynamics, we can figure that our final solution must containsome units of temperature, specifically we expect to get something like inverse Kelvins. Looking atthe units of the 4 different variables, we get

P = atmV = m3

S = J/KU = JNote that only entropy, S, has any temperature units and the only way to isolate the tempera-

ture unit is to do (∂S

∂U

)V

(2.184)

Correct Answer

(E)

Alternate Solution

Recall, from thermodynamics, the equation

dU = T dS − PdV (2.185)

you should never forget this equation because chemistry, more than any other subject, is T dS(T dS = tedious..... GET IT!). Temperature is only well-defined for a system in equilibrium, so welet dV = 0 and we get,

101


dU = T dS (2.186)

1

T=

(dS

dU

)V

(2.187)

Correct Answer

(E)

102


2.67 PGRE8677 #67


Recall from thermodynamics that with excessive amounts of energy at the disposal of a system,particles will tend to equally populate all energy levels. Since the question tells us that temperature(i.e. thermal energy) is significantly larger than the energy levels, we know that every energy levelmust have an equal likelihood of being populated. From this, calculate the average of the 3 energylevels as

0 + ε+ 3ε

3=

4

3ε (2.188)

Correct Answer

(C)

103


2.68 PGRE8677 #68


This problem and the solution should be one of those ideas you know by heart. If you don’t, stopwhat ever you are doing and commit this to your memory. As an object with mass moves faster andapproaches extremely high velocities, the mass of the object begins to increase. As you approachthe speed of light, the mass approaches infinity. The only way a particle can achieve a velocity atthe speed of light is if the particle has a rest mass of 0. In case the rule of thumb isn’t enough toconvince you, recall

E =√p2c2 +m2

0c4 (2.189)

let m→ 0, giving

E =√p2c2 (2.190)

= pc (2.191)

then, by the Planck relationship,

pc = hν (2.192)

p =hν

c(2.193)

=h

λ(2.194)

which we know, experimentally, is the momentum of the photon.

Correct Answer

(A)

104


2.69 PGRE8677 #69


In this problem, you should immediately eliminate (D) because we know the car can’t move fasterthan the speed of light. Next, eliminate (E) because there is sufficient information in this problemto answer the question explicitly. Lastly, look for the speed which justifies our use of relativisticequations to account for the contraction of a 2 meters, which would be 0.8c.

105


Correct Answer

(C)

Alternate Solution

Length contraction from relativistic effects is related to the Lorentz factor by,

L′ =L

γ= L

√1− v2

c2(2.195)

where L′ is the length of a contracted object in the rest frame and L is the rest length. Plug inthe given values to solve

3 m = 5 m

√1− v2

c2(2.196)

9

25= 1− v2

c2(2.197)

16

9=

v2

c2(2.198)

v =4

5c (2.199)

Correct Answer

(C)

106


2.70 PGRE8677 #70


Now, we are talking about the moving car being the rest frame and the garage being the contractedframe. We use the same contraction equation used in the previous problem, however we let L′ bethe contracted garage reference frame and let L be the rest frame for the garage. In the previousequation, we calculated the speed of the car as v = 0.8c, so we plug everything into our contractionequation to get

107


L′ = 4 m

√1− v2

c2(2.200)

= 4 m√

0.36 (2.201)

= 2.4 m (2.202)

Correct Answer

(A)

108


2.71 PGRE8677 #71


I’m generally not a big fan of these multi-part GRE questions, however I really like this one becauseit illustrates one of the classic paradoxes, or at least apparent paradoxes, in relativity. In the vehiclesreference frame, the garage is too short to contain the vehicle. However, in the garage referenceframe the car should have no problem fitting in the garage. The primary qualitative lesson welearned from Einstein is that we can’t prefer one reference frame to another. Based on this, wearen’t justified in saying that one solution is more “true” than the other. The only solution which

109


alludes to this inherent duality of solutions is (E), which describes another classical qualitativeresult from Einstein. Specifically, I’m referring to our inability to achieve truly instantaneousevents between different reference frames.

Correct Answer

(E)

110


2.72 PGRE8677 #72


This problem confuses me because it has 1 correct choice and 4 absurd choices. This means youdon’t need to know why the correct answer is correct, only that the other 4 are ridiculous.

(A) What? Since when? No it doesn’t!

(B) What? Of course they can transmit signals! How else would we get any useful informationout of x-rays in, for example, medical applications.

(C) WHAT? Imaginary mass? When? Where? NO!

(D) And? Why would the order in which physical theories were discovered determine the accuracyof Relativity?

(E) The phase velocity and group velocities indeed are different. This is important because therefractive index of light through a medium is the ratio of the phase speed of light through themedium to that of light through a vacuum. It is, therefore, significant that the phase speednever be faster than the speed of light, which is frequently true of the group velocity. If theyweren’t different, we might get n > 1. However, you don’t really need to know any of that tosolve this problem, do you!

Correct Answer

(E)

111


2.73 PGRE8677 #73


If you’ve ever seen the rainbow patterns that form on water bubbles and oil slicks, then you’ve seenthin film interference. This phenomena occurs as a result of light waves reflecting off of the topsurface of a medium with a higher index of refraction than that from which the light came fromand, additionally, light waves reflecting off of the bottom of that surface as another transition to ahigher index of refraction occurs (Figure 2.10)

Figure 2.10: Thin film interference to create a non-reflecting layer

The same phenomena occurs whether the light is incident on the top surface at an angle ororthogonal to it, however I’ve intentionally drawn it at an angle to make it easier to observe thephase shift between the two reflected rays. For every reflection of a light wave off a medium ofhigher index of refraction, we get a shift of half the wavelength of the light,

∆air =λ

2(2.203)

∆coating =

(λ

2+ 2t

)(2.204)

112


In the previous equations, note that ∆coating has an additional 2t of phase shift because itmust also pass both ways through the thickness of the coating. In order for the coating to eliminateall reflections, we need the outgoing waves to experience destructive interference, so we find therelative shift of the system

∆ =(∆coating

)−(∆air

)(2.205)

=

(2t+

λ

2

)−(λ

2

)(2.206)

= 2t (2.207)

since we only get destructive interference with waves of differing phase shift at half-odd integervalues , set the relative shift to

2t =

(m+

1

2

)λ (2.208)

t =λm

2+λ

4(2.209)

where m = 0, 1, 2, 3, . . . . Since we are only concerned with the thinnest possible coating whichwill give us deconstructive intereference, we choose m = 0 and t = λ/4.

Correct Answer

(A)

113


2.74 PGRE8677 #74


Consider a beam of unpolarized light headed towards you and you have three ideal polarizers inhand

Figure 2.11: Different orientations of a polarizing filter

In Figure 2.11, grey vectors indicate individual directions of oscillation, black vectors indicatethe net polarization, greyed out areas indicate areas of absorbed polarization and white areasindicate unaffected polarization. If we wanted to completely eliminate all light using two of thepolarizers, we could place them in series with a rotation of exactly π/2 between them, for exampleusing the horizontal and vertical polarization. In this arrangment, anything that survived throughthe horizontal polarizer would be caught by the vertical polarizer and no light would be transmittedon the other side. However, imagine we were to place another polarizer between the horizontal andvertical polarizers such that this third polarizer is rotated π/4 or 45 with respect to the othertwo. As the light first passes through the horizontal polarizer, only half of the photons that hitthe polarizer will pass through. These photons will continue to the 45 polarizer where, again,half of the remaining photons get absorbed and the other half pass through. After the photonspass through the 45 polarizer, the remaining photons will spread out from -22.5 to +67.5. Thisoccurs because linearly polarized light will always completely fill a full 90 of angular spread, whichI didn’t mention previously because the previous instances came out with a 90 spread. Finally, theremaining photons will pass through the vertical polarizer giving a final total of 1/8 the originalnumber of photons (Figure 2.12)

Correct Answer

(B)

114


Figure 2.12: Remaining light after passing through the three polarizer system

2.75 PGRE8677 #75


Recall Kepler’s third law of motion

”The square of the orbital period of a planet is directly proportional to the cube of thesemi-major axis of its orbit”

so we have

T 2 = βr3 (2.210)

The period is given as T = 80 minutes, so

(80 min)2 = βR3e (2.211)

β =6400 min2

R3e

(2.212)

using the same proportionality constant, β, but for 24 hours (i.e. 1440 minutes), we get

115


T 2 = βr3 (2.213)

(1440 min)2 =

(6400 min2

R3e

)r3 (2.214)

r3 =1440 min2 R3

e

6400 min2 (2.215)

For a quick and easy approximation, pretend that 1440 = 1500 and 6400 = 6000, so then weget

r3 =(1440 · 1440)R3

e

6400(2.216)

=1440

4R3e (2.217)

= 360 (2.218)

Finally, take the cube root of both sides to get

r =3√

360 (2.219)

≈ 7 (2.220)

Correct Answer

(B)

116


2.76 PGRE8677 #76


Recall the angular momentum equation

L = Iω (2.221)

in which the moment of inertia, I, of a ring is

Iring = MR2 (2.222)

which gives us

L = MR2ω (2.223)

To solve for ω sum the total energy of the rolling ring

Utot = Uroll + Utrans (2.224)

Mgh =

(1

2Iω2

)+

(1

2Mv2

)(2.225)

but since ω = v/R, we get

117


Mgh =

(1

2MR2ω2

)+

(1

2MR2ω2

)(2.226)

gh = R2ω2 (2.227)

ω =

√gh

R2(2.228)

Then combining ω into the angular momentum equation, we get the solution

L = MR2ω (2.229)

= MR2

√gh

R2(2.230)

= MR√gh (2.231)

Correct Answer

(B)

118


2.77 PGRE8677 #77


Any object with a net force of F = −kx is a simple harmonic oscillator (via Hooke’s Law). Theposition equation for a SHO is

x = A cos(ωt) (2.232)

take the derivative of position to get the velocity equation

dx

dt= −A sin(ωt)ω (2.233)

If we square both equations and get their trig functions by themselves, we can apply the identitycos2(θ) + sin2(θ) = 1,

x2

A2= cos2(ωt) (2.234)

v2

A2ω2= sin2(ωt) (2.235)

using our trig identity and letting our position be x = A/2,

cos2(ωt) + sin2(ωt) = 1 (2.236)

x2

A2+

v2

A2ω2= 1 (2.237)

1

4+

v2

A2ω2= 1 (2.238)

Now solve for v

v2

A2ω2=

3

4(2.239)

v2 =3

4A2ω2 (2.240)

119


Finally, since ω = 2πf ,

v =

√3

4A24π2f2 (2.241)

=√

3πfA (2.242)

Correct Answer

(B)

120


2.78 PGRE8677 #78


Total energy of the system must be the sum of the kinetic and potential energies

Unet = T + V (2.243)

We are told that the initial potential is 0 and one of the particles has an initial velocity and,therefore, an initial kinetic energy

U0 =1

2mv2

0 (2.244)

Since the problem tells us to neglect radiation, we assume energy is conserved so the final energyneeds to equal the initial energy

Uf = U0 = T + V =1

2mv2

0 (2.245)

which tells us that the total energy of the system is constant over time and also that the energyis positive (assuming neither particle has negative mass).

Correct Answer

(C)

121


2.79 PGRE8677 #79


Recall that Maxwell’s eqution ∇ · ~B = 0 tells us that divergence of the magnetic field must alwaysbe 0 and that divergence is defined as the amount of outward flux of a vector field. With thisdefinition, we know that we need to look for any vector field that has some non-zero flux leavingthe area.

(A) 5 flux lines come in and 5 flux lines go out. Net flux leaving the area is 0.

(B) 5 flux lines come in and 5 flux lines go out. Net flux leaving the area is 0.

(C) 5 flux lines come in and 5 flux lines go out. Net flux leaving the area is 0.

(D) 0 flux lines come in and 8 flux lines go out. Net flux leaving the area is NOT 0.

(E) 0 flux lines come in and 0 flux lines go out. Net flux leaving the area is 0.

Correct Answer

(D)

122


2.80 PGRE8677 #80


From Gauss’s law, we get

∇ · ~E =ρ

ε0(2.246)

with no charges in the region, ρ = 0 and

∇ · ~E = 0 (2.247)

take the derivative of each term with respect to its appropriate variable (i.e. x→ i, y → j andz → k)

(A) A(2y − x) 6= 0

(B) A(−x+ x) = 0

(C) A(z + 0) 6= 0

(D) Ayz +Axz 6= 0

(E) Ayz

Correct Answer

(B)

123


2.81 PGRE8677 #81


Recall from Faraday’s law of induction,

ε = −dφ ~Bdt

(2.248)

since the magnetic field is proportional to the current, ~B ∝ I, we know we will eventually haveto take the derivative of I and this means we should end up with sin(ωt) rather than cos(ωt) so wecan eliminate (A) and (D). Next, since we know we want to end up with a units of volts for ourEMF, check the units of (B), (C) and (E). µ0 is given in the front of the test booklet and as longas you recall that weber = volt · sec,

(B)(volt·secAmp·m

)(Amp)

(m2

m

) (1sec

)= volt

(C)(volt·secAmp·m

)(Amp)

(mm2

) (1sec

)= volt

m2

(E)(volt·secAmp·m

)(Amp)

(mm

) (1sec

)= volt

m

124


Correct Answer

(B)

Alternate Solution

From Faraday’s law of induction,

ε = −dφ ~Bdt

(2.249)

where the magnetic flux, φ ~B, is

φ ~B =

∫~B · dA = ~Bπa2 (2.250)

From the Biot-Savart law, we get the magnetic field from a current passing through a loop ofwire as

~B =

∫µ0Idl × r

4πR2(2.251)

=

∫µ0Idl sin(θ)

4πR2(2.252)

since all of the potential solutions have no θ dependence like they should, we’ll assume that theapproximate solution the problem refers to is simply finding the resulting EMF at the optimizedangle, θ = π/2, with radius b,

~B =

∫µ0Idl sin(π/2)

4πb2(2.253)

=µ0I

4πb2

∫dl (2.254)

=µ0I

4πb2(2πb) (2.255)

=µ0I

2b(2.256)

Plug this all back into the magnetic flux to get

φ ~B =µ0Iπa

2

2b(2.257)

=µ0π

2

a2

b(I0 cos(ωt)) (2.258)

Finally, take the derivative of φ ~B with respect to time and get

V =µ0π

2

a2

bI0d

dt(cos(ωt)) (2.259)

=µ0π

2

a2

bI0ω sin(ωt) (2.260)

Correct Answer

(B)

125


2.82 PGRE8677 #82


This problem is describing a phenomena called the Zeeman effect, in which an atom with electronsin a degenerate state (i.e. electrons with equivalent energies but differing electron configurations)is passed through a magnetic field. Since the affect of a magnetic field is related to the electronconfiguration, the degenerate electrons will have their energies altered by the magnetic field indifferent ways and the single emission line which was initially composed of multiple degenerateelectrons becomes visible as different emission lines (Figure 2.13),

Figure 2.13: The Zeeman effect for degenerate energy states

This tells us that the possible solutions is either (D) or (E) but since there is no specificrestriction which says the emission lines must always be doubled, we choose (E).

Correct Answer

(E)

126


2.83 PGRE8677 #83


The phrasing of this problem tells us everything we need to know in order to figure out the differencebetween a low-density and high-density state of the same gas. For a high density gas the mean lifeof an atomic state is significantly longer than the amount of time it takes for atomic collisions tooccur, so by the time the atomic life is up, multiple collisions have occurred and energy has beenswapped around a bunch of times. In a less dense state, not as many of these energy exchangeshave occurred so there are fewer distinct energies and the ones that are there are more well-definedthan the high-density gas. This is all you need to know to tell you that the high-density gas willhave a broader spectral line than the low-density gas. If you want something a bit more rigorous,recall Heisenberg’s Energy-Time uncertainty principle

∆E ∆t ≥ h

2(2.261)

For the higher-density gas, as compared to a low-density gas, the decrease in time uncertaintyrequires an increase in energy uncertainty so we get a broader (i.e. less precise) spectral line.Alternatively, the increase in time uncertainty for the less-dense gas will give require that theenergy uncertainty be less than the high-density gas and so we expect a more well-defined (i.e.thinner) spectral line.

Correct Answer

(C)

127


2.84 PGRE8677 #84


Sodium has 11 electrons, just as the problem so kindly gives us, so we construct our energy leveldiagram (Figure 2.14)

Figure 2.14: Energy level diagram of Sodium in its ground state

With our one lone electron, we expect to get a spin of S = 1/2. This tells us that our termsymbol component of 2S + 1 will become 2(1/2) + 1 = 2, so our term symbol will be of the form

2LJ (2.262)

which allows us to eliminate all but (B) and (D). Since the final electron is in the 3s shell, L = 0and since J = L+ S, J = 1/2. From this, our term symbol should be

2S 12

(2.263)

Correct Answer

(B)

128


2.85 PGRE8677 #85


As a general rule,

Photoelectric effect: Low Energy (eV Range)

Compton Scattering: Mid Energy (KeV Range)

Pair Production: High Energy (> 1 MeV)

Based on this general rule, we know that line #2 should correspond to pair production because itonly exists at higher energy ranges, which eliminates (A), (D) and (E). Between (B) and (C), it ismore likely that line #1 corresponds to the Photoelectric effect because the interaction peaks at alower energy range.

Correct Answer

(B)

129


2.86 PGRE8677 #86


In Faraday’s Ice Pail experiment, Michael Faraday took an insulated pail which he connected to anelectrometer. A brass ball with a charge on its surface was lowered down into the pail. Because thebrass ball surface has a net negative charge, the pails positive charges were attracted to the innerwalls of the pail and negative charges were pushed to the outside of the pail and also pushed downthe wire to the electrometer. As the charged brass ball was lowered enough to touch the inner floorof the pail, the charge was completely discharged to the pail and the ball could be removed withthe electrometer reading a complete transfer in charge. From this experiment, Faraday was able todetermine that the inverse square law was accurate to at least one part in a billion.

Correct Answer

(E)

130


2.87 PGRE8677 #87


Recall that the specific heat for any atom/molecule is proportional to the degrees of freedom of themolecule. In other words, the more degrees of freedom a molecule has the higher the specific heat.

(A) a specific heat of 3/2 Nk corresponds to a monotomic atom with only translational degreesof freedom, so we know that these diatomic molecules must be larger than that.

(B) Model I has translational and rotational degrees of freedom while Model II has tranlational,rotational and vibrational degrees of freedom. We would expect the specific heat to be largerin Model II than Model I, not the other way around.

(C) At high energies (i.e. high temperatures) vibrational effects must be considered, so we can’talways treat a molecule as a rigid rotor.

(D) At low energies (i.e. low temperatures) vibrational effects are minimal, so we don’t alwayshave to treat a molecule as a springy dumbbell.

(E) At low energies (i.e. low temperatures) there is minimal vibrational effect so we can treatthe molecule as if it is a rigid rotor. However, at higher energies (i.e. higher temperatures)vibrational effects are involved so we want to treat it as a springy dumbbell. In other words,the model we use on a molecule is dependent on its temperature.

Correct Answer

(E)

131


2.88 PGRE8677 #88


Recall that fermions must obey the Pauli Exclusion principle which restricts the particles fromoccupying the same energy levels. Bosons, on the other hand have no such restriction. This tells usthat in its lowest energy state, N bosons will all be able to occupy the minimal energy state whileN fermions will be forced to occupy higher energy levels after the ground level has been filled. Thistells us that the pressure, which is proportional to temperature (energy), should be greater for thefermions than the bosons, which eliminates all but (B) and (C). Since the boson pressure is the lowpressure extreme and fermion pressure is the high pressure extreme, it is only logical to concludethat the classical result (think PV = nRT ) will be somewhere in between.

Correct Answer

(B)

132


2.89 PGRE8677 #89


For two identical particles, the wave function for those particles can only be either symmetric oranti-symmetric under interchange of x1 and x2 so our two options are

ψ(x1, x2, t) = ψ(x2, x1, t) (2.264)

or

ψ(x1, x2, t) = −ψ(x2, x1, t) (2.265)

As it turn out, whether or not the wave function is symmetric or anti-symmetric is determinedby the type of particle. Specifically, symmetric wave functions obey Bose-Einstein statistics andcorrespond to bosons. Anti-symmetric wave functions obey Fermi-Dirac statistics and correspondto fermions. From this, our two potential wave functions are

ψboson =1√2

[ψα(x1)ψβ(x2) + ψβ(x1)ψα(x2)] (2.266)

ψfermion =1√2

[ψα(x1)ψβ(x2)− ψβ(x1)ψα(x2)] (2.267)

Clearly, the wave function referred to in the problem corresponds to a boson and not a fermion.Electrons, positrons, protons and neutrons are all fermions while the deuteron is a boson, so wechoose (E).

Correct Answer

(E)

133


2.90 PGRE8677 #90


Without solving for anything explicitly, you can be sure that the ground state energy of any systemwill never be 0 eV and that the ground state energy, which is the lowest energy state, will be lessthan the total energy of the particle at 2 eV, so eliminate (A) and (E). Next, recall that the energyof the infinite square well is proportional to the nodes of its wave function by

En =n2h2π2

2ML2(2.268)

where the zero point energy, or equivalently the minimum energy value E1, is

E1 =h2π2

2ML2(2.269)

so the zero point energy is related to the rest of the energy levels by

En = n2E1 (2.270)

The problem tells us that the energy, En = 2 eV so with n = 2, we get

134


E1 =Enn2

(2.271)

E1 =2 eV

22(2.272)

=1

2eV (2.273)

Correct Answer

(C)

135


2.91 PGRE8677 #91


Without doing much work, you can eliminate (A) and (B) because they are much too slow to bethe velocity of an electron. We can also eliminate (E) because the electrons shouldn’t be movingfaster than light. As a general rule, expect electron velocities around 105 m/s which, if you arefeeling lucky and a bit adventurous, then you could risk it and guess (D). To be more rigorous, youwill have to recall the Bragg condition from Bragg diffraction optics

2d sin(θ) = nλ (2.274)

and recall the de Broglie relation

λ =h

p(2.275)

combining the two gives you

2d sin(θ) = nh

mv(2.276)

Since we are told that first-order reflection occurs, n = 1. Using approximate values for theremaining constants, we can solve for v,

v =h

md(2.277)

≈ 6× 10−34 m2kg/s

(10× 10−31 kg)(3× 10−10 m)(2.278)

≈ 5× 107 m/s (2.279)

which is closest to (D).

Correct Answer

(D)

136


2.92 PGRE8677 #92


Keep in mind that we are looking for a selection rule which is NOT compatible with electric dipoleemission. You don’t need to recall what any of the specific selection rules are as long as you recallwhat type of selection rules we have. There most definitely are selection rules for ∆j, ∆l, ∆ml

and ∆n but there are no selection rules for spin, ∆s. Thus, without knowing if any of the otherselection rules are correct, we can be sure that the proposed spin rule is NOT correct.

Correct Answer

(D)

137


2.93 PGRE8677 #93


Recall that the frictional force f , is equal to the product of the coefficient of friction, µ and thenormal force Fn = 100 newtons

f = µ(100 newtons) (2.280)

µ =f

(100 newtons)(2.281)

Next, recall that power, P , is

P = IV = Fv (2.282)

where I is current, V is voltage, F is force and v is velocity. Solve for F with the problemsgiven values for velocity, v = 10 m/s and current and voltage I = 9 Amps and V = 120 volts

F =IV

v(2.283)

=(9 Amps)(120 volts)

(10 m/s)(2.284)

= 108 N (2.285)

This force is the applied force of the sander which we need for the coefficient of friction equation,so make the substitution to get

138


µ =108 N

100 N(2.286)

= 1.1 (2.287)

Correct Answer

(D)

139


2.94 PGRE8677 #94

140



Before switch S is closed, the voltage through the circuit should be 0. As soon as we throw theswitch at time t = 0, the maximum voltage is reached and it continually decreases as it reachesequilibrium while passing through resistors R2 and R1. From this, we know that our solution shouldstart at a maximum VA and approach 0 which is not true of (C), (D) and (E). Once the switchis opened again at time t1, the current is able to flow again and since the potential collected bythe inductor only has to pass current through one of the resistors, R2 to be specific, we expect theinitial potential at t1 to be greater than at t0 so (B) is correct.

Correct Answer

(B)

141


2.95 PGRE8677 #95


The Carnot cycle is, theoretically, perfectly efficient and so we should expect in a Carnot engine,the entropy of the system should be perfectly conserved, dS = 0. (C) is in disagreement with thisaspect of the Carnot cycle so it is incorrect.

Correct Answer

(C)

142


2.96 PGRE8677 #96


The perturbation described in this problem should look a little something like Figure 2.15From this, we see that the maximum impact of the perturbation will be at the center, x = 0 and

so wave functions with a node at that location will be relatively unaffected (i.e. wave functions withodd-values for n). However, because all wave functions with odd-valued n are also odd functions,

143


Figure 2.15: Infinite square well subject to a perturbation

we know that the total area under the curve will sum to 0 (which is something a perturbation couldhave corrupted but, in our case, didn’t). So, for all odd values of n, the perturbed wavefunctionwill still have its coefficient a0,n = 0 and we choose (B).

Correct Answer

(B)

144


2.97 PGRE8677 #97


First off, eliminate (E) because it should have no dependence on time. Next, eliminate (B) and (D)because they are the same answer and so if they are both correct, we have a problem. Now, whenchoosing between (A) and (D), consider that the total angular momentum will be the sum of thetranslational component and the rotational component

L = Ltrans + Lrot (2.288)

Recall that angular momentum is

145


L = Iω0 = ~R× ~p = ~R×M~v (2.289)

For the translational component of angular momentum, we use the fact that L = ~R×M~v withour initial velocity ~v0

Ltrans = ~R×M~v0 (2.290)

= RM

(1

2ω0R

)(2.291)

=1

2MR2ω0 (2.292)

For the rotational component, we use the moment of inertia given in the problem

Lrot = Iω0 (2.293)

=1

2MR2ω0 (2.294)

However, because the rotational component is rotating in the opposite direction to that of thetranslational component, we must take the difference of the components to get

L = Ltrans − Lrot (2.295)

=1

2MR2ω0 −

1

2MR2ω0 (2.296)

= 0 (2.297)

Correct Answer

(A)

146


2.98 PGRE8677 #98


Start with Coulomb’s Law for a point charge

V =kQ

l(2.298)

from the perspective of point P , each differential piece of the glass rod will apply a force and itwill do so proportional to the charge density ρ = Q/l. Thus, we integrate the differential pieces ofglass rod,

V =

∫ 2l

l

kρ

ldl (2.299)

= kρ

∫ 2l

l

dl

l(2.300)

= kρ [ln(2l)− ln(l)] (2.301)

= kρ ln(2) (2.302)

=kQ

lln(2) (2.303)

Correct Answer

(D)

147


2.99 PGRE8677 #99


Recall that the ground state energy of Hydrogen is equal to 1 Rydberg, −13.6 eV . Positroniuminvolves an electron-positron pair while hydrogen involves a proton-electron pair. There is nodifference between the two in terms of charge but there is a significant difference in mass. SinceRydberg’s constant is mass dependent, we have to alter the original Rydberg constant

Rhydrogen =memp

me +mp

e4

8cε20h3

(2.304)

Which becomes,

Rpositronium =meme

me +me

e4

8cε20h3

(2.305)

me

2

e4

8cε20h3

(2.306)

To convince yourself that this makes the Rydberg constant half as large, consider that the ratioof the proton mass to electron is approximately 1836:1. Calculating the original effective mass withthis, you get

mpme

mp +me=⇒ 1× 1836

1 + 1836≈ 1 (2.307)

Using the same values for our new effective mass

148


me

2=⇒ 1

2(2.308)

Since the energy is proportional to the Rydberg constant, the ground state energy of positroniummust be half of the hydrogen ground state energy

Epositronium =Ehydrogen

2= −6.8eV (2.309)

again, if you aren’t convinced, consider the Rydberg equation for hydrogen

1

λ= R

(1

n21

− 1

n22

)(2.310)

and since E = hν = hcλ

E =hc

λ= hc

R

2

(1

n21

− 1

n22

)(2.311)

Correct Answer

(B)

149


2.100 PGRE8677 #100


The quickest solution to this problem (which you probably want because it is the last problemand you are likely running out of time) is to consider which of these solutions would actuallymake for a reasonable “pinhole” camera. For some sample values, let’s pick λ = 400 nm andD = 50 cm = 5.0× 108 nm to satisfy our condition that λ << D

(A)√

(400 nm)(5.0× 108 nm) ≈ 450, 000 nm = 0.045 cm: This doesn’t seem unreasonable to me.

(B) λ: The pinhole is the size of our wavelength, in our case 400 nm, and it is very unlikely youare going to pull off a pinhole this small.

(C) λ10 = 400 nm

10 = 40 nm: This is even smaller than the pinhole in (B) and is even less likely tobe useful.

(D) λ2

D = (400 nm)2

5.0×108 nm = 0.00032 nm: Ha! Good luck with that.

(E) D2

λ = (5.0×108 nm)2

400 nm = 6.25× 1014 nm = 625, 000 m: That’s a HUGE PIN!

Correct Answer

(A)

150

Chapter 3

PGRE9277 Solutions

151


3.1 PGRE9277 #1


The momentum operator from quantum mechanics is

P =h

i∇ψ (3.1)

If we substitute in the wave function ψ = ei(kx−ωt),

P =h

i

∂

∂x

(ei(kx−ωt)

)(3.2)

=hki

iei(kx−ωt) (3.3)

= hkψ (3.4)

Correct Answer

(C)

152


3.2 PGRE9277 #2


Bragg diffraction describes the phenomena by which specific angles of incident and wavelengths ofx-rays will generate a peak in reflected radiation. From Bragg diffraction we get Bragg’s law,

2d sin(θ) = nλ (3.5)

From Bragg’s law, it’s clear that the wavelength for any given n will be maximized whenθ = 90 = π/2 making the LHS 2d.

Correct Answer

(D)

153


3.3 PGRE9277 #3


From the Bohr Model, we get the approximation of any Hydrogen like atoms as

En = −Z2Ren2

(3.6)

for the ratio between carbon and magnesium, the only component of our approximation thatwill change is Z, so take the ratio of the 2 values,

ECEMg

=Z2C

Z2Mg

(3.7)

=62

122(3.8)

=1

4(3.9)

Correct Answer

(A)

154


3.4 PGRE9277 #4


Recall that the force due to gravity between two objects of mass m1 and m2 is proportional to theinverse squared value of the radius,

F = Gm1m2

R2(3.10)

thus, if we double the radius (i.e. R→ 2R) then we get

F (R)

F (2R)=

1/R2

1/(2R)2(3.11)

=4R2

R2(3.12)

= 4 (3.13)

Correct Answer

(C)

155


3.5 PGRE9277 #5


In this problem, the point mass is located inside of the earth, specifically half way between theearth center and its surface. We can’t just use the inverse square law in this form so utilize thepart of the problem that tells us to assume the planet is homogenous. From this, we can calculatethe mass of the earth, and the point mass, as being proportional to its density, ρ, by

M =4

3πR3ρ (3.14)

so the gravitational force it wields is

F (R) =GM

R2=

G

R2

(4

3πR3ρ

)=

4

3πRρ (3.15)

compare this to the case of of R/2,

156


F (R/2) =G

(R/2)2

(4

3π(R/2)3ρ

)=

4

3πR

1

2ρ (3.16)

finally, take the ratio of the two equations to get

F (R)

F (R/2)=

43πRρ

43πR

12ρ

= 2 (3.17)

Correct Answer

(C)

157


3.6 PGRE9277 #6


For starters, throw out option (E) as it can’t be true that the system is in balance for all conceivablevalues of M of the block. second, get rid of (A) because it doesn’t account for the coefficient offriction. If you aren’t convinced we need it, consider that when µ = 0, we should see our equationgo to 0, which isn’t true of (A). Next, eliminate (B) because when we maximize the coefficient offriction at µ = 1, then no amount of mass should move the wedges and the equation should blowup. We can’t make any more reasonable simplifications so if you struggle with the mechanics, atleast you can guess. However, to solve between (C) and (D), let’s consider the influence of the blockon just one wedge. The block has a force downward which, because of the 45 angle between itand the block, generates a vertical force and horizontal force on the wedge (in fact it is the normalforce of the block at its angle of incidence on the wedge). Since the angle is 45, we can find theamount of force the block is putting out by

tan(θ) =FGFG−x

(3.18)

FG−x = FG tan(45) (3.19)

FG−x = FG (3.20)

158


which tells us that the horizontal force is equivalent to the vertical force of the block. Now,since half of the force will be used on each block, if we are only considering one block, the horizontalforce generated by the block will be

FG−x =1

2Mg (3.21)

As the block applies the force, the frictional force of the wedge will try to resist it. From this,we know that the wedge will begin to move when the applied force over powers the frictional force,

FG−x + f > 0 (3.22)

and since the frictional force is f = µFN , we find the normal force of the wedge by summingthe vertical forces

FN = −FG (3.23)

= −(m+

M

2

)(3.24)

where the mass on the wedge is the wedges mass plus half of the blocks mass (i.e. M/2).combine our equations and solve to get

FG−x + f > 0 (3.25)

1

2Mg − µ

(m+

M

2

)> 0 (3.26)

M − 2µm− µM > 0 (3.27)

M(1− µ) > 2µm (3.28)

M >2µm

(1− µ)(3.29)

Correct Answer

(D)

159


3.7 PGRE9277 #7


For the given apparatus, there are 3 possible modes. The first one they give to us is a normalmode of 0 (i.e. no frequency). The next mode given represents the 2 masses swaying in the samedirection. Finally, we need to consider the last mode which occurs when masses sway in oppositedirections, in which case it doesn’t matter what the masses are and we can choose (A).

Correct Answer

(A)

160


3.8 PGRE9277 #8


The description in this problem is a little bit ridiculous but once you figure out what is going on,the problem is relatively easy. Torque is positive or negative based on the right hand rule andfrom this, we know that we want the cone to rotate in a clockwise direction about the z-directionwhen viewing the cone from above (+k). Any force in the k isn’t going to get our cone spinning soeliminate (A), (B) and (E). Next, looking at (C) and (D) it should be apparent that (C) will giveus a negative torque (which is what we want) while (D) gives us a positive torque.

Correct Answer

(C)

161


3.9 PGRE9277 #9


The intent behind coaxial cable shielding is to eliminate (at least in theory) the presence of anE&M field outside of the cable to reduce interference with other electronic equipment. This leadsus to choice (A).

Correct Answer

(A)

Alternate Solution

As our distance from the cable blows up to infinity (r →∞) we would expect the magnetic field togo to 0, which eliminates (B), (D) and (E). Next, recall that the magnetic field of a single, infinitelylong cable can be found from Amperes law

162


~B =µ0i

2πR(3.30)

which is identical to (C). It is unreasonable to assume that adding a shielding element won’talter this equation with some dependence on a & b, so we are left with (A).

Correct Answer

(A)

163


3.10 PGRE9277 #10


First, let’s recall the inverse square law for the 2 charges q1 and q2

F =1

4πε0

q1q2

r2r (3.31)

because of the infinite, grounded conducting plane at x = 0, we will get image charges of −qand −2q at x = −0.5 a and x = −1.5 a respectively.

This tells us that we will get three charges pushing on charge q at x = 0.5 a. The first chargewill be 2q at x = 1.5 a which will oppose q to the left. The other two charges, −q and −2q willattract q to the left as well. Sum all of the forces on q to get

F =q

4πε0

[2q

a2(−x) +

−qa2

(x) +−2q

(2a)2(x)

](3.32)

164


Figure 3.1: Mirror (image) charges induced as a result of an infinite grounding plate

=q

4πε0

[−4q

2a2− 2q

2a2− q

2a2

](3.33)

=1

4πε0

7

2

q2

a2(3.34)

Correct Answer

(E)

165


3.11 PGRE9277 #11


Recall that the energy of the capacitor is

U =1

2CV 2 (3.35)

Next, use Kirchhoff’s second law which tells us that the sum of all voltages about a closedcircuit is zero, to get

VC + VR = 0 (3.36)

Q

C+ IR = 0 (3.37)

Q

C+ QR (3.38)

where I = Q because current is defined as a moving charge, Q. Rearrange the previous equationand integrate to get

Q

C= −dQ

dtR (3.39)

166


−∫

dt

RC=

∫dQ

Q(3.40)

− t

RC= ln

(Q

Q0

)(3.41)

Q = Q0e−t/RC (3.42)

From this, we also conclude that

I = I0e−t/RC (3.43)

V = V0e−t/RC (3.44)

From our initial energy equation, U = 12CV

2, we get a voltage equation

V =

√2U

C(3.45)

and so our voltage equation is

√2U

C=

√2U0

Ce−t/RC (3.46)

U = U0e−2t/RC (3.47)

since we are concerned with the point at which half of the energy has dissipated, substitute inU = U0/2

U0

2= U0e

−2t/RC (3.48)

1 = 2e−2t/RC (3.49)

e−2t/RC = 2 (3.50)

− 2t

RC= ln(2) (3.51)

t =RCln(2)

2(3.52)

Correct Answer

(E)

167


3.12 PGRE9277 #12


Recall LaPlace’s equation

∇2V = 0 (3.53)

since the problem only tells us to concern ourselves with the φ component, we can integrateLaPlace’s equation to get

d2V (φ)

dφ2= 0 (3.54)

168


dV (φ)

dφ= A (3.55)

V (φ) = Aφ+B (3.56)

since we have the initial condition V (0) = 0, we know that B = 0

V (φ) = Aφ (3.57)

since V (a) = V0,

V0 = Aa (3.58)

A =V0

a(3.59)

compare this to the equations general form, V = Aφ, to get

V0

a=

V

φ(3.60)

V =V φ

a(3.61)

Correct Answer

(B)

169


3.13 PGRE9277 #13


If you are like me, you probably learned and memorized Maxwell’s equation, ∇ · ~B = 0 as the“There ain’t no magnetic monopoles” law. For this reason, you know you can immediately get IVas one of the laws that becomes INCORRECT. Next, consider that if we don’t require the magneticfield to curl back on itself in order to force the divergence of the magnetic field to zero, then it ispossible to get the electric field to not curl which tells us II could also be INCORRECT and wechoose (D).

Correct Answer

(D)

170


3.14 PGRE9277 #14


From Stefan-Boltzmann’s Law, we get the power radiation of a black body as

j∗ = uT 4 (3.62)

which tells us that doubling the temperature of the black body will alter the power proportionalto the fourth power

j∗ = u(2T )4 (3.63)

= 16(uT 4) (3.64)

since power is energy over time and heat energy is

Q = mc∆T (3.65)

we get that a unit increase in energy will increase temperature by 0.5C and, therefore, 16 units ofenergy increase will get a change in temperature of 8C.

Correct Answer

(C)

171


3.15 PGRE9277 #15


Heat capacity of a molecule is determined by the number of degrees of freedom of the molecule.For example, in a monatomic gas, the heat capacity is

CV =3

2R (3.66)

where the 3 comes from our 3 degrees of translational freedom (x, y, z). For a springy, diatomicmolecule, we have to then include additional degrees of freedom for its rotation and its vibration

CV =3

2R+Rrot +Rvib =

7

2R (3.67)

Correct Answer

(C)

172


3.16 PGRE9277 #16


The maximum efficiency of a Carnot engine (a theoretically, perfectly efficient heat engine) is

η =W

QH= 1− TC

TH(3.68)

where η is the maximum efficiency, W is the work done by the system, QH is the heat input,TC is the absolute temperature of the cold reservoir and TH is the absolute temperature of the hotreservoir. Convert temperatures to units of Kelvin to get

η = 1− TCTH

(3.69)

= 1− 800 K

1000 K(3.70)

= 1− 0.8 (3.71)

= 0.2 (3.72)

equate η to the work over heat equation with a heat of QH = 2000 J to get

W

QH= 0.2 (3.73)

W = (0.2)(2000 J) (3.74)

= 400 J (3.75)

Correct Answer

(A)

173


3.17 PGRE9277 #17


The problem tells us that the frequency in the x-direction is twice that of the y-direction, so weknow that the oscilloscope will be plotting

f(t) = sin(ωt) + sin(2ωt) (3.76)

Using your copy of Mathematica provided in the front of your test booklet, plot the functionto get the figure below

However, in the case that your test booklet doesn’t have Mathematica, we can eliminate options(E) and (D) because the superposition of two sine waves shouldn’t give us either of the two curves.

174


Next, eliminate (B) and (C) because both represent just a function of sin(ωt) or cos(ωt), not asuperposition of trig functions.

Correct Answer

(A)

175


3.18 PGRE9277 #18


With coaxial cables, impedance matching is necessary because differences in characteristic impedancecan result in signal reflection, particularly in the case of a damaged/kinked line or an incor-rectly/damaged termination to the cable.

Correct Answer

(C)

176


3.19 PGRE9277 #19


Use Newton’s second law and our gravitational force law,

F = ma (3.77)

F = GmM

r2(3.78)

combine the two and cancel m to get

ma = GmM

r2(3.79)

a = GM

r2(3.80)

M =gr2

G(3.81)

to simplify the mental math, assume thatG = 6×10−11 m3/kg·s2, g = 10 m/s2 and r = 6×106 m

M =gr2

G(3.82)

=

(10 m/s2

) (6× 106 m

)2(6× 10−11 m3/kg · s2)

(3.83)

=6× 1013

6× 10−11(3.84)

= 1× 1024 kg (3.85)

which is closest to (A).

Correct Answer

(A)

177


3.20 PGRE9277 #20


based on the given diagram, we know it can’t be true that d < ω so (C) and (E) can be eliminated.Next, recall that the equation for constructive interference in double slit diffraction is

d sin(θ) = m1λ (3.86)

additionally, we know that we will get a “missing” interference maximum when the constructivedouble slit equation coincides with the single slit diffraction, so we use

ωsin(θ) = m2λ (3.87)

get both equations equal to sin(θ) and set them equal to one another and solve for d,

ω

m2=

d

m1(3.88)

d =m1

m2ω (3.89)

178


comparing this to (A) and (B), we aren’t going to get an irrational number (i.e.√

2 or√

3)with some fraction of integers so we can confidently choose (D).

Correct Answer

(D)

179


3.21 PGRE9277 #21


In thin film optics, and most optics in general, I is unequivocally silly so eliminate all choices whichinclude it, i.e. (A) and (D). Next, consider III and IV and recognize that both are correct. Morespecifically, it is true that we will get a phase change as the light transitions from a lower indexof refraction to a higher one (as it enters the bubble) and no phase change as it transitions froma higher index of refraction to a lower one (as it exits the bubble). Eliminate any options thatdon’t include both of these choices, specifically (C). Finally, when considering option II, recall thatequations for thin film optics

Constructive Interference 2t = λ/2

Destructive Interference 2t = λ

which tells us that in either case, the thickness of the bubble is generally less than the wavelength(i.e. half or a quarter)

Correct Answer

(B)

180


3.22 PGRE9277 #22


The quick and easy solution, dare I say the “Plug-n-chug” method, is to use the magnificationequation for a convex lens

M =f0bj

feye(3.90)

We are given the objective focal length, fobj = 1.0 meter and the magnification M = 10 so wesolve for feye,

feye =fobjM

(3.91)

=1.0 m

10(3.92)

= 0.1 m (3.93)

Lastly, we get the total distance from the sum of the two focal lengths,

d = fobj + feye (3.94)

= 1.0 m + 0.1 m (3.95)

= 1.1 m (3.96)

Correct Answer

(D)

181


3.23 PGRE9277 #23


The average speed of a conduction electron is described by the Fermi velocity, with equation

vf =

√2Efm

(3.97)

where the Fermi energy, Ef , is related to the Fermi temperature, Tf , by

Ef = kTf (3.98)

where k is Boltzmann’s constant. Plug everything into our equation to get

vf =

√2Efm

(3.99)

=

√2kTfm

(3.100)

simplify the values for the electron mass, fermi temperature and Boltzmann’s constant

vf =

√2kTfm

(3.101)

=

√2(1× 10−23 J/K)(80, 000 K)

10× 10−31 kg(3.102)

≈√

10× 1031

16× 1019 m2/s2(3.103)

≈ 1× 106 m/s (3.104)

Correct Answer

(E)

182


3.24 PGRE9277 #24


Argon, like the other noble gases, has a full valence shell. Ionic bonding is bonding between ametal and a non-metal and also requires that one atom lack electrons from its valence shell andthe other have an excess charge (think along the lines of the salt molecule with Na+ + Cl−) sowe can eliminate (A). Covalent bonding, on the other hand, involves the sharing of electrons tofill out the valence shell when an atom is lacking electrons in its valence shell, but again Argonisn’t missing any electrons so we eliminate (B) and then eliminate (C). Finally, since Argon isn’ta metal, eliminate (D). As it turns out, argon bonds to other argon atoms by induced dipoles viathe Van der Waals force.

Correct Answer

(E)

183


3.25 PGRE9277 #25


The ability of a particle to pass through solid material is increased as the size of the particledecreases and as the speed of the particle increases. From this, we would expect larger particleslike protons and neutrons to struggle to pass “deep underground” and we can eliminate (A), (B)and (C). Between (D) and (E), all of the listed particles are sufficiently fast and small but neutrinoshave no charge, unlike positrons, electrons and muons so we should expect them to not be caughtor repelled away from electrons in the matter they are trying to pass through and we should expectthem to be in the final answer.

Correct Answer

(D)

184


3.26 PGRE9277 #26


At time t = 0, the number of counts is at 103.5 counts. We can approximate this value by

103.5 = 107/2 = (107)1/2 =√

100 · 100 · 100 · 10 = 10 · 10 · 10 ·√

10 (3.105)

if we approximate√

10 ≈ 3 then we get

103.5 = 3000 (3.106)

This means that half the counts will be 1500 and so at about 17 min, where the counts is103 = 1000 counts, we’ve passed our half way point and we can eliminate (D) and (E). For (A),(B) and (C), the counts are approximately

185


(A) 103.4 ≈ 2500 counts

(B) 103.2 ≈ 1600 counts

(C) 103.1 ≈ 1300 counts

so we choose (B) which is the closest

Correct Answer

(B)

186


3.27 PGRE9277 #27


The width of the wave-function is determined by the relative size of ∆p and ∆x in the Heisenberguncertainty principle

∆x ∆p ≥ h

2(3.107)

Recalling our momentum equation

p = hk (3.108)

where h is a constant, we get ∆p = h∆k. Combine this with the Heisenberg uncertaintyprinciple and solve for ∆k

∆x ∆p ≥ h

2(3.109)

(∆x) (h∆k) ≥ h

2(3.110)

∆k ≥ 1

2(∆x)(3.111)

which gives us the inverse relationship between ∆k and ∆x like in (B).

Correct Answer

(B)

187


3.28 PGRE9277 #28


In quantum mechanics, the probability of finding the system in a certain state is given by theintegral over the squared wave function, or in our case Y m

l (θ, φ). The problem asks us about thestate with azimuthal orbital quantum number, m=3, so we take the squared values of the formY 3l (θ, φ). Since our wave function is already normalized, we just need to square the values of the

first two terms of ψ

ψm=3 = 52 + 12 = 26 (3.112)

ψtotal = 52 + 12 + 22 = 30 (3.113)

therefore, the probability of ψm=3 out of the total ψtotal is

ψm=3/ψtotal = 26/30 (3.114)

= 13/15 (3.115)

Correct Answer

(E)

188


3.29 PGRE9277 #29


Questions regarding the infinite square well (particle in a box) and its related plot show up onnearly every test. For this reason, and simply because it is something you oughta know, you shouldmemorize some of the fundamental aspects of the infinite square well graph. First, the solution tothe infinite square well is sinusoidal, which allows us to eliminate (A), (C) and (D). Next, when thefunction impinges on the infinite barrier at x1 and x2, the amplitude continually decreases toward0 and stops oscillating, which then allows us to eliminate (E).

Correct Answer

(B)

189


3.30 PGRE9277 #30


Considering how often the ground state energy of positronium comes up in the GRE, and becauseit’s trivial to memorize, recall that its value is half that of hydrogen’s ground state

E0,pos =E0,hyd

2=

13.6 eV

2= 6.8 eV (3.116)

Keep in mind that this is for the ground state but we need the binding energy in state n = 2.Using the Bohr equation, we see that the energy is inversely proportional to the squared value ofthe fundamental quantum number, n

En =Z2E0,pos

n2(3.117)

since Z = 1, the final answer is

E2 =E0

8(3.118)

Correct Answer

(E)

190


3.31 PGRE9277 #31


Recall our generic form for the Russel-Saunders term symbol

2s+1LJ (3.119)

the problem specifies that the helium atom has term symbol

3S (3.120)

so we know that 2s+ 1 = 3, J = 1 and L = S. Solve for s to get s = 1 and then recalling theangular momentum quantum number can be found by

j = l + s (3.121)

since S = L, and S corresponds to 0, (i.e. (S, P,D, F, . . .)→ (0, 1, 2, 3, . . .)), we finally get

j = 0 + 1 = 1 (3.122)

Correct Answer

(B)

191


3.32 PGRE9277 #32


First, recall that the equivalent resistance of resistors in parallel can be found by

1

Req=

1

R1+

1

R2+ · · · (3.123)

From Equation 3.123, we can see that the equivalent resistance of the R3-R4 system and R2-R5

system will have less resistance than R1 on its own. Additionally, we know that power is relatedto voltage and current by

P = I2 ·R =V 2

R(3.124)

Which tells us that in addition to R1 having the highest resistance, it will also be the case thatthe highest current will interact with this resistor and so we should expect R1 to have the biggestamount of current to dissipate.

Correct Answer

(A)

192


3.33 PGRE9277 #33


Using the rules for resistors and in parallel, find the equivalent resistance of all resistors in thecircuit as Req = 75 Ω. This tells us that the current of the entire circuit should be

I =V

R=

3.0 V

75 Ω(3.125)

Next, finding the equivalent resistance of resistors R3, R4 and R5, you’ll find that you have thesame resistance in the RHS of the circuit as you do in the LHS (i.e .just resistor R2). This tellsus that half the current (3/150 A) will go through resistor R2 and the other half will pass throughR3,4,5. Finally, find your equivalent resistance between R3 and R4 to get 2o Ω and then solve forthe voltage

V = IR =

(3

150A

)(20 Ω) = 0.4 V (3.126)

Correct Answer

(A)

193


3.34 PGRE9277 #34


Because we are talking about electromagnetic waves in a waveguide, we can throw out all conditionswhich don’t give the results as being orthogonal for the electric and magnetic fields separately, i.e.(A) and (C). Then, we can eliminate (B) and choose (C) because the electric field doesn’t propagatein the same direction as the direction of current so the transverse electric field, Et, should be 0.

Correct Answer

(D)

194


3.35 PGRE9277 #35


An optical diffraction grating works in effectively the same way that a double slit, triple slit, etcinterference except with more slits. Based on this similarity, we should try to use our genericequation for double slit diffraction

d sin(θ) = mλ (3.127)

θ = arcsin

(mλ

d

)(3.128)

where d is the distance between slits, θ is the angle of incidence, λ is the wavelength andm = 0, 1, 2, . . .. The wavelength is given as 5200 angstroms and we can find the distance betweenslits by assuming that the 2000 slits are evenly spaced across each centimeter of the diffractiongrating. Plug this all into Equation 3.128 to get

θ = arcsin

(mλ

d

)(3.129)

= arcsin

((1)(5200angstroms)

(0.0005 cm)

)(3.130)

= arcsin(0.1) (3.131)

≈ 6 (3.132)

Correct Answer

(B)

195


3.36 PGRE9277 #36


The problem tells us, quite explicitly might I add, that the surface which the E&M wave interactswith is a perfect conductor. This tells us that the net electric field must go to 0 as a result of theinteraction. Since the electric field impinges on the surface with some value E0, we know that thevalue afterwards must be equal and opposite,

E1 = −E0 (3.133)

or, in words, its direction must reverse while maintaining the same magnitude. The magneticfield, however, won’t change direction because of the conductor so (C) becomes the obvious choice.

Correct Answer

(C)

196


3.37 PGRE9277 #37


If you get this problem wrong, you probably should give up any aspirations you’ve ever held whichinvolve you being a physicist. The problem tells us that the π0 meson decays into 2 photons whichhead off in opposite directions. Arguably, the most fundamental and important aspect of relativityis that photons travel at the speed of light, 1 C, in all reference frames. From this, we know theonly solutions can be (A) or (D). Since the problem tells us that the photon, γ2, proceeds in the“backwards” direction, the sign should be negative and we choose (A).

Correct Answer

(A)

197


3.38 PGRE9277 #38


Before starting, let’s take a look at the general time dilation equation

∆t′ = γ∆t =∆t√

1− v2/c2(3.134)

from this, we can see that ETS has, quite rudely, written things in terms of inverse Lorentzfactors. At this point, I highly recommend that you quietly curse ETS under your breath and thenre-write the equations in a more standard form,

(A) ∆t1 = γ12∆t2

(B) ∆t3 = γ13∆t1

(C) ∆t3 = γ23∆t2

(D) ∆t2 = γ23∆t3

(E) ∆t2 = γ23∆t1

Right off the bat, eliminate (E) because it incorporates a Lorentz factor with frame 3 in it whenthe time for frame 3 isn’t even represented. Next, we know from the time dilation effect that the

198


time of a moving frame in relation to a stationary frame will appear to be longer in the stationaryframe. If we let the stationary frame be S1, then we see that (A) incorrectly concludes that timein the stationary frame would be longer in the stationary frame than the moving frame. (C) and(D) don’t involve the stationary frame at all and we aren’t given enough information to concludeanything about the relation between the two inertial frames, so those are both likely to be wrong.Only (B) correctly predicts the stationary vs inertial relationship

Correct Answer

(B)

199


3.39 PGRE9277 #39


Recall that the sine function is an odd function, just like the plot given in the problem, while thecosine function is an even function. Based on this, we should eliminate all solutions which utilizecosine functions, i.e. (C), (D), and (E). I should point out that there is always a possibility thatthe cosine function could be shifted to produce and sine-esque plot, however in our case none ofthe solutions feature the necessary shift. Next, to choose between (A) and (B), plug in t = π/ωwhich should give us an amplitude of V (t) = −1. in (A), we get

V (t) =4

π

∞∑1

1

nsin

(nπω

ω

)(3.135)

=4

π

∞∑1

1

nsin (nπ) (3.136)

From Equation 3.136, it should be clear that V (π/ω) will be 0 for all values of n, so our solutionmust be (B).

Correct Answer

(B)

200


3.40 PGRE9277 #40


The acceleration at any point on the cylinder will be equal to the sum of all its accelerations. Sincethe problem explicitly specifies that the cylinder doesn’t slide, we know that there are no lateralforces to contribute. The only acceleration we have is the centripetal acceleration from its rotationwhich will be pointing toward the center of the cylinder. When the point under consideration istouching the surface of the plane, the acceleration must point up to point towards the cylindercenter.

Correct Answer

(C)

201


3.41 PGRE9277 #41


The kinetic energy of a rotating object is related to angular frequency, ω, and moment of inertia,I, by

EK =1

2Iω2 (3.137)

The moment of inertia is given as I = 4 kg·m2 and we are told that the initial angular frequencyof 80 rad changes down to 40 rad. Using Equation 3.137 and accounting for the change in ω, wecalculate the kinetic energy as

EK =1

2I(ω2f − ω2

i

)(3.138)

=1

2

(4 kg ·m2

) ((80 rad/s)2 − (40 rad/s)2

)(3.139)

=1

2

(4 kg ·m2

)(4800 rad/s) (3.140)

= 9600 J (3.141)

Correct Answer

(D)

202


3.42 PGRE9277 #42


Recall our equation for torque

τ = Iα (3.142)

where α is the angular acceleration. We are given the change in angular velocity as ω =40 rad/sec. Since the rate of change of angular velocity is given, we can find the average angularacceleration as

α =∆ω

∆t=

40 rad/s

10 s= 4 rad/s2 (3.143)

Plug our angular acceleration value from Equation 3.143 into Equation 3.142 to get

τ = Iα (3.144)

=(4 kg ·m2

) (16 rad/s2

)(3.145)

= 16 N ·m (3.146)

Correct Answer

(D)

203


3.43 PGRE9277 #43


This problem is one of those infuriating exam questions that you either know, or you don’t. In thisinstance, you can only be sure you’ve got the right answer if you recall that Noether’s theorem tellsus that pn is a constant under the condition

pn =∂L

∂qn(3.147)

Even if you don’t know this, we can try to eliminate some of the options based on some commonsense

(A) An ignorable coordinate is a coordinate that doesn’t show up in the Lagrangian which is notthe case

(B) I can’t think of a compelling reason to eliminate this one

(C) There is no reason to assume that differentiating the Lagrangian with respect to qn will beundefined except when ∂qn = 0 which would be a poor assumption

(D) There is no mention made of a time dependence so it is unlikely that ∂L∂qn

= ddt

(∂L∂qn

).

(E) Keep in mind that the Lagrangian and Hamiltonian are both measures of Energy and it is notlikely that you can differentiate only one but have them each keep the same units.

Correct Answer

(B)

204


3.44 PGRE9277 #44


Recall that the Langrangian is the difference between the kinetic and gravitational potential energy,

L = T − V (3.148)

We can first eliminate (E) because the potential term most oppose the kinetic term and we needat least one minus sign. Next, we know the solution must have some potential energy term mgy sothat all of the energy is kinetic at bottom of the parabola, so we eliminate (C) and (D). Finally,we know that the kinetic energy pieces should be adding together, not fighting one another, so wechoose (A).

Correct Answer

(A)

205


3.45 PGRE9277 #45


Before the ball is dropped, the net energy of the system is all potential equal to mgh. Once theball is released, the potential is converted to kinetic until the ball hits the ground and all mgh ofthe energy is now kinetic equal to 1/2mv2

0. The problem tells us that the velocity after collision isonly 4/5 of its initial velocity, v0, so the kinetic energy on its way back up is

Tf =1

2m

(4

5v2

0

)(3.149)

=1

2m

(16

25v2

0

)(3.150)

If we compare this to the original kinetic energy, it is clear that the final kinetic energy and,therefore, potential energy is 16/25 = 0.64 times as big as its initial energy.

Correct Answer

(D)

206


3.46 PGRE9277 #46


The critical isotherm refers to a curve that has the property that the derivative of the pressurewith respect to the volume is 0

207


∂P

∂V= 0 (3.151)

Of the curves shown, only curve 2 has a point where taking the tangent to the curve results ina horizontal line (i.e. the derivative of the curve is 0). This occurs at precisely the point where thevertical and horizontal dashed lines cross.

Correct Answer

(B)

Alternate Solution

Without knowing anything about isotherms, we can eliminate some options through a bit of rea-soning. First, eliminate curves 3, 4 and 5 because they are all effectively the same, especially whendiscussing everything in qualitative terms. Next, you can eliminate curve 1 because, unlike curve2, it is the same as all other similar curves above the horizontal dashed line and we would expectthe solution to be unique.

Correct Answer

(B)

208


3.47 PGRE9277 #47


This problem asks us which region will have vapor and liquid in equilibrium which tells us that weexpect to have both states present in the region. The quickest way to determine the answer is to

209


consider each region in its limit

(A) In region A, the volume extends to extremely small sizes and this would cause vapor to getcompressed to liquid. This likely won’t support vapor and liquid phases at the same time.

(B) Region B represents a middle ground of pressure and volume so there are no glaring limitissues.

(C) In region C, the volume can blow up to infinity which would likely force everything to a vaporphase.

(D) Region D allows the volume and the pressure to blow up to infinity, meaning there willabsolutely be states which either force everything to vapor or everything to liquid.

(E) Region E will allow the pressure to blow up to infinity which will force any vapor present intoliquid.

Only (B) lacks a limiting value that could potentially ruin our equilibrium.

Correct Answer

(B)

210


3.48 PGRE9277 #48


At some point in your undergraduate career, you were probably forced to calculate some standarddeviations by hand. If you did, then you likely used this equation

σ =

√√√√ N∑i=1

pi (xi − x)2 (3.152)

From Equation 3.152, we know to look for a square root and we can then eliminate (A), (D)and (E) based on this condition. Next, we should expect to see σ2 values rather than simply σ sowe eliminate (B) and choose (C).

Correct Answer

(C)

211


3.49 PGRE9277 #49


Considering the amount of variation between the possible solutions, let’s do an approximation.Muons move very quickly and the problem tells us that the motion is relativistic, let’s approximatethe speed of a muon going nearly the speed of light as just the speed of light

vµ = 3.0× 108 m/s (3.153)

Since the scintillators are 3.0 meters apart, we can find the time scale as

∆t =∆x

vµ(3.154)

=3.0 m

3.0× 108 m/s(3.155)

= 1× 10−8 s (3.156)

= 10 nanoseconds (3.157)

so we will want to choose the nanosecond range, i.e. solution (B).

Correct Answer

(B)

212


3.50 PGRE9277 #50


First of all, eliminate (E) because we would never be so lucky that we could expand a wave functioninto basis states under any and all circumstances. Next, eliminate (C) and (D) because A and Bas well as α and β are qualitatively identical to one another so if (C) was true, (D) should also betrue and we can’t choose both. Finally, eliminate (A) because whether or not the two eigenvaluesare non-degenerate should have nothing to do with the basis functions of the wave function.

Correct Answer

(B)

213


3.51 PGRE9277 #51

214



Starting with classical momentum, P = mx, recall that the analog for the expectation value ofmomentum is

〈p〉 = md〈x〉dt

(3.158)

in the infinite square well, the expectation value for position is

〈x〉 =a

2(3.159)

since 〈x〉 is nothing but constants, if we take the derivative of it then it goes to 0 and so does〈p〉.

Correct Answer

(B)

Alternate Solution

More rigorously, we can use the general equation for expectation value

〈p〉 =

∫ ∞−∞

ψ∗h

iψ dx (3.160)

we are given the wave function so if we plug everything in, we get

〈p〉 =

∫ ∞−∞

ψ∗h

iψdx (3.161)

=2nπh

a2i

∫ ∞−∞

sin

(nπx

a

)cos

(nπx

a

)(3.162)

and since cosine and sine are orthogonal with respect to each other, integrating over all of xwill result in each one canceling out the other and the total area is 0.

Correct Answer

(B)

215


3.52 PGRE9277 #52

216



The condition for orthonormality,

〈n|m〉 = δnm (3.163)

is a function of the Kronecker delta type, i.e.

δnn = 1 (3.164)

δnm = 0 (3.165)

This is precisely the description in the problem so we choose the orthonormality condition.

Correct Answer

(B)

217


3.53 PGRE9277 #53

218



The energy of the infinite square well isn’t constant so we can immediately eliminate (C) and (E).Next, we can eliminate (A) because, in theory, there shouldn’t be an upper bound to the energy and(A) suggests that there is such a maximum energy. Lastly, you can quickly check the coefficientsfor (B) and (D) by recalling the energy equation derived form the Schrodinger equation,

E =p2

2m+ V (3.166)

since V = 0 inside the potential, we get E = p2/2m which tells us the coefficient should be 1/2rather than 1/8 and so we can choose (B).

Correct Answer

(B)

219


3.54 PGRE9277 #54


This problem is a left/right hand rule paradise (or nightmare depending on your familiarity withthe rules). First of all, start by applying the right hand rule to the vertical wire with your thumbin the up direction. This tells us that the magnetic field is pointing into the plane on the side ofthe loop of wire. Now, apply the right hand rule of a magnetic field into the plane and into theloop such that your thumb is pointing into the plane and your hands are looping clockwise (Figure

220


3.2), allowing us to eliminate (A), (B) and (C).Lastly, use the left hand rule (Figure 3.3) with your thumb in the direction of the current,

pointer finger into the plane and middle finger in the direction of the resulting force to find thatthe left side of the loop goes to the left and the right side of the loop goes right, leaving us with(E).

Correct Answer

(B)

I I

B

B

Figure 3.2: Right hand rule for a magnetic field passing through a loop of wire

Figure 3.3: Left hand rule for a current through a magnetic field

221


3.55 PGRE9277 #55


The quickest way to solve this problem is to consider the limits of the lengths a and b. If either ofthese go to 0 then the flux goes to 0 and so too does the force. For this reason, we can eliminateany solution that doesn’t have some dependence on both a and b, i.e. (A) and (B). Next, note that(C) and (E) blows up to infinity when a→ 0 so eliminate both of these.

222


Correct Answer

(D)

223


3.56 PGRE9277 #56


The general equation for energy of a quantum harmonic oscillator in state n is

E =

(1

2+ n

)hν (3.167)

In its ground state, n = 0 so the solution should be

E =

(1

2+ 0

)hν (3.168)

=1

2hν (3.169)

Correct Answer

(D)

224


3.57 PGRE9277 #57

225



Recall Faraday’s law which states that a current will be induced in a conductor due to a change inmagnetic flux

|ε| =∣∣∣∣dφBdt

∣∣∣∣ (3.170)

from the description, we know that the half circle is rotating “uniformly” so the induced currentshould be constant and we can eliminate (C), (D) and (E). As the half circle begins to enterthe rectangle, it will have a constantly increasing induced current and once it begins to exit therectangle, it should have a constantly decreasing induced current. Option (A) gives us this featurebut (B) suggest a constantly increasing increase in induced current, which is not what we want.

Correct Answer

(A)

226


3.58 PGRE9277 #58


In ground state, the number of electrons on the atom should be the same as Z = 11. The quickestway to figure out the number of electrons proposed in each of the 5 options is to sum up all of thesuperscripts in each configuration. Doing so will eliminate option (E). Next, we can eliminate (A)because we should completely fill 2p to 2p6 before moving to the next energy level. Next, eliminate(B) because the s level can’t have 3 electrons in it. Finally, recall your energy level diagrams (Figure3.4) to see that we should progress to 3s after 2p as opposed to going from 2p to 3p

Figure 3.4: Energy level diagram of Sodium

Correct Answer

(C)

227


3.59 PGRE9277 #59


In its ground state, the helium atom has 2 electrons in the first shell which, by the Pauli ExclusionPrinciple, has one spin up and one spin down. The spin multiplicity, which is 2S + 1, is whatdetermines whether an atom is a singlet, doublet, triplet, etc.

Singlet 2(0) + 1 = 1

Double 2(1/2) + 1 = 2

Triplet 2(1) + 1 = 3

In the case of helium, since we have two electrons with opposite directions of spin, they cancelto give us S = 0 which is a singlet

Correct Answer

(A)

228


3.60 PGRE9277 #60


If you recall the equation for cyclotron resonance frequency, this problem is a quick plug-n-chugproblem

ωc =eB

m(3.171)

from the problem description, we are given values for B and m and we can get the elementarycharge value e from the front of the test booklet. Plug these values into Equation 3.171 and solve

ωc =

(1.6× 10−19 coulombs

)(1 tesla)

(0.1) (9× 10−31 kg)(3.172)

≈ 2× 1012rad/s (3.173)

Correct Answer

(D)

229


3.61 PGRE9277 #61


Start by recalling the frequency equation for a pendulum,

ω =

√mgrcom

I(3.174)

where the moment of inertia will always be that of a point mass, I = mr2. for the firstpendulum, all of the mass is located at the bottom of the pendulum which makes our center ofmass at a distance r, making our frequency equation

ωI =

√2mgr

2mr2=

√g

r(3.175)

for the second pendulum, however, the masses are separated at a distance of r/2 and r whichforces the center of mass to be 3/4 r, so the frequency equation becomes

ωII =

√2mg(3/4)r

(m(1/2 r)2) + (mr2)(3.176)

230


=

√(6/4)g

(5/4)r(3.177)

=

√6

5

g

r(3.178)

at which point it should be clear that pendulum II has a frequency of√

6/5 that of pendulum I

Correct Answer

(A)

231


3.62 PGRE9277 #62


Anytime you get an increase in volume, you will be doing work so we should first eliminate (A). Bythe same type of reasoning, if V1 = V0 then the work should be 0 so we can eliminate (B). Next,because the problem tells us that we can treat the gas as an ideal gas, which should mean that thetype of gas is irrelevant, we should be able to ignore specific heats because they are dependent onthe type of gas. From this, eliminate (D). Lastly, we need to use the thermodynamic work equationand ideal gas law to see that the solution should have a natural log,

W = −∫ V1

V0PdV (3.179)

P =nRT

V(3.180)

Combine Equation 3.179 and Equation 3.180, and integrate to get

W = −∫ V1

V0

(nRT

V

)dV (3.181)

= nRT [ln(V1)− ln(V2)] (3.182)

= nRT ln

(V1

V0

)(3.183)

so we should choose (E)

Correct Answer

(E)

232


3.63 PGRE9277 #63


If you’re clever, you’ll notice that (A) and (B) are exactly opposite so they can’t both be wrongand we know it must be one or the other. In order to choose between the two, recall that a systemof maximal probability is in its most stable state so we would expect no spontaneous changes andwe choose (D).

Correct Answer

(D)

233


3.64 PGRE9277 #64


Immediately eliminate (D) because the presence of Ez = kz guarantees we will have some sort ofelectric field. Next, eliminate (C) because there is nothing about (Ex, Ey, Ez) that forces the electricfield to vary, especially considering that the only non-zero component is scaled by a constant k andz can be constant as well. Next, eliminate (A) because nothing about what is given demonstratesany time dependence and, for that matter, it says essentially the same thing as (C) which wealready decided wasn’t correct. If (B) is also untrue, then we choose (E). However, recall Gauss’law

∇ · ~E =ρ

ε0(3.184)

which clearly states that we should get a charge density in the region of the electric field.

Correct Answer

(B)

234


3.65 PGRE9277 #65


We are looking for angular frequency so we should expect to get units of inverse time, recall the SIunits for each variable/constant used in the solutions

Q/q = C (3.185)

ε0 = C2/N ·m (3.186)

m = kg (3.187)

R = m (3.188)

check each of the potential solutions

235


(A) Qq2πε0mR2 = m

sec2

(B) Qq4πε0mR2 = m

sec2

(C) Qq2πε0mR3 = 1

sec2

(D)√

Qq4πε0mR2 =

√m

sec

(E)√

Qq2πε0mR3 = 1

sec = sec−1

Correct Answer

(E)

Alternate Solution

If you insist on doing this problem in the rigorous fashion, start with Coulomb’s law

1

4πε0

qQ

R2(3.189)

since there are two charges pushing on the central charge q, we change this to account for bothwith

F =1

4πε0

2qQ

R2(3.190)

=1

2πε0

qQ

R2(3.191)

Next, recall that the force for an oscillating spring is

mx = −kx (3.192)

and has angular frequency

ω2 =k

m(3.193)

re-arrange Equation 3.192 to get everything in terms of k/m to get

ω2 =k

m= − x

x(3.194)

or equivalently,

ω2x = x (3.195)

Finally, since F = mx, use Equation 3.195 with Equation 3.191 (and let x = R) to solve forangular frequency

236


mx =qQ

2πε0R2(3.196)

mω2x =qQ

2πε0R2(3.197)

ω2 =qQ

2πε0mR3(3.198)

ω =

√qQ

2πε0mR3(3.199)

Correct Answer

(E)

237


3.66 PGRE9277 #66


Recall from the Work-Energy theorem

W = ∆EK (3.200)

Since energy is conserved, the kinetic energy used to move the chain up will be equal to thetotal potential energy at the top of the axle. The potential energy can be found by

W = EG = mgh = (10 kg)(10 m/s2)(10 m) = 1000 J (3.201)

Correct Answer

(C)

Alternate Solution

You could, if you aren’t as industrious as I, set up a differential to account for the changing mass ofthe chain as it is lifted up. However, to do this quickly let’s first consider our simple work equation

W = F∆x (3.202)

To make a quick approximation, assume that g = 10 m/s2 and make measurements for every1 m of change in the chain which will account for a decrease in 20 N. The initial change of of 1 mwith a 10 meter long chain with 2 kg per meter is 200 N ·m. The next bit of work will be 180 N ·mand then 140 N ·m and so on to get

Wnet = 200 + 180 + 160 + 140 + 120 + 100 + 80 + 60 + 40 + 20 = 1060 N ·m (3.203)

which is closest to (C).

Correct Answer

(C)

238


Alternate Solution

If you insist on doing things the hard way, start out with the integral form of work

W = −∫F · dx (3.204)

In our problem the force changes as the length changes and it changes proportional to

F = mg = (20 kg− 2l)g (3.205)

that is to say that the mass is initially 20 kg and then decreases by 2 times the length of thechain. Plug Equation 3.204 into Equation 3.205 to get

W = −∫ 0

10(20 kg− 2l)gdl (3.206)

=

∫ 10

020g − 2gl dl (3.207)

=[20gl − gl2

]10

0(3.208)

= (2000 N ·m)− (1000 N ·m) (3.209)

= 1000 N ·m (3.210)

Correct Answer

(C)

239


3.67 PGRE9277 #67


The intensity of the light that gets transmitted through the polaroid is given as

I = A+B cos(2θ) (3.211)

Which tells us that one term can go to 0 when cos(2θ) goes to 0 while the other term, the Aterm, can’t. This tells us that the light is composed of two different types of polarizations and weeliminate (A), (B) and (E). Lastly, to distinguish between (C) and (D), recall Malus’ law whichstates that plane polarized light has intensity proportional to

I = I0 cos2(θ) (3.212)

which we can re-write in a similar form as Equation 3.211 by the double angle identity

2 cos2(a) = 1 + cos(2a) (3.213)

which gives us (C).

Correct Answer

(C)

240


3.68 PGRE9277 #68


To calculate optical resolution, we need to use the Rayleigh Criterion,

sin(θ) = 1.22λ

d(3.214)

the angle and wavelength are given so we can re-arrange Equation 3.214 to solve for d,

d =1.22λ

sin(θ)(3.215)

By a small angle approximation, which we can make because the angle is in microradians, letsin(θ) = θ and then convert all of the values into the same units to get

d =1.22λ

θ(3.216)

=1.22(5.5× 10−7 m)

8.0× 10−6 m(3.217)

≈ 1× 10−1 m (3.218)

≈ 10 cm (3.219)

which is closest to (C).

Correct Answer

(C)

Alternate Solution

Even without remembering the necessary equation, you eliminate some choices by a bit of commonsense. Because we are talking about a telescope reflecting mirror, we can probably eliminate (A)and (B) as being ridiculously small to be a reflecting mirror on a telescope. On the other end of thespectrum, a 100 m reflecting mirror would be ridiculously too big. In fact, the largest telescopeson earth peak at or just sightly above 10 m so 100 m is very unlikely and we can eliminate (E). Atthis point, you can now guess between (C) and (D).

241


Correct Answer

(C)

242


3.69 PGRE9277 #69


A photon travels through a medium with an index of refraction, n, according to the equation

v =c

n(3.220)

The index of refraction of the glass is, n = 1.5, so we plug that in and solve

v =c

3/2(3.221)

=2

3c (3.222)

Correct Answer

(D)

243


3.70 PGRE9277 #70


Start off with the relativistic equation

E2 = p2c2 +m2c4 (3.223)

The problem tells us that the energy is E = 100mc2 so we plug that into Equation 3.223,

(100mc2)2 = p2c2 +m2c4 (3.224)

10000m2c4 = p2c2 +m2c4 (3.225)

We could then combine terms with m2c4 but doing so will make almost no change to the10000m2c4 so let’s just ignore it. Finally, solve for the p in Equation 3.225 to get

p2c2 = 10000m2c4 (3.226)

p2 = 10000m2c2 (3.227)

p = 100mc (3.228)

which is (D).

Correct Answer

(D)

244


3.71 PGRE9277 #71

245



Start off by recalling that as temperature and, therefore, net energy of a system blows up toinfinity, energy levels will start to become equally populated. Based on this, get rid of any solutionthat doesn’t account for a temperature dependence, specifically (A). Also from this fact, we caneliminate any solution that doesn’t give the average number as N0/2 when T → ∞, which wouldbe (C). Finally, when the temperature is minimized (i.e. let T = 0), we would expect all N0 of theparticles to be at energy level E1 so plug this into the remaining options to find

(B) N0

1+e−k/ε(0)= N0

(D) N0

1+ek/ε(0)=∞

(E) N0eε/kT

2 =∞

So we choose (B).

Correct Answer

(B)

246


3.72 PGRE9277 #72

247



Recall that heat capacity is the derivative of energy with respect to temperature,

CV =∂U

∂T(3.229)

the problem gives us the energy so we take the derivative of it

CV =dU

dT(3.230)

=d

dT(E1N0) +

d

dT

(N0ε

1 + eε/kT

)(3.231)

= N0k

(ε

kT

)2 eε/kT

(1 + eε/kT )2(3.232)

which is (A). The worst part of this problem is doing the quotient rule under pressure but youcan recognize certain pieces that should be there and only do part of the derivative to get the rightanswer.

Correct Answer

(A)

248


3.73 PGRE9277 #73

249



We can immediately eliminate (B) because it is generally true that entropy increases as temperatureincreases. We can also eliminate (D) because entropy should approach 0 as temperature approaches0, not approach some non-zero value. Eliminate (E) because we have more than enough informationto pick a choice. Finally, we need to decide whether or not entropy has an upper limit (i.e. option(C)) or goes off to infinity (i.e. option (A)). Because there is some temperature at which all energylevels get equally populated, we also have a temperature at which any further increases in tempwon’t result in a wider dispersion of the particles so we should choose (C).

Correct Answer

(C)

250


3.74 PGRE9277 #74


We start off with the angular frequency equation,

ω =

√mgR

I(3.233)

The moment of inertia can be found by using the parallel axis theorem

I = ICOM +mR2 (3.234)

where ICOM is the moment of inertia of the object about an axis passing through its center ofmass. In the case of a loop, that moment of inertia about the center of mass is the same as a pointparticle at distance R so we get

I = ICOM +mR2 (3.235)

= mR2 +mR2 (3.236)

= 2mR2 (3.237)

so the moment of inertia for rings X and Y is

IX = 2(4m)(16R2) (3.238)

IY = 2mR2 (3.239)

plugging these into the angular frequency, ω, gives

ωX =

√(4m)g(4R)

2(4m)(16R2)(3.240)

=

√4g

32R(3.241)

and

251


ωY =

√mgR

2mR2(3.242)

=

√g

2R(3.243)

comparing the two, we get

ωXωY

=

√g/8R√g/2R

(3.244)

=

√1

4(3.245)

=1

2(3.246)

which is (B).

Correct Answer

(B)

252


3.75 PGRE9277 #75


By conservation of momentum, we should have the original momentum equal to the sum of theindividual momentums of the two atoms,

Pnet = PT + PH (3.247)

0 = mT vT +mHvH (3.248)

vH = −mT

mHvT (3.249)

If we combine Equation 3.249 with the kinetic energy equation for helium, we get

KH =1

2mHv

2H (3.250)

=1

2mH

m2T

m2H

v2T (3.251)

=1

2

m2T

mHv2T (3.252)

=mT

mH

(1

2mT v

2T

)(3.253)

and since mT > mH , the kinetic energy of the Helium atom must be larger than the kineticenergy of the Thorium atom.

Correct Answer

(E)

253


3.76 PGRE9277 #76


The total angular momentum quantum number, j, is the sum of the spin angular momentum, s,and the orbital angular momentum number, l,

j = l + s (3.254)

Since we have three electrons and all electrons have a spin of 1/2, the total spin angularmomentum must be

s =1

2+

1

2+

1

2(3.255)

s =3

2(3.256)

Then, recalling the orbital angular momentum rules (S, P,D, F, . . .) → (0, 1, 2, 3, . . .), we get

l = 0 + 1 + 1 (3.257)

l = 2 (3.258)

Sum the values from Equation 3.256 and 3.258 to get

s = 2 +3

2(3.259)

s =7

2(3.260)

Correct Answer

(A)

254


3.77 PGRE9277 #77


If we keep in mind that the magnetic moment is a measure of the tendency of an object to alignitself with a magnetic field. Although the magnetic moment of the nucleus and electrons are bothnon-zero (which let’s us eliminate (A)), we can determine just from common sense that a verysmall and light weight particle will more easily change alignment to conform to the magnetic fieldthan will a “heavy nucleus”. This means that the ratio of magnetic moment between nucleus andelectron should be less than 1, which eliminates all but (D) and (E). Between the two, we cancomfortably choose (E) because, as I said previously, the mass of the nucleus is what resists thechange in alignment.

Correct Answer

(E)

255


3.78 PGRE9277 #78


Note that at time t = 0, the velocity should be equal to the initial velocity 2v0 that it had rightbefore grabbing the pole. To check this limit, take the derivative of all of the possible solutions attime t = 0 to see which of these correctly predicts the condition.

(A) x′ = v = 2v0

(B) x′ = v +(

0.5b(3v0)b

)cos(0) = 2.5v0

(C) x′ = 0.5v0 +(

0.5b(3v0)b

)cos(0) = 2v0

(D) x′ = v0 +(

0.5b(6v0)b

)cos(0) = 4v0

(E) x′ = 0.5v0 +(

0.5b(6v0)b

)cos(0) = 3.5v0

Only (A) and (C) meet this criteria and since (A) doesn’t properly account for the rotation with asine function, choose (C).

Correct Answer

(C)

256


3.79 PGRE9277 #79


The group velocity is the velocity with which the overall wave travels while the phase velocity is therate at which the phase propagates. The relevant equations for group velocity and phase velocityare

vg =∂ω

∂k(3.261)

vp =ω

k(3.262)

take the derivative (tangent on the curve) between k1 and k2 to get a roughly constant negativevalue for the group velocity. The phase velocity, however, has a positive value because it’s a negativeslope with an inverse relationship. Since one is positive and the other is negative, they should bein opposite directions and we choose (A).

Correct Answer

(A)

257


3.80 PGRE9277 #80


Start with the equation for the energy of an electromagnetic wave,

E =hc

λ(3.263)

and now adjust it to solve for the wavelenth

λ =hc

λ(3.264)

we know the energy is 25 kilovolts and can utilize Planck’s constant and the speed of lightconstant from the front of our test booklet. Plug everything in to get

λ =(4× 10−15 eV · s)(3.0× 108 m/s)

2.5× 104 eV(3.265)

=12× 10−7 eV ·m

2.5 × 104 eV(3.266)

= 4× 10−11 m (3.267)

= 0.4 Angstroms (3.268)

which is nearly (B).

Correct Answer

(B)

258


3.81 PGRE9277 #81


In electronics, we will reach the max steady-state amplitude at the point when impedence is matchedbetween

ZL = ZC (3.269)

The inductor impedance is

ZL = jωL (3.270)

and the capacitor impedance is

259


ZC =1

jωC(3.271)

so applying Equations 3.270 and 3.271 to Equation 3.269 gives us

ZL = ZC (3.272)

jωL =1

jωC(3.273)

= ω2j2 (3.274)

=1

LC(3.275)

ωj =

√1

LC(3.276)

which gives the same inverse LC dependence that (C) suggests.

Correct Answer

(C)

260


3.82 PGRE9277 #82


First, recall the angular impulse H is proportional to the moment of inertia by

H = Iω (3.277)

the moment of inertia of a plate through its center is (1/12)mL2, so with our length of 2d, weget

261


I =1

12mL2 (3.278)

=1

12m(2d)2 (3.279)

=1

3md2 (3.280)

now solve for ω in Equation 3.277 with Equation 3.277 plugged into it to get our final answer

ω =H

I(3.281)

=H

(1/3)md2(3.282)

=3H

md2(3.283)

Correct Answer

(D)

262


3.83 PGRE9277 #83


We can get the right proportionality to figure out the solution by making a few small angle approx-imations. Start by summing the forces in both dimensions

FA = −FT−x (3.284)

FT−y = −FG (3.285)

by geometry, we can figure the length of d as

sin(θ) =d/2

L(3.286)

d = 2L sin(θ) (3.287)

263


we can also use some trigonometry to a relationship between tensions

sin(θ) =FT−xFT

(3.288)

plug Equation 3.288 into Equation 3.287 to get

d = 2LFT−xFT

(3.289)

Finally, get a relationship between FT−y and FT and apply the small angle approximationcos(θ) ≈ 1 to get

cos(θ) =FT−yFT

(3.290)

1 =FT−yFT

(3.291)

FT = FT−y (3.292)

but since FT−y = −FG = −mg and FT−x is the force from Coulomb’s law, substitute thesevalues into Equation 3.289,

d = 2LFT−xFT

(3.293)

=2Lkq2

d2mg(3.294)

which matches (A).

Correct Answer

(A)

264


3.84 PGRE9277 #84


For choices (A) and (B), we can see that these are both true by the Larmor formula

P =e2a2

6πε0c3(3.295)

Next, we know that (C) must be true by the LinardWiechert potential which states, in a bighot mess,

~E(~x, t) = q

(~n− ~β

γ2(1− ~β · ~n)3R2

)+q

c

~n× [(~n− ~β)× ~β]

(1− ~β · ~n)3R

(3.296)

of which, we only car that ~E ∝ 1/R2. Finally, we know that (E) is true because as we go off toinfinity, both fields tend to 0. We are left with (D) so that must be our correct answer.

Correct Answer

(D)

265


3.85 PGRE9277 #85


Start of with the relativistic energy equation

E = γm0c2 (3.297)

the energy is given to us as E = 1.5 MeV and the mass is me = 0.5 MeV/c2,

1.5 MeV = γ(0.5 MeV/c2)c2 (3.298)

γ = 3 (3.299)

now, we want to use the relativistic momentum equation but to do so, we need the velocity ofthe electron. Using our result from Equation 3.299, solve for v,

γ =1√

1− v2

c2

(3.300)

3 =1√

1− v2

c2

(3.301)

9

(1− v2

c2

)= 1 (3.302)

v2

c2=

8

9(3.303)

v =

√8

3c (3.304)

Plug results from Equation 3.299 and 3.304 into the relativistic momentum equation to get thefinal answer

P = γm0v (3.305)

266


= 3(0.5 MeV/c2)(√

8/3 c (3.306)

= (0.5)√

8 MeV/c (3.307)

= 1.4 MeV/c (3.308)

Correct Answer

(C)

267


3.86 PGRE9277 #86


First, eliminate any option that includes V0 (i.e. (A) and (C)) as the oscilloscope provides this dataand therefore wouldn’t be dependent on any of the other pieces of data being known. Next, recallthat a capacitor discharges according to

V = V0e−t/RC (3.309)

which tells us that we will need R and we can eliminate (C) and (E). Additionally, we will needthe time which we can derive from the sweep rate, s, so we get our final solution as (B).

Correct Answer

(B)

268


3.87 PGRE9277 #87


To find the net energy, we need to sum the kinetic energy and potential energy of the particle

H = T + V (3.310)

since the orbit is circular, we know that the centripetal force must be equivalent to the attractiveforce

Fc = FK (3.311)

mv2

r=

K

r3(3.312)

In Equation 3.312, if we just multiply a 1/2 to the LHS and cancel out the r, this becomes ourkinetic energy equaiton

1

2mv2 =

K

2r2(3.313)

Now for the potential energy, we use

269


V = −∫F · dr (3.314)

= −∫K

r3dr (3.315)

=K

2r2(3.316)

but because the potential is attractive, it becomes negative. We then sum the 2 potentials fromEquation 3.313 and 3.316 to get

H = T + V (3.317)

=K

2r2− K

2r2(3.318)

= 0 (3.319)

which is (C).

Correct Answer

(C)

270


3.88 PGRE9277 #88


According to the problem, this parallel plate capacitor is connected to a batter. As long as it isnot removed, the voltage and electric field should not be altered, even if a dielectric is put in place.This tells us that (A), (B) and (D) must all be wrong. Next, we can find the before and aftercharge as

Q0 = C0V0 (3.320)

Qf = κC0V0 (3.321)

so (C) must be wrong. This only leaves (E).

Correct Answer

(E)

271


3.89 PGRE9277 #89


Recall the solution to the Infinite square well, which gives us a set of sinusoidal waves (Figure 3.5)The first plot given represents n = 0, the second is n = 1, the third is n = 2 and so on. It

should become clear that all even values for n peak at x = 0 and so these will be disrupted by theinfinite potential at this point. The odd valued quantum numbers won’t so they will remain. Thisdescription only matches (E).

Correct Answer

(E)

272


Figure 3.5: Plots of the solution to the infinite square well

3.90 PGRE9277 #90


It’s worth memorizing the scale of energy spacing for the different energy levels

E = Etrans + Erot + Evib + Eelec (3.322)

those being

Erot ≈ 0.001 eV (3.323)

Evib ≈ 0.1 eV (3.324)

Eelec ≈ 1 eV (3.325)

from which we see that the rotational energy level should be around 10−3 which is (B).

Correct Answer

(B)

273


3.91 PGRE9277 #91


Eliminate (A) because the pion isn’t a lepton. Next, eliminate (B) because the Λ particle is abaryon so it must have spin 1/2. We can also quickly eliminate (D) because angular momentumis conserved. Lastly, eliminate (C) because it isn’t true that all interactions that don’t produce aneutrino are weak.

Correct Answer

(E)

274


3.92 PGRE9277 #92


For a single coil of wire, it is relatively clear that a rotating magnet with frequency of 10 Hz willgive us an alternating voltage of 10 Hz. However, for three coils, for every third of a rotation 10 Hzwill have been generated for a single coil and a full rotation will have done 3 of these, making thenet frequency 40 Hz

Correct Answer

(D)

275


3.93 PGRE9277 #93


When the weight is released at time t = 0 and angle θ = 0, the weight is essentially in free fall soit should have an acceleration of a = g. Plug in the θ = 0 condition to check the limit

(A) g sin(0) = 0

(B) 2g cos(0) = 2g

(C) 2g sin(0) = 0

(D) g√

3 cos2(0) + 1 = 2g

(E) g√

3 sin2(0) + 1 = g

and only (E) meets our criteria.

Correct Answer

(E)

276


3.94 PGRE9277 #94


The Lorentz transformation always takes the form of

t′ = γ (t− vx) (3.326)

x′ = γ (x− vt) (3.327)

y′ = y (3.328)

z′ = z (3.329)

which tells us that whatever the coefficients we have on x′ and t′, they should be the same withthe variables swapped. This is only true of (C) so this must be our solution.

Correct Answer

(C)

277


3.95 PGRE9277 #95


Assuming that ETS hasn’t given us a bunch of useless information in this problem, which is probablya pretty good assumption, we can arrive at the answer quickly with a bit of dimensional analysis.We are given, and should try to use,

1012 proton/sec (3.330)

1020 nuclei/cm2 (3.331)

102 proton/sec (3.332)

10−4 steradians (3.333)

The only way to arrange these 4 values to get a final unit of cm2/steradian is by

(102 protons/sec)

(1020 nuclei/cm2)(1012 proton/sec)(10−4 steradians)= 10−26 cm2/steradian (3.334)

Correct Answer

(C)

278


3.96 PGRE9277 #96


If you recognize that this setup is precisely the setup used to measure the index of refraction of airand you recall that the index of refraction of air is 1.000293, then you can quickly see that (C) isthe correct answer.

Correct Answer

(C)

Alternate Solution

In a gas interferometer, a beam of light is passed to a partially silvered mirror which splits thebeam into two parts. One part continues through the mirror and the other is reflected at a rightangle. Ultimately, both beams arrive at the observer and create an interference pattern. We know

279


that the optical path length is related to the index of refraction by nL, but since the light travelsthe distance L twice, we re-write it as 2nL. Now, if we remove the gas from the system, our indexof refraction must change (∆n) and the interference pattern will its fringes shift according to

∆n =mλ

2L(3.335)

Since the index of refraction of most gases is nearly 1, we typically define the index of refractionof any gas as the index of refraction of a vacuum (i.e. n = 1) plus some additional factor kp, wherek is some constant and p is the air pressure.

ngas = 1 + kp (3.336)

changes in index of refraction are proportional to changes in air pressure by

∆p = ∆nk (3.337)

so we combine Equations 3.335, 3.336 and 3.337 to get

n = 1 +mλp

2Lδp(3.338)

to make a quick approximation, and because this information is given, lose the dependence onp (i.e. p = ∆p) and solve to get

n = 1 +mλ

2L(3.339)

= 1 +(40 fringes)(500 nm)

2(5 cm)(3.340)

= 1 + 0.0002 (3.341)

= 1.0002 (3.342)

which is (C).

Correct Answer

(C)

280


3.97 PGRE9277 #97


To start off, check the units of each potential solution to see that only (A) and (D) give some sortof mass

(A) m∗ = 12 h

2k(dkdE

)= kg

(B) m∗ = h2k

( dkdE )= m6 kg4

sec4

(C) m∗ = h2k(d2kdE2

)1/3= A big mess that clearly has extra units

(D) m∗ = h2(d2Edk2

) = kg

(E) m∗ = 12 h

2m2(d2Edk2

)= kg5 m8

sec4

At which point you can either make an educated guess or recall that you should be differentiatingthe energy with respect to wave number, as in (D), rather than differentiating wave number withrespect to energy, like in (A).

Correct Answer

(D)

281


3.98 PGRE9277 #98


Recall that we find the eigenvalues from a matrix by finding the determinant of the characteristicequation

det

0− λ 1 00 0− λ 11 0 0− λ

You can, and for speed you should, use the quick method of finding the determinant of a 3

dimensional matrix (which I once heard called the “shoe string method”) −λ 1 0 −λ 10 −λ 1 0 −λ1 0 −λ 1 0

multiplying the diagonals and summing them (via “shoe string method”), you get

(−λ3 + 1 + 0)− (0 + 0 + 0) = 0 (3.343)

−λ3 + 1 = 0 (3.344)

λ3 = 1 (3.345)

Then from complex analysis, since we clearly only have 1 real solution (i.e. λ = 1) the rest ofthe solutions must be complex and, therefore, (B) must be a false statement.

Correct Answer

(B)

282


3.99 PGRE9277 #99


For this problem, you should be able to immediately recognize that the correct answer is (A). Thisis because the Hydrogen atom, unlike nearly every other problem in all of quantum mechanics, isan ideal and exactly solvable system. For this reason, there is no correction factor for the hydrogenatom in its ground state.

Correct Answer

(A)

283


3.100 PGRE9277 #100


In order to figure out the balancing point of the system, we need to find its net center of mass.

m1x1 +m2x2

mtotal(3.346)

We know the masses of each block are m1 = 20 kg and m2 = 40 kg and the positions arex1 = −5 m and x2 = 5 m, with respect to the center. We also know the total mass is the sum ofthe 2 blocks and the mass of the rod (i.e. 20 kg + 40 kg + 20 kg = 80 kg) so we can plug everythingin and solve to get

COM =(20 kg)(−5 m) + (40 kg)(4 m)

80 kg(3.347)

=100

80m (3.348)

= 1.25 m (3.349)

which is (C).

Correct Answer

(C)

284

Chapter 4

PGRE9677 Solutions

285


4.1 PGRE9677 #1


First, consider the state of this circuit before the switch, S, is changed from point a to point b.When connected to the potential, V , a constant current is passed through the capacitor. Thecapacitor will continue to gain a potential until the potential difference on the capacitor is equalto that of the potential. Once the switch is moved over to b, that potential is gone and what isleft is the potential stored on the capacitor, which will proceed through the resistor, R. Unlikethe potential, the capacitor won’t be able to maintain a constant current so we would expect thecurrent to decline as the resistor “resists” the current and energy dissipates. At this point you cancross off any curve on the plot which isn’t decreasing over time.

Now, all you have to do is decide whether the initial potential energy provided by V will be V/Ror V/(R + r). Considering that resistor r is isolated from resistor R, it is reasonable to concludethat only resistor R will have an influence on the current.

Correct Answer

(B)

Alternate Solution

Using the same reasoning as in the “Recommended Solution”, eliminate choices (A),(C) and (E)due to the fact that a current supported only by a charged capacitor must decline as it is forcedto pass through a resistor. Again, we need to figure out what the initial current must be for thecharged capacitor-resistor circuit. Recall Kirchhoff’s first rule (also commonly known as the LoopRule)

Kirchhoff’s First Rule The sum of the changes in potential energy encountered in a completetraversal of any loop of a circuit must equal zero

286


Applying the Loop Rule to the second stage of our circuit (that is to say, once the capacitorhas reached an equal potential to that of V and the switch, S, has been thrown to point b) we mustaccount for two potentials summing to 0

Vcap + Vres = 0 (4.1)

iR+ q/C = 0 (4.2)

R and C are both constants but i and q are functions of time and so with 1 equation and 2variables, we are stuck. And yet, we are too industrious to leave things at that. We can relate thecharge, q, to the current, i, because current is merely the change in current.

i =dq

dt(4.3)

which gives us

dq

dtR− q/C = 0 (4.4)

Now that the only variable is q, we can solve this ordinary homogenous differential equation.Doing so, we get

q = q0e−t/RC (4.5)

Where e is Euler’s number, not an elemental charge number. Taking the derivative of q thengives us the current as

dq

dt= i = −

(q0

RC

)e−t/RC (4.6)

Setting t = 0 for the initial time and utilizing our equation for capacitance (q0 = CV0) we canfind the initial current as

i0 = −(CV0

RC

)e−0/RC = −V0/R (4.7)

Which states that the initial current of the final stage of the circuit is V/R. We can ignore thenegative on the initial current as it is merely indicating that the capacitors charge is decreasing.

Correct Answer

(B)

287


4.2 PGRE9677 #2


In this problem we are asked to determine the reading of an ammeter attached to a circuit witha potential (V) of 5.0 V , a resistor of 10 Ω while sitting in a uniform but changing magnetic fieldof 150 tesla/second. In this scenario we have to account for two different phenomena which aregenerating a current. The first phenomena generating a current is the 5.0 V potential which willcreate a current according to Ohm’s Law

I =V

R(4.8)

I =5.0 volts

10 Ω= 0.5 A (4.9)

This current will be moving in a counterclockwise direction because the “positive flow of current”moves in the opposite direction of the electron flow. In other words, the current will be movingfrom the positive terminal to negative terminal (The larger bar of the potential V to the smallerbar of the same potential V).

The second source of current comes from the changing magnetic field being directed “into thepage”. We know that a changing magnetic field generates a current thanks to Maxwell’s/Faraday’slaws, and in particular we know from Faraday’s law of induction that a changing magnetic flux willinduce a potential (emf) ε

ε =

∣∣∣∣dφBdt∣∣∣∣ (4.10)

where φB is

φB =

∫ ∫S

~B(~r, t) · d ~A (4.11)

288


which simply states that the magnetic flux (φB) is equal to the magnetic field (B) through sometotal surface area (i.e. the surface integral over differential pieces of area dA )

Taking the surface integral for this simple loop of wire is the same as just calculating the areaof the square with sides of 10 cm× 10 cm, or in more standard units, .10 m× .10 m which gives atotal surface area of A = .01 m2. This makes the flux equation

φB = ~B(~r, t) · ~A (4.12)

Since the vector ~A is the vector perpendicular to the surface area and it is in the same directionas the magnetic field lines,

~B · ~A = B A (4.13)

substituting that in for our equation calculating the emf (ε)

ε =

∣∣∣∣d(φB)

dt

∣∣∣∣ =

∣∣∣∣dBdt∣∣∣∣A (4.14)

ε =

(150

tesla

sec

)(0.01 m2

)= 0.15 A (4.15)

Finally, calculate the difference between the 2 potentials (Equations 4.9 and 4.15) we found andyou have

0.5 A− 0.15 A = 0.35 A (4.16)

Correct Answer

(B)

289


4.3 PGRE9677 #3

Figure 4.1: Electric potential at point P in relation to ring of radius R


Point p in Figure 4.1 represents a “test charge” location and we want to know what the electricpotential is at that point due to a uniformly charged ring with charge Q. Using Coulombs Law inGaussian units, we have

U =

∫s

1

4πε0

dQ

r(4.17)

where r is the distance between any point on the ring and point P, and dQ is a differential pieceof the ring. because the distance from every point on the ring to our point P is the same, r is aconstant. We can pull all constants out from the integral, giving

U =1

4rπε0

∫sdQ (4.18)

U =Q

4rπε0(4.19)

Lastly, we need to solve for r. Drawing a line between the ring and P gives us a right trianglewhich allows us to use the Pythagorean theorem (Figure 4.2)

290


Figure 4.2: Constructing a right triangle between points P and R

r =√R2 + x2 (4.20)

and substituting Equation 4.20 and 4.19 gives us

U =Q

4πε0√R2 + x2

(4.21)

Correct Answer

(B)

Alternate Solution

The trick to this alternate solution involves manipulating the variables involved. No specific R or xvalues are given, so the correct solution must work for any choice of R and x. For example, lettingx go to infinity, we would expect the electric potential to go to 0. For solutions (C) and (D), lettingx go to infinity would make the potential go to infinity. Similarly, if we let R go to infinity, againpotential should go to zero. This isn’t the case for solution (E), so cross it off. Finally, you justhave to decide whether or not the distance between the ring (or rather differential pieces of thering, dQ) and point P is in agreement with (A), in which r = x, or if it is in agreement with (B),in which r =

√R2 + x2. By the Pythagorean theorem, the hypotenuse can’t be the same length as

one of its sides so r 6= x.

Correct Answer

(B)

291


4.4 PGRE9677 #4


To solve this problem we will need Coulomb’s law, Hooke’s Law and the general equation for angularfrequency

Coulomb’s Law: F = 14πε0

q1q2r2

Hooke’s Law: F = −kx

Angular Frequency: ω =√k/m

In our situation we have a charged surface, specifically a ring, so we will need to replace one ofour point charges with this value, which can be found by integrating a differential piece of chargeover the entire ring.

F =−q

4πε0r2

∫SdQ (4.22)

F =−qQ

4πε0r2(4.23)

292


Figure 4.3: Component forces at point P

Equation 4.23 represents the equation for the net force. However, we will need to decomposethe net force into its horizontal and vertical components. Fortunately, for every force in the verticaldirection from one piece of the ring, the piece on the opposite side of the ring will cause an equaland opposite force.

The vertical forces cancel each other out and we are left with the horizontal forces. Hopefullyit is clear from Figure 4.3 that the horizontal force component should be

Fx = Fcos(θ) = F (x/R) (4.24)

substitute 4.23 with 4.24 to get

F =−qQ

4πε0r2

x

R(4.25)

Finally, we can calculate the distance between P and any point on the ring using the pythagoreantheorem

r2 = R2 + x2 (4.26)

combining 4.26 with 4.25 gives us

F =−qQx

4πε0(R2 + x2)R(4.27)

Now taking Hooke’s law, solve for k to get

k = −F/x (4.28)

and substitute our force equation into it. After simplifying, you should get

k =qQ

4πε0(R2 + x2)R(4.29)

and then substitute this in for the k in our angular frequency equation and simplify to get

ω =

√qQ

4πε0(R2 + x2)Rm(4.30)

Finally, take into account that the problem asked you to consider when R >> x, meaning xeffectively goes to 0, giving

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ω =

√qQ

4πε0mR3(4.31)

Correct Answer

(A)

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4.5 PGRE9677 #5


To start off with, figure out every force that will be on this car as it travels through the arc. Theproblem identifies for us the force from air resistance, Fair. Additionally, we can identify thatthere will be a net centripetal force pointing towards the center of the arc. Keep in mind that thecentripetal force is not one of the component forces but the sum total of all the forces. drawingout the force diagram gives us Figure 4.4

Figure 4.4: Net force diagram on a turning vehicle

Now we must ask ourselves which direction the horizontal force must point such that when itis added to Fair, we will get a net centripetal force pointing down. Check each of the 5 choices inFigure 4.5 and you will quickly see that only (B) can be correct.

Correct Answer

(B)

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Figure 4.5: Potential force combinations

Alternate Solution

A quick and qualitative way of figuring out the solution is to consider why the car is turning at all.The force from air resistance cant be responsible for the motion and the centripetal force is merelya description of the net force (i.e. the sum of the component forces) acting on the car. As such,the force that the road is applying to our car tires must have some vertical component of forcepointing down. You can eliminate choices (C), (D), and (E) based on this. Next, consider thatthe air resistance force has a horizontal component of force (technically it only has this horizontalforce) and so for our net centripetal force to have no horizontal force, the allusive force we arefinding must have an equal and opposite horizontal component. (A) has no horizontal componentso cross it off.

Correct Answer

(B)

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4.6 PGRE9677 #6


This problem can be solved quickly by utilizing conservation of energy laws. At the top of theincline, the only energy for the system is gravitational potential energy

UG = mgh (4.32)

According to the description, the block slides down the incline at a constant speed. This meansthat the kinetic energy Uk = 1

2mv2, is the same at the beginning of the blocks motion as it is at

the end and thus none of the potential energy we started with changed to kinetic energy. However,because energy must be conserved, all of the energy that didn’t become kinetic energy (i.e. all mghof it) had to have dissipated from friction between the incline and the block.

Correct Answer

(B)

Alternate Solution

Start by drawing out a force diagram for the block at the top of the slide (Figure 4.6)take the sum of all of the forces in the x and y direction. Note that the acceleration in the x

direction is 0 because the speed is constant, therefore the net force is 0.

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Figure 4.6: Force diagram of a block/incline system

∑Fnet,x = FGX − f = 0 (4.33)

∑Fnet,y = FN − FGY = 0 (4.34)

Using trigonometry, you should see that FGX = FG sin(θ) and FGY = FG cos(θ). Since F = mg,FGX = mg sin(θ) and FGY = mg cos(θ). Additionally, f = µFN so we get

f = FGX = µFN = mg sin(θ) (4.35)

and

FN = mg cos(θ) (4.36)

Combine Equations 4.35 and 4.36 and get

µmg cos(θ) = mg sin(θ) (4.37)

µ =sin(θ)

cos(θ)= tan(θ) (4.38)

Now, recall that work, which is equivalent in magnitude to energy, is

W = F (∆X) (4.39)

and since we are concerned with the energy (work) generated from friction, the force in 4.39must be f . Make the substitution to get

Wf = f(∆X) (4.40)

Wf = mg sin(θ)(∆X) (4.41)

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where ∆X is the length of the ramp. Since sin(θ) = h/∆X, we know that

∆X = h/sin(θ) (4.42)

finally, combine 4.41 and 4.42

Wf =mgsin(θ)h

sin(θ)(4.43)

Wf = mgh (4.44)

Correct Answer

(B)

299


4.7 PGRE9677 #7


Because the collision is elastic, we know that energy and momentum are conserved. From this, weknow that the initial potential at height h will equal the kinetic energy immediately before the ballstrikes the brick. Additionally, we know that the momentum of the ball/brick system must be thesame before collision as after. This gives us the equations

1

2mV 2

b,0 =1

2mV 2

b,f −mV 2B,f (4.45)

and

mVb,0 = mVb,f + 2mVB,f (4.46)

With two equations and three unknowns, we can get a relationship between any of the twovariables. Combine 4.45 and 4.46 in order to get a relationship for Vb,0 with Vb,f and Vb,0 with VB,f .

VB,f =2

3Vb,0 (4.47)

Vb,f =1

3Vb,0 (4.48)

This tells us that the final velocity of the ball b is 1/3 of its initial velocity and the block Bleaves the collision at 2/3 the initial velocity of the ball. Once the ball leaves at its velocity of

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Vb,f , it will move up to its final height as the kinetic energy becomes potential energy. The kineticenergy is

Uf =1

2mV 2

b,f =1

2m

(1

3Vb,0

)2

(4.49)

Uf =1

2m

(1

9V 2b,0

)=

1

9

(1

2mV 2

b,0

)(4.50)

Uf =1

9mgh (4.51)

Comparing the final energy to the intial, we get

mghf =1

9mgh0 (4.52)

hf =1

9h0 (4.53)

Correct Answer

(A)

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4.8 PGRE9677 #8

The quickest way to solve this problem is to recognize that the problem is describing a dampedharmonic oscillator. The object, in this case a particle with mass m, oscillates with a damping forceproportional to F = −bv applied to the particle. We know that the force applied to the particleis fighting against the oscillation because it is always in the opposite direction of the velocity andthis confirms that this is a damped harmonic oscillator. If the force was positive and adding to theforce of the oscillation, then we would have a driven harmonic oscillator. Now, consider the periodof damped harmonic oscillator in comparison to an unhindered SHO without the opposing force.Strictly speaking, a damped oscillator doesn’t have a well-defined period and without knowing thespecific values for mass, spring constant, etc we don’t know whether we are talking about a systemthat is underdamped, over damped or critically damped. Nevertheless, we can see that all of thedamped oscillations (Figure 4.7) will experience an increase in the period.

Figure 4.7: Damped harmonic oscillators

Correct Answer

(A)

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4.9 PGRE9677 #9


The Lyman and Balmer series both refer to different types of transitions of an electron in a hydrogenatom from one radial quantum level (n) to another. The Lyman series is a description of allsuch transitions from n=r to n=1, such that r ≥ 2 and is an integer. The first Lyman transition(commonly called Lyman-alpha) is n=2 going to n=1, the second (Lyman-beta) involves a transitionof n=3 to n=1, etc. The Balmer series, on the other hand, involves transitions from some n=s ton=2, such that s ≥ 3 and is an integer. The longest wavelength for both series involves the smallesttransition, i.e. n=2 going to n=1 for the Lyman Series and n=3 going to n=2 for the Balmer. TheRydberg formula can then be used to find the wavelength for each of the two transitions

1

λ= R

(1

n2f

− 1

n2i

)(4.54)

For this problem we won’t need to compute anything, just compare λL and λB. Doing this forthe shortest Lyman transition gives

1

λL= R

(1

12− 1

22

)(4.55)

1

λL=

3

4R (4.56)

λL = 4/(3R) (4.57)

and for the Balmer transition

1

λB= R

(1

22− 1

32

)(4.58)

1

λB=

5

36R (4.59)

λB = 36/(5R) (4.60)

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making the ratio

λL/λB =4/(3R)

36/(5R)= 5/27 (4.61)

Correct Answer

(A)

304


4.10 PGRE9677 #10


In internal conversion, one of the inner electrons of the molecule is ejected. Because of this, oneof the outer electrons will drop down a level to fill the space and some form of electromagneticradiation is released. From this, you can immediately remove (A), (D) and (E). Now you simplyhave to decide whether the electromagnetic radiation will correspond to the energy level of X-Raysor γ rays. As it turns out, this electronic transition will release X-rays. However, if you don’t knowthis, you may be able to reason through to the answer. Consider that γ rays are the highest energyform of electromagnetic radiation and these will generally be a result of an energy transition fora nucleus. An electron transition, on the other hand, involves smaller exchanges of energy andcorrespond to the lower energy X-rays.

Correct Answer

(B)

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4.11 PGRE9677 #11


In this problem, ETS is testing your knowledge of physics history. If you recall the Stern-Gerlachexperiment (1922), this problem becomes quite easy. The Stern-Gerlach experiment involved firingneutral silver atoms through an inhomogenous magnetic field. The classical understanding (i.e.before Stern-Gerlach) would have suggested no deflection because the atoms are neutral in chargeand have no orbital angular momentum and thus generate no magnetic dipole. However, thisexperiment showed that the beam split into two distinct beams, adding evidence to the ultimateconclusion that electrons have a spin property of 1/2. Keep in mind that, in general, the number ofdistinct beams generated after hitting the magnetic field will be equal to 2S+ 1. Silver has a singleunpaired electron and so S = 1/2, giving the original Stern-Gerlach experiment its two beams. Inthis problem, ETS went easy on us and gave us hydrogen to work with, which also has S = 1/2and thus will split into two beams.

Correct Answer

(D)

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4.12 PGRE9677 #12


When positronium questions pop up on the GRE, they generally can and should be solved by itsrelation to the hydrogen atom. In this case, you have to recall that the ground state energy ofHydrogen is equal to 1 Rydberg = −13.6 eV. Positronium involves an electron-positron pair whilehydrogen involves a proton-electron pair. There is no difference between the two in terms of chargebut there is a significant difference in mass. Since Rydberg’s constant is mass dependent (Equation4.62), we have to alter the original Rydberg constant

Rhydrogen =memp

me +mp

e4

8cε20h3

(4.62)

Which becomes,

Rpositronium =meme

me +me

e4

8cε20h3

(4.63)

=me

2

e4

8cε20h3

(4.64)

To convince yourself that this makes the Rydberg constant half as large, recall that the ratioof the proton mass to the electron mass is approximately 1836:11. Calculating the effective masswith an electron and proton, we get

mpme

mp +me=

1 ∗ 1836

1 + 1836≈ 1 (4.65)

Calculating the effective mass for the electron/positron pair, with their equivalent masses, givesus

meme

me +me=me

2(4.66)

Since the energy is proportional to the Rydberg constant, the ground state energy of positroniummust be half of the hydrogen ground state energy

1It isn’t necessary to know this quantity in order to arrive at this simplification. All that is really important isthat the difference between the two masses is significant

307


Epositronium =Ehydrogen

2= −6.8eV (4.67)

again, if you aren’t convinced, consider the Rydberg equation for hydrogen

1

λ= R

(1

n21

− 1

n22

)(4.68)

and since E = hν = hcλ

E =hc

λ= hc

R

2

(1

n21

− 1

n22

)(4.69)

Correct Answer

(C)

308


4.13 PGRE9677 #13


The problem gives us the specific heat of water as 4.2 kJ/kg which should be a strong hint thatyou will need to use the equations for heat absorption into a solid,

Q = cm∆T (4.70)

We are also given the power of the heating element, meaning we know the amount of energyinput. Additionally, because the problem states that the water never manages to boil, even if itcomes close, the water must be outputting energy at an equal rate or at least very close to it. Wethen just need to know how long it will take for energy to dissipate from the system equivalent toa change in temperature of 1C. Using the definition of power,

P = W/∆t (4.71)

Or equivalently

W = P∆t (4.72)

We can then combine the two equations (Q=W) to get

cm∆T = P∆t (4.73)

When making substitutions, keep in mind that 1 L of water in mass is 1 kg, giving(4.2

kJ

kg

)(1 kg)(1C) = (100 watts)(∆t) (4.74)

We can get everything into the same units by converting watts to kJ. Specifically, 1 watt =1 J/s = 0.001 so 100 watt = 100 J/s = 0.1 kJ/s.

309


4.2kJ = 0.1kJ

s∆t (4.75)

Then solving for ∆t gives

∆t ≈ 40 sec (4.76)

Correct Answer

(C)

310


4.14 PGRE9677 #14


This is one of those rare “Plug-n-Chug” problems on the PGRE. Cherish it! In this problem wehave 2 copper blocks in an insulated container. This tells us that the total energy of the systemis conserved. Since both blocks are of the same mass, we know the final temperature of bothblocks will reach equilibrium at 50 kcal each. Since heat travels from high temperatures to lowtemperatures, all energy transfer will take place from the block with T2 = 100C to the blockT1 = 0C and so we only need to consider this one direction of energy transfer. Using our equationfor heat absorption for a solid/liquid body, we can plug in all known values

Q = cm∆T (4.77)

Q = (0.1kCal/kg K)(1kg)(50 K) (4.78)

Q = 5 kCal (4.79)

Correct Answer

(D)

Additional Note

In this specific problem ETS has been very nice to us by making both blocks of the same mass. Inthe case that they gave us blocks of different masses, we wouldn’t be able to easily conclude thatthe final temperature of each block would be at 50C. If, for example, all other values were thesame but the block at T1 = 0C had a mass of m1 = 1 kg and the block at T2 = 100C had a massof m2 = 2 kg, then we would have to solve for the final temperature. To do so, consider that the

311


system is insulated so no energy can leave. This means that the total energy transfer will be 0 andthe sum of the energy transfers between the two blocks will sum to 0. This gives us

Q1 +Q2 = cm1(Tf − Ti,1) + cm2(Tf − Ti,2) = 0 (4.80)

c is the same for both so cancel it out and solve for Tf ,

Tf = 33.3C (4.81)

312


4.15 PGRE9677 #15


The thermodynamic process goes through its entire cycle, so energy is conserved, ∆U = 0. Fromthis and the first law of thermodynamics, ∆Q = ∆U + ∆W , we know that the total heat will justbe the sum of its work terms.

∆Qnet = ∆Wnet = ∆WAB + ∆WBC + ∆WCA (4.82)

The work equation for a thermodynamic system is W =∫PdV and we have the ideal gas law,

PV = nRT , so

WAB =

∫ V2

V1PdV (4.83)

313


=

∫ V2

V1

nRT

VdV (4.84)

= nRTln(V2

V1) (4.85)

Next, moving from B to C, we get

WBC = P∆V = nR∆T (4.86)

Finally, taking C to A, there is no change in volume so we would expect to get a work of 0, i.e.

WCA = P∆V = P (0) = 0 (4.87)

Adding up all of the components (Equations 4.85, 4.86 and 4.87), we get

Wnet = WAB +WBC +WCA (4.88)

= nRTln(V2/V1) + nR∆T + 0 (4.89)

Since the problem specifies we have one mole of gas, Equation 4.89 becomes

Wnet = RTln(V2/V1) +R∆T (4.90)

Finally, consider the work equation for path BC to realize that ∆T = (Tc − Th). Reversing thetwo heats, as we see in all of the possible solutions, will result in a negative sign coming out, givinga final result of

Wnet = RTln(V2/V1)−R(Th − Tc) (4.91)

Correct Answer

(E)

314


4.16 PGRE9677 #16


The problem gives us the equation for the mean free path as 1ησ . To get the density of air, use the

ideal gas law and use the gas constant R = 8.2× 10−5 m3 AtmK Mol

PV = nRT (4.92)

n

V=

P

RT(4.93)

Which should give you

6× 1023

2.4× 10−2

Mol

m3= 3× 1025 Mol

m3(4.94)

Approximate the size of any given air molecule as being about 1 nm so the collision cross sectionis 1 nm2 which is 1× 10−18 m2. Put this into the mean free path equation provided and you have

1

ησ=

1

(3× 1025 Mol/m3)(1× 10−18 m2)≈ 1× 10−7 (4.95)

Correct Answer

(B)

315


4.17 PGRE9677 #17


Recall from quantum mechanics that we can get the probability of finding a particle in any positionby taking the integral of the squared wave function,

Pab =

∫ b

a|Ψ(x)|2dx (4.96)

The integral of a curve is just the area underneath it and since a plot of the function is provided,we can quickly find the area. However, be aware that the curve represented is Ψ, and we want thearea under the curve for Ψ2. For this reason, we must square every piece of the plot and then takethe area under the curve. Since we are concerned with the probability of the particle being locatedbetween x = 2 to x = 4, we need to compare that with the total area of the squared wave function.Doing so, from left to right, we have

Area2→4 = (2)2 + (3)2 = 13 (4.97)

Area0→6 = (1)2 + (1)2 + (2)2 + (3)2 + (1)2 + (0)2 = 16 (4.98)

thus the probability is 13/16

316


Correct Answer

(E)

317


4.18 PGRE9677 #18


Recall the “infinite square well” that everybody does as their first, and frequently only, exactlysolvable quantum system. You should recall that the wave function with infinite potential barriers

318


on each side restricted the wave function to the area between the potential walls, with the exceptionof a small bit of tunneling on each side of the infinite potentials. This should tell you that no matterwhat the wave function looks like, it better have an amplitude that is lessened by interacting with agreater potential wall than it would in the open space. From this, eliminate (A), (D) and (E). Now,you just need to decide if the the wavefunction will be able to maintain its amplitude, frequencyetc as its trying to tunnel through the wall. Again, you should recall from the infinite square wellthat this wasn’t the case. Instead, the amplitude of your wave function continually dropped andapproached Ψ(x) = 0. (C) shows this characteristic drop but (B) does not.

Correct Answer

(C)

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4.19 PGRE9677 #19


This problem is a simpler case of Ruterford scattering. We are told that the scattering angle is 180

which is the maximum possible scattering angle. This means that the alpha particle is contactingthe silver atom head on and all of the kinetic energy of the electron is becoming potential energy,resulting in Equation 4.99

1

2mv2 = 5 MeV =

1

4πε0

q1q2

r(4.99)

ε0 is given in the list of constants in the beginning of the test. q for an alpha particle is2(1.6 × 10−19) C and q for silver is 50(1.6 × 10−19) C. Converting the known kinetic energy intomore convenient units gives, 5 MeV ≈ 8 × 10−13 J and plugging everything in and solving for r,you get

r =2(1.6× 10−19 C) 50(1.6× 10−19 C)

4π(8.85× 10−12)(8× 10−13 J)≈ 2.9× 10−14 m (4.100)

Correct Answer

(B)

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4.20 PGRE9677 #20


In this problem we are told that an elastic collision occurs, which tells us that energy and momentumwill be conserved. From this, write the equation for each.

Momentum: Ptotal = mV0 = −mV1 + µV2

Energy: Etotal = 12mV

20 = 1

2mV2

1 + 12µV

22

Substitute in 0.6V0 = V1 and simplify to get the momentum equation

mV0 = −0.6 mV0 + µV2 (4.101)

1.6 mV0 = µV2 (4.102)(1.6 mV0

µ

)= V2 (4.103)

and the energy equation

mV 20 = 0.62 mV 2

0 + µV 22 (4.104)

0.64 mV 20 = µV 2

2 (4.105)(0.64 mV 2

0

µ

)= V 2

2 (4.106)

For your momentum equation, square both sides and set the resulting equation equal to yourenergy equation

2.56 m2V 20

µ2=

0.64 mV 20

µ(4.107)

Now simplify and you are left with only m and µ. Solving should give you

321


µ = 4 m (4.108)

and since m = 4 u,

µ = 16 u (4.109)

Correct Answer

(D)

Additional Note

The method by which you solve this problem is identical to that used in problem #7 (See Section4.7) on this same test (PGRE9677). It’s all a matter of recognizing that energy and momentumare conservered in an elastic collision.

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4.21 PGRE9677 #21


From the Parallel-Axis theorem, we know that the moment of inertia of any object is equal to thesum of the objects inertia through its center of mass and Mh2

I = Icom +Mh2 = MR2 +MR2 = 2MR2 (4.110)

plug the provided values into Equation 4.110 to get

T = 2π

√2MR2

MgR= 2π

√2R

g(4.111)

T = 2π

√2(0.2 cm)

10 m/s2≈ 1.2 (4.112)

Correct Answer

(C)

323


4.22 PGRE9677 #22


The problem tells us that the golf ball is orbiting mars, which tells us that the height of the golfball relative to the surface of the planet is constant. The easiest way to deal with the providedinformation is to utilize the kinetmatic equation

y − y0 = v0t−1

2at2 (4.113)

We can assume the initial velocity is 0 and the vertical change is 2 m, giving

2 m = −1

2(−0.4g)t2 (4.114)

Solving for t to figure out how much time passes for the orbital motion gives

t =√

4 m/0.4g ≈ 1 sec (4.115)

Since velocity is change in position over time, which is 1 second, we get a final velocity of

v =∆X

1 sec= 3600 m/s = 3.6 km/s (4.116)

Correct Answer

(C)

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4.23 PGRE9677 #23


The problem specifies that ε, what ever value it might be, is small. Assume that the value of epsilonis so small as to be effectively irrelevant. Under this condition, look for statements which conformto known orbital phenomena.

(A) There’s no reason to assume that energy isn’t conserved in this scenario. Mechanical energy(Work) is W =

∫F ·ds and this won’t interfere with the conservation of energy. Additionally,

we should only expect conservation of energy to fail if there is some fricitional force appliedto the object and the problem never mentions such a force.

(B) Angular momentum is conserved as long as there are no external torque on the system. εwouldn’t impose an external torque.

(C) This is consistent with Kepler’s third law of planetary motion (i.e. P 2 ∝ a3)

(D) Noncircular orbits are rare and from process of elimination (see (E)) we can see that this isthe only false proposition

(E) Circular type orbits are relatively common and so we should expect this to be true.

Correct Answer

(D)

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4.24 PGRE9677 #24


This problem can be solved with nothing more than Coulomb’s law,

F =1

4πε0

q1q2

r2(4.117)

We start with two spheres, each with a charge of q. When the uncharged sphere touches sphereA, electrons are passed to the uncharged sphere until they each reach equilibrium. In this case,equilibrium involves half of the charge ending up on each sphere. Now, the initially unchargedsphere has a charge of 1

2q and sphere B has a charge of 1q. When these two come in to contact,they equilibrate again. The average for these two charges is

(12 + 1)

2q = 3/4q (4.118)

The initially uncharged sphere leaves itself and sphere B having a charge of 3/4 q. Knowingthe charge for sphere A and sphere B, plug this into the equation for Coulomb’s Law to get

1

4πε0

(1/2 q)(3/4 q)

r2=

3

8

(1

4πε0

qAqBr2

)(4.119)

i.e. the force is now 3/8 of its original charge.

Correct Answer

(D)

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4.25 PGRE9677 #25


Generally, I would recommend going through each possible solution to make sure you are findingthe “best solution”. However, in this problem one of the options stood out as being clearly false.Before the switch is thrown, current flows through the path of C1 and charges it to energy U0.After the switch has been thrown, C1 will still charge to energy U0 so we know U0 = U1. Withoutknowing, U2, we can see that (E) violates this result (unless U2 = 0 which is clearly not the case).I’d also like to point out that (D) and (E) disagree with one another (again, unless one of theenergies is 0) so you can immediately determine that your answer must be one of these two.

Correct Answer

(E)

Alternate Solution

Before the switch is thrown, current is flowing through the path of C1 and charges it to U0 whichmeans that the initial capacitance referred to in the problem is C0 = C2. Also, from the problem,

327


C1 = C2 so all three capacitors are equivalent. Additionally, all potentials are equal in a parallelcircuit (i.e. V0 = V1 = V2), giving Q0/V = Q1/V = Q2/V

(A) Q0 = Q1 = Q2 (which will be shown in part (B)) and half of the sum of two identical thingswill equal itself.

(B) Since Q0/V = Q1/V = Q2/V , multiply the V out and get Q0 = Q1 = Q2

(C) As was mentioned previously, the potential in a parallel circuit is always equivalent across allcapacitors in the circuit. V0 = V1 = V2

(D) Since all capacitors have the same capacitance and the same voltage, by U = 12CV

2, we getU0 = U1 = U2

(E) As was demonstrated in (D), U0 = U1 = U2 so it has to be the case that U1 + U2 = 2U0. (E)is false.

Correct Answer

(E)

328


4.26 PGRE9677 #26


If you know the equation for frequency in an RLC circuit, then this problem is relatively straight-forward. This question is primarily testing your ability to do unit conversions and multiply verylarge and/or very small numbers. The equation for the resonance frequency in an RLC circuit is

f =1

2π√LC

(4.120)

f2 =1

4π2LC(4.121)

We know that the final frequency should be 103.7 MHz which we can simplify as 100 MHz.Squaring this value gives us 1.0 × 104 MHz. Converting this to hz, we get 1 × 1010 hz. Theinductance is 2.0 microhenries which is equivalent to 2.0 Ω · s. Rearranging the previous equationto solve for capacitance, gives

C =1

4 π2(2.0 Ω · s)(1.0× 1010 hz)(4.122)

To make things easier, let’s set π2 = 10, which makes 4π2 = 40. Substituting everything intoEquation 4.122,

C =1

8× 1011Farads (4.123)

Finally, convert to pF by recalling that 1 × 1012 pF = 1 F, giving a final value of C = 0.8 pFwhich is closest to (C) 1 pF.

Correct Answer

(C)

329


4.27 PGRE9677 #27


Logarithmic scaling is great for exponential curves and inverse curves (i.e. negative exponentials).From this, you know that (A), (C), (D) and (E) all should be either log-log or semilog. (A), (C),and (E) fulfill this requirement but (D) doesn’t. Just to be sure that (D) is the “Best Solution”check (B) to see that we would expect it to utilize a linear graph just as the solution suggests.

Correct Answer

(D)

330


4.28 PGRE9677 #28


The wave on this oscilloscope clearly displays the sum of two separate waves with different frequen-cies. The wave with the larger frequency has a wavelength of about 6 cm. The speed of the wave isgiven as v = 0.5 cm/ms. Using the relationship of a linearly traveling wave through a homogeneousmedium, we can calculate the frequency

λ =v

f(4.124)

f =v

λ(4.125)

Plug in the values from the problem into Equation 4.125

331


f =(0.5 cm/ms)

(6 cm)=

1

12 ms(4.126)

However, we want the frequency in Hz, so convert Equation 4.126 to get

f =1

12 ms

(1000 ms

1 s

)= 83 Hz (4.127)

Which agrees with option (D).

Correct Answer

(D)

332


4.29 PGRE9677 #29


The units for the Planck length should be a length (probably meters). The units for the threeconstants are

Gravitational Constant: G ≡ m3

kg·s2

Reduced Planck’s Constant: h ≡ m2·kgs

Speed of Light: c ≡ ms

We know that there can’t be a unit of seconds in the final result and since all instances of secondsshows up as a denominator, we know at some point we will have to do some division. This onlyhappens in option (E) .

Correct Answer

( E )

Alternate Solution

Using the units for the three constants, multiply out each one to figure out which of these gives usa result in meters

(A) Ghc =(m3

kg·s2) (

m2·kgs

) (ms

)= m6

s4

(B) Gh2c3 =(m3

kg·s2) (

m2·kgs

)2 (ms

)3= m10·kg

s7

(C) G2hc =(m3

kg·s2)2 (m2·kg

s

) (ms

)= m9

kg·s6

333


(D) G1/2h2c =(m3

kg·s2)1/2 (m2·kg

s

)2 (ms

)= m13/2kg3/2

s4

(E)(Gh/c3

)1/2=

√(m3

kg·s2) (

m2·kgs

)/(ms

)3= m

Correct Answer

( E )

334


4.30 PGRE9677 #30


This problem can be solved with just a bit of clever reasoning. First, consider that the initial stateof the system is water at equilibrium and at 20 cm from the bottom of the curve. We can get agood approximation of the total mass of the water by assuming that all of the water is accountedfor by the 40 cm of vertical tube length (i.e. ignoring the curve). There is likely not much of adifference between the two values and as long as we are consistent with this assumption, deviationswon’t present themselves in the final ratio. So, we assume that with 40 cm of water at 1 g/cm3,we have roughly 40 g of water (Technically we would have 40 g/cm2 but because the tube is thesame size throughout, the cross-sectional slices with units of cm2 will be the same for all parts andwe might as well treat it as 1). Now, we can add to that the more dense liquid which we knowaccounts for 5 cm of the tube and has a density 4 times greater than the water. The total mass ofthe more dense liquid is 20 g giving a grand total of 60 g of liquid. The system will be at equilibriumwhen pressure on both sides of the tube is equal. Keeping in mind that pressure is proportional todensity (i.e. P = ρgh) and density is proportional to mass (i.e. ρ = m/V ) we can conclude, andit should seem reasonable that, the system will be at equilibrium when we have equal amounts ofmass in each side of the tube. On the left side, we have 20 g of the denser liquid, leaving the 40 gof water. Leaving 10 g of water on the left side and the remaining 30 g of water on the right, weget 30 g of liquid on each side. Now that it is in equilibrium, figure the amount of height taken upby each liquid. On the left, 20 g of the dense liquid is taking up 5 cm of space (as marked on the

335


diagram) and the 10 g of water is adding an extra 10 cm, giving 15 cm on the left. On the right,we just have 30 g of water which makes 30 cm of liquid. The ratio of the right side to the left sideis 30 cm/15 cm or 2/1.

Correct Answer

(C)

Alternate Solution

A more rigorous method of tackling this problem is to use the equation for pressure as P = ρghwith the condition that the system is at equilibrium when the pressure on the left side is the same ason the right. Setting P1 = P2, we have to note that P1 is actually composed of two different liquidswith different heights and different densities, i.e. P1 = Pdense + Pwater,1 = Pwater,2. Substituting invalues, we have

(ρdense g (5 cm)) + (ρwater g (h1 − 5 cm)) = ρwater g h2 (4.128)

(4 g/cm3)(5 cm) + (1 g/cm3)(h1 − 5 cm) = (1 g/cm3)h2 (4.129)

Then using the fact that h1 + h2 = 45 cm, you will have 2 equations and 2 unknowns. Whenyou solve for h1 and h2, you should get h1 = 15 cm and h2 = 30 cm, making the ratio h2/h1 =30 cm/15 cm = 2/1.

Correct Answer

(C)

336


4.31 PGRE9677 #31


(A) Any object which falls through a medium which resists its motion will continually increase itsvelocity until it reaches some terminal velocity. If it helps, think of the mass falling throughair resistance as the ”viscous medium”. In order for the retarding force to decrease kineticenergy, velocity would have to decrease at some point, which it doesn’t.

(B) As in part (A), the mass will reach some terminal velocity, however after it has done that itwill maintain that velocity, not slow down and stop.

(C) The terminal velocity of an object, by definition, is the maximum speed which the mass canreach in a given medium. In other words, the maximum speed and the terminal speed arethe same, so the object can’t decrease its speed from a maximum speed to a terminal speed.

(D) For (D) and (E) we finally have an accurate description of terminal velocity, however we nowneed to decide whether the speed of the object is dependent on b and m or just b. You mightbe inclined to think that speed isn’t dependent on mass because of the classical, well knownresults of Galileo which demonstrates that objects fall at the same speed regardless of theirmass. Keep in mind that this is only true in a vacuum and drawing out a force diagramshould make it quite clear that the speed is dependent on mass (i.e. Fnet = ma = mg − bv).If you still aren’t convinced, ask yourself why a feather falls more slowly than a brick in areal world scenario.

(E) This solution is the same as (D) except that it correctly identifies b and m as being variablesof velocity

Correct Answer

(E)

337


4.32 PGRE9677 #32


Rotational kinetic energy is Uk = 12Iω

2. The moment of inertia for a point mass about a radius Ris I = mR2. For the rotation about point A, we need to determine the length between each massand point A. Since the line between B and A bisects a 60 angle, we can make a right triangle anduse trigonometry to find the length as

cos(30) =l/2

R=⇒ RA =

l√3

(4.130)

Which gives us the rotational kinetic energy equation as

Uk−A =1

2(3m)R2ω2 =

1

2(3m)

(l2

3

)ω2 =

1

2ml2ω2 (4.131)

The rotational kinetic energy around point B is easier to calculate because we have just twomasses rotating at length l, giving us

Uk−B =1

22ml2ω2 = ml2ω2 (4.132)

Comparing Equations 4.131 and 4.132, we can see that Uk−B is twice as big as Uk−A

Correct Answer

(B)

338


4.33 PGRE9677 #33


From quantum mechanics, recall that the probability of finding the state of any given operator canbe found by Equation 4.133

P =

∫〈ψ|A|ψ〉 (4.133)

There are two terms with quantum number l = 5 which have coefficients of 2 and 3. This givesa total of 32 + 22 = 13. Thus, the probability is 13 out of a total sum of 38 (i.e. 22 + 32 + 52 = 38)so the probability is 13/38.

Correct Answer

(C)

339


4.34 PGRE9677 #34


(A) Gauge invariance deals with an invariance of charge. This isn’t violated simply due to apreferential direction.

(B) Time invariance doesn’t deal with spin or Electromagnetic interactions. Besides, time invari-ance is rarely violated in any context.

(C) Translation invariance deals with the invariance of system equations under any translationalframe. This isn’t related to this problem.

(D) Reflection invariance is violated in this instance because the preferential direction in one framechanges if we were to create a reflection of that frame. Consider, for example, the way in whichthings will look reversed when viewed in a mirror. If particles are moving in a preferentialdirection, say +x, then there exists some reflected frame in which the particles move in thepreferential direction of -x.

(E) Rotational invariance deals with maintaining system equations under any rotation of the sys-tem frame. This isn’t violated in this example.

Correct Answer

(D)

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4.35 PGRE9677 #35


(A) Recall from doing atomic energy level diagrams that the Pauli exclusion principle tells us thatelectrons in an atom can’t have the same set of 4 quantum numbers: n, l, ml and ms. Ingeneral, it also tells us that fermions may not simultaneously have the same quantum stateas another fermion. More rigorously, it tells us that for two identical fermions, the totalquantum state of the two is anti-symmetric.

(B) The Bohr correspondence principle says that quantum mechanical effects yield classical resultsunder large quantum numbers. Definitely not the answer.

(C) The Heisenberg uncertainty principle tells us that the information of two related aspects ofa quantum state can only ever be known with inverse amounts of certainty. Said plainly,the more you know about one component of the state of a system, the less you know aboutanother2. This isn’t related to the quantum state of fermions.

(D) A Bose-Einstein condensate is a gas of weakly interacting bosons which can be cooled suffi-ciently to force them to their ground state energies. This doesn’t have anything to do withfermions.

(E) Fermi’s Golden Rule involves the rate of transition from one energy eigenstate into a continuumof eigenstates. This is clearly not the right answer.

Correct Answer

(A)

2The typical example is momentum and position

341


4.36 PGRE9677 #36


When an object with mass moves at relativistic speeds, the mass of the object increases. You canalways remember this because as an object approaches the speed of light, the mass approachesinfinity, hence why massive objects can’t travel at the speed of light. When the two lumps of clayhit one another the sum of the masses in non-relativistic terms would be 8 kg and thus at relativisticspeeds, the mass must be higher. We can eliminate (A), (B) and (C) from this fact and, if youcan’t figure out the next step, then you at least have the problem down to two solutions. Relatingthe rest mass of the combined lumps (M) to the total energy of the system for the two massesseparately (2m)gives

Enet = 2γmc2 = Mc2 (4.134)

21√

1− v2

c2

m = M (4.135)

Plug all of the values provided in the problem into Equation 4.135 to get

M =1√

1− (35c)

2/c2(4kg) = 10kg (4.136)

Correct Answer

(D)

342


4.37 PGRE9677 #37


Hooray for plug-n-chug physics problems. Relativistic addition of velocities is solved with

u′ =u+ v

1 + vuc2

(4.137)

Plug in the values given and, shortly thereafter, chug to get

u′ =0.3c+ 0.6c

1 + (0.3c)(0.6c)c2

=0.9c

1.18= 0.76c (4.138)

Correct Answer

(D)

Alternate Solution

Velocity addition “sort of” still works in relativistic terms but it isn’t quite as simple as findingthe sum of velocities. We know that the if we could just add up the velocities, then the velocityof the particle would be 0.9c. However, our ability to continually gain additional velocity drops offas we approach the speed of light, so the speed must be less than 0.9c and we eliminate (E). Sinceaddition still “sort of works”, we would expect the speed to at least be greater than the speed ofjust the particle at 0.6c, so we can eliminate (A) and (B). Finally, you just have to decide whether(C) or (D) is a more reasonable speed. 0.66c is barely larger than the speed of the particle on itsown so (D) is a more reasonable solution.

Correct Answer

(D)

343


4.38 PGRE9677 #38


Consider Einstein’s equations for relativistic energy and relativistic momentum,

Relativistic energy: Erel = γmc2

Relativistic momentum: Prel = γmv

From these 2 equations, it should be clear that the only way we will get velocity from them isto divide Prel over Erel, which results in

PrelErel

=γmv

γmc2=

5 MeV/c

10 MeV=

1

2c (4.139)

Correct Answer

(D)

344


4.39 PGRE9677 #39


Ionization potential is lowest for atoms with full valence shells or nearly full valence shells becausethey generally don’t want to lose electrons. However, atoms with 1 or 2 additional electrons willbe very likely to lose an electron and will have high ionization potential.

(A) Full valence shell

(B) Nearly full valence shell

(C) Nearly full valence shell

(D) Full valence shell

(E) Cs has one additional electron, so it is most likely to lose that electron

Correct Answer

(E)

345


4.40 PGRE9677 #40


From the Rydberg Formula,we get

E = E0

(1

λ21

− 1

λ22

)(4.140)

Finding the values needed for the equation gives us,

λ2 = 470 nm (4.141)

E0 = 4× 13.6eV ≈ 55eV (4.142)

E =hc

λ≈ 2 eV (4.143)

Then, plug in our recently calculated values (Equations 4.141 4.142 and 4.143) to solve for λ1

2eV = 55eV

(1

λ21

− 1

16

)=⇒ λ1 ≈ 3 (4.144)

Which is only true of solution (A)

Correct Answer

(A)

346


4.41 PGRE9677 #41


You should be able to immediately eliminate (D) and (E). Option (D) clearly can’t be correctbecause it suggests that the electron has a spin quantum number of s = 3/2 which is never true(it must always be s = ±1/2). Option (E) can’t be true because you can’t, by definition, moveto a different energy level and maintain the same energy. Next, recall that the angular quantumnumber for P corresponds to L = 1 and, in general

(S, P,D, F, . . .) −→ (0, 1, 2, 3, . . .) (4.145)

so (C) should have L = 1, not L = 3. Finally, between (A) and (B), you must recall yourquantum number selection rules, specifically that transitions are allowed for ∆L = ±1 so (A) isallowed because L moves from P = 1 to S = 0 while (B) violates this allowed transition.

Correct Answer

(A)

347


4.42 PGRE9677 #42


From the photoelectric effect, the equation for maximum kinetic energy is

UK = hν − φ =hc

λ− φ (4.146)

where φ is the Work Function. Plug all of your known values in and round everything to 1significant figure to simplify things,

UK =(4× 10−15 eV · s)(3× 108 m/s)

5× 10−7 m− φ (4.147)

=12× 10−7 eV ·m

5× 10−7 m− φ (4.148)

=12

5− φ (4.149)

= 2.4− φ (4.150)

Finally, Plug in your value for the work function into equation 4.150 and solve

UK = 2.4− 2.28 = 0.12 (4.151)

which is closest to (B).

Correct Answer

(B)

348


4.43 PGRE9677 #43


The line integral in this problem moves about a circle in the xy-plane with its center at 0. Sincewe will be doing a line integral, we should probably know the general solution for a line integralabout a curve, C ∮

Cf(s) =

∫ b

af(t)

∣∣f ′(t)∣∣ (4.152)

So in our case, we need to solve the integral∫ b

af(x, y, z)

∣∣f ′(x, y, z)∣∣ (4.153)

Since the circle is contained entirely in the xy-plane, z = 0. Additionally, since we are dealingwith a circle, we will want to convert our values for x and y into their parametric equivalents.Specifically, x = R cos(θ) and y = R sin(θ). Replacing and x and y with our parametric equationsgives us 4.155 ∫ b

a

(ydx

dθ− xdy

dθ

)dθ (4.154)

∫ 2π

0[(R sin(θ))(−R sin(θ))− (R cos(θ))(R cos(θ))] dθ (4.155)

Multiplying out and simplifying, you should get∫ 2π

0−R2

[sin2(θ) + cos2(θ)

]dθ (4.156)

Since sin2(θ) + cos2(θ) = 1, Equation 4.156 becomes∫ 2π

0−R2dθ = −2πR2 (4.157)

The negative sign in 4.157 (and in fact the sign in general) is dependent on the direction inwhich you traverse the curve so we are primarily concerned with the magnitude of the solution.

Correct Answer

(C)

349


Alternate Solution

Kelvin-Stokes theorem tells us that the surface integral of the curl over a vector field is equivalentto the line integral around that vector field in a Euclidean 3-space. Mathematically, that is∫

(∇× f) · da =

∮f · dl (4.158)

Which means that if we can take the curl of our function, u, and the integral of it over ourarea, then we have the line integral of the same function. Recall that the curl of a function can becalculated by taking the determinant of the matrix∣∣∣∣∣∣∣

i j k∂∂x

∂∂y

∂∂z

Fx Fy Fz

∣∣∣∣∣∣∣Applying this to our function, u and taking the determinant gives

i

(∂z

∂y+∂x

∂y

)+ j

(∂z

∂x− ∂y

∂z

)+ k

(−∂x∂x− ∂y

∂y

)(4.159)

Since x, y and z are not functions of one another, everything in Equation 4.159 goes to 0 withthe exception of the final term, which becomes −2. Substituting ∇× u into Equation 4.158,∫ 2π

0−2 · dA = −2A (4.160)

where the area of the circle is πR2, so∮u · dl = −2A = −2πR2 (4.161)

Correct Answer

(C)

350


4.44 PGRE9677 #44


Velocity is given and since the derivative of velocity is acceleration, take the derviative. Take note,however, that v is a function of position (x) and position is a function of time (t) so use chain ruleto get

a =dv

dx

dx

dt(4.162)

=dv

dx(v) (4.163)

=(−nβx−n−1

) (βx−n

)(4.164)

= −nβ 2x−2n−1 (4.165)

Correct Answer

(A)

351


4.45 PGRE9677 #45


A simple method of generating a low-pass filter in a circuit involves placing a resistor in series witha load and a capacitor in parallel with that same load. low-frequency signals that attempt to passthrough the circuit will be blocked by the capacitor and will be forced to pass through the load

352


instead. Meanwhile, high-frequency signals will be able to bypass the capacitor with little to noeffect. Of the possible solutions, only (D) provides us with a resistor in series with a load and acapacitor in parallel with that load.

Correct Answer

(D)

353


4.46 PGRE9677 #46


From Faraday’s law of induction, we know that the potential (Electromotive Force) for a loop ofwire is

|ε| = N

∣∣∣∣dφBdt∣∣∣∣ (4.166)

Since there is only one loop, it simplifies to


∣∣∣∣ (4.167)

The problem tells us that the potential is, ε = ε0 sin(ωt), so we can set that equal to Equation4.167 and substitute in φB = B · dA, to get

ε0 sin(ωt) =d

dt(B · dA) (4.168)

move dt over and integrate both sides∫ε0 sin(ωt) dt =

∫B · dA (4.169)

− ε0 cos(ωt)/ω = B ·A = BπR2 (4.170)

Finally, rearrange to solve for ω,

354


ω =−ε0 cos(ωt)

BπR2(4.171)

and consider when the angular velocity is maximized, t = 0, to get

ω =−ε0

BπR2(4.172)

Correct Answer

(C)

355


4.47 PGRE9677 #47


Faraday’s law gives us potential for a changing magnetic field as


∣∣∣∣ (4.173)

Where φB is the flux of the magnetic field through some area which is

φB =

∫B · dA (4.174)

The area through which the flux passes is just the area of a circle with radius R so φB = BπR2

|ε| = d

dt

(BπR2

)(4.175)

and since the flux is changing from the the rotation and the loops rotate at N revs ,

356


|ε| = NBπR2 (4.176)

Correct Answer

(C)

Alternate Solution

Alternatively, you can go through a process of elimination by removing unlikely or impossiblechoices. (A) is clearly wrong because, by Faraday’s law, a changing magnetic field will generatea potential. (D) is also clearly wrong because it suggests that potential decreases as the rate ofvariance in the magnetic field increases. For (B), (C) and (E), check the units.

(B) 2πNBR ≡ (rev)(kg/s2A)(m) ≡ revA

kg ms2

(C) πNBR2 ≡ (rev)(kg/s2A)(m2) ≡ revA

kg m2

s2= rev

A J

(E) NBR3 ≡ (rev)(kg/s2A)(m3) ≡ revA

kg m3

s2

Of these, only (C) has the right units.

Correct Answer

(C)

357


4.48 PGRE9677 #48


Mesons are hadrons so they have mass, which means they can’t exceed the speed of light and alsocan’t match the speed of light (i.e. the mass-less photon) so we eliminate (D) and (E). Next, let’stry to treat the motion as if it is non-relativistic. Only half of the π+ mesons make it through the15 meters, thus the amount of time to move the 15 meters is 1 half life or 2.5×10−8 seconds. Fromthis, try

v =∆X

∆t=

15 m

2.5× 10−8≈ 5× 108m/s (4.177)

The value gives us a speed faster than light, meaning that what ever the actual speed is, itneeds to be analyzed using relativistic equations. This tells us that the velocity must be very nearc but less than it.

(A) 12 C: Not nearly fast enough for relativistic influence to take effect

(B)√

25 C ≈ 0.6 C: Very similar to (A) and likely not fast enough to have significant relativistic

effect.

(C) 2√5C ≈ 0.9 C: This value is the closest to justifying relativistic influence so it is the most

likely to be correct.

Correct Answer

(C)

358


Alternate Solution

To calculate the value exactly, start with your equation for “proper time”,

∆τ2 = ∆t2 −∆X2 (4.178)

∆t2 = ∆τ2 + ∆X2 (4.179)

We know, from the half-life given in the problem and the fact that only half of the sample makesit through the 15 meters, that it takes one half-life of time to move 15 meters. Substitute in 15 forthe position and 15/2 for the time,

∆t2 = (15/2)2 + 152 = 152

[(1

2

)2

+ 1

]= 225

(1

4+ 1

)(4.180)

Take the square root of both sides to get

∆t =

√225

(1

4+ 1

)= 15

(√5/4

)(4.181)

Now using the basic definition for velocity (i.e. v = ∆X/∆t), with distance ∆X = 15, we get

∆X

∆t=

15

15√

5/4=

1√5/4

=2√5

(4.182)

Correct Answer

(C)

359


4.49 PGRE9677 #49


Relativistic influences on Electricity and Magnetism occurs proportionally to the lorentz factor, γ.Since Ez = σ

2ε0, we would expect the Electric field to be influenced by

Ezγ =σ

2ε0

1√1− v2/C2

(4.183)

Correct Answer

(C)

360


4.50 PGRE9677 #50


The space time interval equation is

∆S2 = −(c ∆t)2 + ∆X2 (4.184)

The space time interval is S = 3C · minutes and the position interval is ∆X = 5c · mintues.Plug these into Equation 4.184

9C2 = −(C ∆t)2 + 25C2 (4.185)

Then, solve for ∆t to get

∆t = 4 minutes (4.186)

Correct Answer

(C)

361


4.51 PGRE9677 #51


The solution to the infinite square well (also known as the particle in a box) is a sine function inwhich the number of nodes on the wave is n + 1. In the ground state, you will have 2 nodes atthe end of the infinite walls with the peak of the wave at exactly the middle. For n = 2, we get 3nodes with two peaks (or a peak and a trough if you insist) and the middle of the wave falls on anode. If you keep checking all of the values for n, you will find that all states with even values forn result in a node in the middle of the well.

Correct Answer

(B)

Alternate Solution

The solution to the infinite square well is

ψn(x, t) = A sin

(nπx

L

)(4.187)

362


If we are only concerned with the middle position, set x = L/2, giving

ψn(x, t) = A sin

(nπ

2

)(4.188)

and this equation will go to zero any time you are taking the sine of integer values of π (i.e.sin(0), sin(π), sin(2π), . . .).

Correct Answer

(B)

363


4.52 PGRE9677 #52


For the spherical harmonics, we get a sine term for the harmonic

Y 11 = −1

2

√3

2πsin(θ)eiπφ (4.189)

and

Y −11 =

1

2

√3

2πsin(θ)eiπφ (4.190)

which means that m = ±1. The eigenvalues of a spherical harmonic can be found with

LZψ = mhψ (4.191)

Plugging in our value for m gives

LZψ = ±1hψ (4.192)

which is (C).

Correct Answer

(C)

364


4.53 PGRE9677 #53


Positronium atoms can only decay into even numbered groupings of photons (to conserve spin) sowe can eliminate choices (B) and (D).

Positronium atoms are very unstable and since energy must be conserved, there is no way thatthe atom will decay without releasing some photons so we can eliminate (A).

Finally, between 2 photons and 4 photons, consider how silly a feynman diagram will look with4 photon lines emanating from the interaction. In case I wasn’t being clear, it is wicked silly.

Correct Answer

(C)

365


4.54 PGRE9677 #54


Electromagnetic waves are typically thought of as being composed of a magnetic and electric wavemoving in the same direction and oscillating orthogonally (Figure 4.8).

The summation of any two or more of these waves will occur in one of two forms. One of theforms involves waves oscillating in the same plane as one another and the other involves oscillationsin different plans. In general, any waves in the same plane will result in field vectors also in thesame plane, resulting in a trajectory that moves in only 2 dimensions. Field vectors in an alternateplane will sum to a 3 dimensional rotational trajectory. For the specific problem here, the secondwave is rotated π or 180 so the field vectors lie in the same plane and the final trajectory mustbe linear. Based on this, you can eliminate (C), (D), and (E). In order to decide between (A) and(B), realize that two waves perfectly in phase (i.e. with no rotation) will have field vectors in thesame quadrant and so the angle will be 45. On the other hand, a wave rotated π from the otherwill be in different quadrants and so the angle will be 135.

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Figure 4.8: Classical view of elctromagnetic wave behavior

Correct Answer

(B)

367


4.55 PGRE9677 #55


By Malus’ Law, the intensity of a an electromagnetic wave after passing through a perfect polarizeris

I =1

2cε0E

20cos2(θ) (4.193)

The optical path difference for the second wave is z = 2π/k. Plugging this into the originalequation and letting z → 0 and t→ 0.

E = xE1ei(kz−ωt) + yE2e

i(kz−wt+π) = xE1e0 + yE2e

i(2π+π) (4.194)

Recall Euler’s identity, eiπ = 1, so the Equation 4.194 becomes

E = E1 + E2 (4.195)

368


Since the two waves are decoupled, the magnitude of the entire wave will be the magnitude ofeach individual wave added separately

I = (E1)2 + (E2)2 (4.196)

Correct Answer

(A)

369


4.56 PGRE9677 #56


Immediately eliminate choice (A) because there is no way you are going to get an angle of 0

between the water surface and the light source. Next, eliminate (E) because an angle of 90 willresult in no bending of the light and so there will be no total internal reflection. As for the last 3,you can probably intuitively figure that it won’t be at 25 simply because that isn’t a very steepangle. However, if we want to be more rigorous, we must use Snell’s law. Recall that total internalinflection occurs in any instance in which using Snell’s law would require you to take the sine of anangle and get a value that isn’t possible. In our scenario, if we apply an angle of 50, we get

sin(θ1)

sin(θ2)=

n2

n1(4.197)

sin(θ2) = 1.33 sin(50) (4.198)

= 1.02 (4.199)

However, there is no angle at which the sine function will give you a value larger than 1 and thismeans you’ve reached total internal reflection. Admittedly, it is a bit rude on the part of ETS tosimultaneously give you relatively complicated decimals and lesser used angles3 with no calculator,but at the very least you could quickly get rid of the angle 25 by using this method. Then, knowingthat ETS is looking for the minimum angle for total internal reflection, do an approximation (e.g.45 as an approximation for θ = 50) to determine which potential angle is closer to pushing ourvalue over 1.

Correct Answer

(C)

3i.e. not one of the angles we’ve all memorized from the unit circle

370


4.57 PGRE9677 #57


The equation relating slit width for a single slit diffraction to wavelength is

d sin(θ) = λ (4.200)

d =λ

sin(θ)(4.201)

Convert the wavelength into meters, giving

d =(4× 10−7 m)

sin(4× 10−3 rad)(4.202)

For small angles, we can make the approximation sin(θ) = θ, which gives us

d =(4× 10−7 m)

(4× 10−3)(4.203)

d = 1× 10−4 m (4.204)

Correct Answer

(C)

371


4.58 PGRE9677 #58


We are trying to convert a “well-collimated” laser with diameter of 1 mm to a “well-collimated”laser of 10 mm, which represents a magnification of 10×. Lens magnification must have the sameproportionality for the focal length as it does for the separation distance, so we would expect afocal length that is ten times larger than the 1.5 cm lens, i.e. 15 cm. We can eliminate (A), (B)and (C) from this. Finally, we need to decide whether the distance for the new lens will be 15 cmor 16.5 cm. Assuming everything is done properly, the first lens will have its focus at 1.5 cm andthe second lens will have its focus at a distance of 15 cm. For ideal collimation, we will want thefoci to be at the same location, so

15 cm + 1.5 cm = 16.5 cm (4.205)

Correct Answer

(E)

372


4.59 PGRE9677 #59


The energy of a single photon with wavelength, λ = 600 nm can be found by

E =hc

λ(4.206)

The total number of photons emitted per second will be equal to the total energy generated persecond divided by the energy of a single photon, i.e.

# of photons =100 watts

hc/λ(4.207)

Plug everything in and rounding all of our numbers to simplify the mental math, you shouldget

# of photons =(1.0× 104 W)(6× 10−7 m)

(6.63× 10−34 J · s)(3× 108 m/s)≈ 3× 1023 1

s(4.208)

However, the question asks for the number of photons per femtosecond, so convert the previoussolution to the right units to get

# of photons ≈ 3× 1023 1

s≈ 3× 108 1

fs(4.209)

Which is closest to (B)

Correct Answer

(B)

373


4.60 PGRE9677 #60


Keep in mind while we work this problem that we don’t necessarily want the most accurate answer,just the quickest method to the correct choice. The Lyman alpha line is the spectral line corre-sponding to a hydrogen atom transition from level n=2 to n=1. The question poses this problemin a way that would indicate that we should use the Doppler shift equation for light. If we wantedto be extremely accurate, and let me stress how much we DON’T want this, then we would need touse the relativistic equations for the Doppler shift. Note, however, that the largest possible speedfor the particle given in the answers, is 2200 km/s. Convert that to m/s and compare it to thespeed of light, for our purposes we will just call it C = 3.0 × 108 m/s. The fastest this particlecould possibly be moving, according to the potential solutions, would be roughly 0.73% the speedof light

2.2× 106 m/s

3.0× 108 m/s= 0.0073 (4.210)

Relativistic effects will be “relatively” negligible at these speeds so let’s just use the non-relativistic equation,

∆λ

λ=v

C(4.211)

With,

λ = 122 nm = 1.22× 10−7 m (4.212)

∆λ = 1.8× 10−12 m (4.213)

C ≈ 3.0× 108 m (4.214)

Since we only care about an approximation, pretend that the two wavelength values of 1.22 and1.8 are just 1. Plug this all in to get

(3.0× 108 m)(1.0× 10−12 m)

(1.0× 10−7 m)= 3.0× 103 m/s (4.215)

374


CAREFUL! The problem asked for the answer in km/s, not m/s like we’ve solved for. Convert3.0× 103 m/s to get 3.0 km/s which is closest to answer (B).

Correct Answer

(B)

375


4.61 PGRE9677 #61


Gauss’s law gives us

~E · d ~A = q/ε0 (4.216)

Taking the integral of both sides and substituting the equation for surface area of a sphere forAs

∫~E · d ~As = EA =

∫ R/2

0

ρAsε0

dr (4.217)

E(4πr2

)=

∫ R/2

0

1

ε0

(A(r)2

) (4π(r)2

)dr (4.218)

E4π

(R

2

)2

=4πA

5ε0

(R

2

)5

(4.219)

Start canceling things out and you get

E =A

5ε0

(R

2

)3

=AR3

40ε0(4.220)

Correct Answer

(B)

376


4.62 PGRE9677 #62


We can calculate Q1 and Q2 when the battery is in the system, using

Q1 = C1V = (1.0 mF)(5.0 V) = 5 mF V (4.221)

Q2 = C2V = (2.0 mF)(5.0 V) = 10 mF V (4.222)

Once the battery is removed the problem tells us that the capacitors are connected to oneanother such that the “opposite charges are connected together”. Doing this, we get Figure 4.9

Figure 4.9: Circuit with opposite charges connected together

The potential for both capacitors will be the same so we can calculate V for both capacitorswith

V =QeqCeq

(4.223)

For a parallel circuit, the orientation of this circuit has its capacitors flipped so Qeq = Q2 −Q1

V =Q2 −Q1

C1 + C2=

(10 mF V )− (5 mF V )

(1 mF ) + (2 mF )=

5

3V ≈ 1.7 V (4.224)

Correct Answer

(C)

377


4.63 PGRE9677 #63


(A) Muon: The Muon is one of the leptons, which are the fundamental particles. Muons aresimilar to electrons in that they have a negative charge and a spin of 1/2. The Muon IS NOTa composite object

(B) Pi-Meson: All Pi-Mesons (also known as a Pion) are composed of some combination of thefirst generation quarks (Up quark and Down Quark). The Pion IS a composite particle

(C) Neutron: Neutrons have a tri-quark arrangement, 1 Up quark and 2 Down quarks. TheNeutron IS a composite particle

(D) Deuteron: A deuteron, the nucleus of a Deuterium atom, is composed of a proton and aneutron as opposed to hydrogen which has just a proton. The Deuteron IS a compositeparticle

(E) Alpha particle: Alpha particles are composed of 2 Neutrons and 2 Protons. The Alphaparticle IS a composite particle.

Correct Answer

(A)

378


4.64 PGRE9677 #64


In symmetric fission, the heavy nucleus is split into two equal halves that we are going to assume areboth an example of a “medium-weight nucleus”, per the problem description. The kinetic energyafter fission will be the difference between the initial total energy and the energy remaining in thetwo ”medium-weight” nuclei.

∆E = Eheavy N − E2 medium N (4.225)

The energy of the heavy nucleus is given in the description as 8 million eV/nucleon and theenergy of the 2 medium nuclei is 7 million eV/nucleon.

∆E = (1 nucleus)

(8

MeV

nucleon

)(N)− (2 nuclei)

(7

MeV

nucleon

)(1

2N

)(4.226)

∆E = (8 MeV− 7 MeV)N = N(1 MeV) (4.227)

Then taking some kind of heavy nucleus, say Uranium-238 (N=238), you get

∆E = 238 MeV (4.228)

Which is roughly (C).

Correct Answer

(C)

379


4.65 PGRE9677 #65


The quickest method for solving this problem is to consider when the mans mass goes to infinity.When this occurs, the man won’t move at all, do to his infinite mass, and so the only energy willinvolve the movement of the boat with kinetic energy Uk = 1

2Mv2. The only choice that gives thisequation at the limit is (D).

Correct Answer

(D)

Alternate Solution

Energy and momentum are conserved, so we get the following equations

Ptot = mv −MV = 0 (4.229)

mv = MV (4.230)

Utot =1

2mv2 +

1

2MV 2 (4.231)

All of the possible solutions involve only M , v and m, so solve for V .

380


V =mv

M(4.232)

Substitute this into Equation 4.231

Utot =1

2mv2 +

1

2M

(mv

M

)2

(4.233)

1

2

(m+

m2

M

)v2 (4.234)

Correct Answer

(D)

381


4.66 PGRE9677 #66


The problem clearly states that the spacecraft is “on a mission to the outer planets” meaning thatit must have an orbit with an escape trajectory. Only Parabolic and Hyperbolic orbits are escapetrajectories so eliminate (A), (B) and (C). Choosing between (D) and (E), go with (E) becausehyperbolic orbits occur at high velocities, like 1.5 times the speed of Jupiter, while parabolic orbitshappen at lower speeds and are generally more rare.

Correct Answer

(E)

382


4.67 PGRE9677 #67


Recall the equation for the event horizon (Schwarzschild radius) of an object is

Re =2GM

c2(4.235)

The mass of the earth is given and G and C can be found in the list of constants in your GREbooklet. Plug these all in to get

Re =2(6.67× 10−11 m3/kg · s2)(5.98× 1024 kg)

(3.0× 108 m/s)2(4.236)

Rounding all of the numbers and multiplying out gives

Re =84× 1013 m3/s2

9× 1016 m2/s2≈ 1× 10−3 m = 1 cm (4.237)

Correct Answer

(C)

383


4.68 PGRE9677 #68


A quick trick to figuring out the solution to this is to consider which variables the Lagrangian mustbe dependent on. Looking over the diagram and considering that the Lagrangian is the differencebetween Kinetic and Potential energy, you should be able to convince yourself that it will have adependence on (m, s, θ, ω). Look through each of the potential solutions to see that only (C) and(E) match this criteria. Now, consider when the angle between the rod and the vertical is 0 (θ = 0).In this case, the potential energy term should go to 0 because the bead won’t be able to move upthe rod if the axis of rotation is parallel to the rod. Thus, we must have a term with a sine ortangent function in it so that the angle is forced to 0 in the potential energy term, which is onlythe case for (E).

Correct Answer

(E)

384


4.69 PGRE9677 #69


Remember our good friend the right hand rule, which gives us, among other things, the directionof magnetic field vectors from a moving current (Note: this is the general principle behind how asolenoid works).

From the diagram given in the problem, we have a current going “into the page” on the rightand coming out of the page on the left. If you’ve done everything right and aren’t too embarrassedto be making hand gestures at your computer screen/test booklet, then you are giving the problema thumbs up (I’m going to assume that this isn’t a case of you approving of the GRE). From this,only (A) and (B) could be correct. The only difference between (A) and (B) is a dependence on r.If we were talking about current I, rather than current density, J , then we would care about theradius (i.e. the size) of the conductive cables. However, since the problem talks about everythingin terms of a constant current density, J , we know there should be no dependence on r. In caseyou aren’t convinced, compare Maxwell’s equations for magnetic fields with dependence on I vsdependence on J .

B =µ0I

4π

∫dl × rr2

(4.238)

385


as compared to

∇×B = µ0J + µ0ε0∂E

∂t(4.239)

Correct Answer

(A)

I I

B

B

Figure 4.10: Magnetic fields generated as the result of a moving electric field

386


4.70 PGRE9677 #70


The Larmor Formula can be used to calculate the power radiated in non-relativistic motion of acharged particle

P =q2a2

6πε0c3(4.240)

The problem states that particle B has half the mass (12m), twice the charge (2q), three times

the velocity (3v) and four times the acceleration (3a). The Larmor Formula isn’t dependent onmass or velocity so we are only concerned with charge and acceleration. Since the denominator ofthe Lamor Formula won’t be altered for either particle, we only care about the numerator.

PBPA

=(q2Ba

2B)

(q2Aa

2A)

=(4q2

A16a2A)

(q2Aa

2A)

= 64 (4.241)

Correct Answer

(D)

387


4.71 PGRE9677 #71


The angle of deflection for this particle can be calculated as

tan(θ) =y

x(4.242)

We already know the velocity in the x direction as v, so

tan(θ) =y

v(4.243)

Force due to gravity will be minimal relative to the Lorentz force so we’ll assume it is 0. Thisgives us a net force of

Fnet,y = my = q[ ~E + (~v × ~B)] (4.244)

The problem says nothing about a magnetic field existing (even though it should) so we’llassume ~B = 0, giving

my = q ~E =qV

d(4.245)

y =qV

dm(4.246)

We know that the velocity in the x-direction is x = L/t and the acceleration in the y-directionis y = y/t. Solve for t in both equations and equate the two, to get

388


L/x = y/y (4.247)

Solve for velocity in the y-direction to get

y = Ly/x (4.248)

We already solved for y in Equation 4.246, so we can plug that in to get

y =LqV

dmx(4.249)

then, plug 4.249 into 4.242 to get

tan(θ) =LqV

dmxv(4.250)

and since x = v, we can solve for θ to get

θ = tan−1(LV q

dmv2

)(4.251)

Correct Answer

(A)

389


4.72 PGRE9677 #72


This solution is far from rigorous but it is the quickest way to solve the problem. Negative feedbackand positive feedback function similarly in an electronic circuit as it does in acoustics. Positivefeedback of an audio wave involves an increase in amplitude for the wave. Think along the linesof placing a microphone too close to a speaker and making that high pitched squeal. Negativefeedback, on the other hand, should cancel out some of the amplitude. From this, you shouldimmediately know that (A) can’t be an aspect of negative feedback.

Correct Answer

(A)

390


4.73 PGRE9677 #73


For a thermodynamic expansion, work done is

W =

∫ Vf

Vi

P dV (4.252)

The problem gives us the ideal adiabatic expansion equation as

PV γ = C (4.253)

P =C

V γ= CV −γ (4.254)

Making the substitution into the work equation and taking the integral should give us

W = C

∫ Vf

Vi

V −γdV =

[C

1− γV 1−γ

]VfVi

(4.255)

At this point, you can see that the denominator must contain a 1− γ term and you can choose(C). However, if you want you can always substitute in C to get

PfVf − PiVi1− γ

(4.256)

Correct Answer

(C)

391


4.74 PGRE9677 #74


Spontaneous events in a thermodynamic system always have positive value changes in entropy soget rid of (D) and (E). Additionally, since we know that the change in entropy is

∆S =

∫ T2

T1

dq

T(4.257)

We are going to get a natural log component (unless one of our temperatures is 0 which is notthe case) so eliminate (A). If you can’t get any farther than this, at least you go it down to 2choices and you can guess. The next step you should take is to use Equation 4.257 to calculate thenet change in entropy as the sum change in entropy of the 2 masses

∆Snet = ∆S1 + ∆S2 (4.258)

Because the masses are of the same size, they will both reach a temperature as an average ofthe two

500 K + 100 K

2= 300 K (4.259)

then, using dq = mCdT and equation 4.259, we get

∆Snet = mC

[∫ 300

100

dT

T1+

∫ 300

500

dT

T2

]= mC[ln(3) + ln(3/5)] = mCln(9/5) (4.260)

Correct Answer

(B)

Additional Note

In the event that ETS didn’t give us objects with the same mass, it should be relatively straightfor-ward to calculate the final heat of the system by the conservation of energy (heat) as Q1 +Q2 = 0.

392


4.75 PGRE9677 #75


Fourier’s law of heat conduction gives

∂Q

∂t= −k

∮S∇T · d ~A (4.261)

Q = −kA∆T

∆X(4.262)

Which tells us that heat transfer is proportional to the thermal conductivity of the material, kand the cross-sectional area of the material but inversely related to the length the heat transfersthrough. In this problem, the cross sectional area is the same for both, so the ratio is

QAQB

=(0.8 watt/mC)/(4 mm)

(0.025 watt/mC)/(2 mm)= 16 (4.263)

Correct Answer

(D)

393


4.76 PGRE9677 #76


Analysis of Gaussian wave packets are fascinating because some of the more interesting and familiarquantum mechanical laws fall out of them. In particular, the Heisenberg Uncertainty principle isone of those results

∆x ∆p ≥ h

2(4.264)

This is relevant because the uncertainty principle tells us that the wavepackets momentum cannever be 0, meaning I is not possible. Eliminate (A), (C), (E). Now, compare II and III. If it’strue that the“width of the wave packet increases with time” then it isn’t possible for the statement“Amplitude of the wave packet remains constant with time” to also be true, since the wave packetstretching in time would alter the amplitude. From this, (D) can’t be true, leaving you with choice(B).

Correct Answer

(B)

394


4.77 PGRE9677 #77


Recall that the expectation value for energy is given by the Hamiltonian operator as

〈H(t)〉 = 〈ψ|H(t)|ψ〉 (4.265)

We are given the Hamiltonian

H = −JS1 · S2 (4.266)

and spin operators

S21ψ1 = S1(S1 + 1)ψ1 (4.267)

S22ψ2 = S2(S2 + 1)ψ2 (4.268)

We can use the polynomial identity

a1 · a2 =1

2

[(a1 + a2)2 − a2

1 − a22

](4.269)

to get

395


〈ψ1|H(t)|ψ2〉 = −J2

[(S1 + S2)2 − S21 − S2

2 ] (4.270)

Substitute in the spin operators (Equations 4.267 and 4.268) to get

〈H(t)〉 = −J2

[(S1 + S2)(S1 + S2 + 1)− S1(S1 + 1)− S2(S2 + 1)] (4.271)

Correct Answer

(D)

396


4.78 PGRE9677 #78


Semiconductors are useful devices because it is relatively straightforward to alter a semiconductorsconductive properties by intentionally adding impurities to the semiconductor lattice. The processof adding these impurities is known as doping. Dopants in the lattice of an n-type semiconductoralter conductivity by donating their own weakly bound valence electrons to the material. This isprecisely the description of (E).

In general, you should try to remember that dopants in an n-type semiconductor always con-tribute electrons to the lattice rather than take from the lattice (those would be p-type semiconduc-tors). Once you’ve concluded that the only solutions could be (D) or (E) it should seem reasonablethat electrons that are donated aren’t going to get donated to a full valence shell.

Correct Answer

(E)

397


4.79 PGRE9677 #79


In its most general form, heat capacity C is given by

C =

(∆Q

∆T

)(4.272)

Which at the very least gives us our temperature dependence for the heat capacity. Specific tothis problem, recall the energy level diagrams from thermodynamics.

the heat capacity of an ideal gas is proportional to the sum of the degrees of freedom for eachof the three energy levels. For a monotonic gas, particles will have three translational degrees offreedom corresponding to the three components of motion (~x, ~y, ~z)

Cv =3

2R (4.273)

For a diatomic molecule, we have to figure into the heat capacity the linear vibrational energyand the rotational energy

Cv =3

2R+Rvib +Rrot (4.274)

From the energy level diagram (Figure 4.11), we can see that small changes in energy levelcorrespond to translational motion. However, large quantities of energy are required for vibrationaland rotational energy to play a part. From this, we get the low temperature (i.e. low energy) heatcapacity as

Cv−low =3

2R (4.275)

and high temperature (i.e. high energy) heat capacity as

Cv−high =3

2R+Rvib +Rrot =

7

2R (4.276)

So the ratio of high temperature heat capacity to low temperature heat capacity is

Cv−high/Cv−low =7/2 R

3/2 R=

7

3(4.277)

398


Energy

RotationalEnergy Levels

VibrationalEnergy Levels

V0

V0

V1

V1

2RR1

3R

2R3R

R1

ElectronicEnergy Levels

Figure 4.11: Energy level diagram and electron transitions

Correct Answer

(D)

399


4.80 PGRE9677 #80


Immediately get rid of (A) and (E) because they suggest no dependence on the mass of eitherstring. Next, consider the scenario in which the string on the right becomes infinitely massive (i.e.µr → ∞). When this occurs it won’t be possible for any amount of energy on the left string tocreate an amplitude on an infinitely massive string on the right and so amplitude should go to 0.Under this condition (B) will become 2, (D) will go become -1 so these can’t be correct. (C) is theonly one which goes to 0 when µr →∞.

Correct Answer

(C)

400


Alternate Solution

Consider the case when µl = µr. In this case, the two part string of different masses becomes asingle string with one mass, call it µ, and the original amplitude of 1 will be maintained. From this(E) and (D) can be eliminated and (A) can be eliminated simply because it doesn’t acknowledgethe dependency on string mass. Finally, eliminate (D) because the amplitude doesn’t go to 0 whenthe mass, µr goes to infinity.

Correct Answer

(C)

401


4.81 PGRE9677 #81


This is one of the few problems I would recommend doing the math in full gritty detail. Thenumber of beats between two waves comes from the difference in frequency between them. Beatscan be observed (heard), for example, when two musical instruments are out of tune. To minimizethe beats with a frequency of 73.416 Hz for D2, we will need the harmonic multiplied by thatfrequency to be very close to 440 Hz. Of the harmonics given, 6 is the most reasonable both fromthe perspective of quick mental math and from the perspective that ETS likes to keep you fromgetting the correct answer by knowing only one of the pieces of information (i.e. there are twosolutions with a harmonic of 6). Multiplying everything out completely, you should get

(73.416 Hz)(6) = 440.49600 Hz ≈ 440.5 Hz (4.278)

Since the number of beats is just the difference between the two frequencies,

440.5 Hz− 440.0 Hz = 0.05 Beats (4.279)

Correct Answer

(B)

402


4.82 PGRE9677 #82


Figure 4.12: Reflection of light on a thin film

The equation for constructive interference of a thin film (Figure 4.12 is

2nd =

(m+

1

2

)λ (4.280)

Plug in n = 1 for air our value for the wavelength as 488 nm,

2d = mλ+1

2λ (4.281)

d = m

(448 nm

2

)+

(448 nm

4

)= m(244 nm) + 122 nm (4.282)

From this, we get

403


When m = 0: d = 122 nm

When m = 1: d = 366 nm

When m = 2: d = 610 nm

Correct Answer

(E)

404


4.83 PGRE9677 #83


Note that when d goes to infinity, every peak and trough will be infinitely tall/short and you won’teven need a velocity to get the ball to free fall (i.e. when d→∞ then v → 0). The only equationwhich fits the bill is (D).

Correct Answer

(D)

405


4.84 PGRE9677 #84


Immediately get rid of (A) as it suggests that the normal mode has no dependence on either of themasses. Next, note that if we let the mass of one of our masses, say m2 go to infinity, then theother mass will oscillate about that mass as if it were connected to a stationary object. When thishappens, dependence on mass m2 should disappear but dependence on m1 should remain. For (B)and (E), allowing m2 → ∞ forces the entire term, including m1 to disappear. Finally, comparing(C) and (D), get rid of (C) because it suggests that the normal mode has no dependence on theacceleration due to g or l, which is not true.

Correct Answer

(D)

406


4.85 PGRE9677 #85


Taking the recommendation at the end of the problem, consider the limiting cases of M → 0 andM → ∞. Looking through the 5 options, you can immediately eliminate (D) and (E) becausethey both suggest that the mass of the string and the mass of the ring have no influence on thewavelength, which is not correct.

Considering the first limiting case, when M → 0 then µ/M → ∞. In the case of (C), asine function is going to limit the maximum and minimum values so we can eliminate (C). since

cot(x) = 1tan(x)

= cos(x)

sin(x)and tan(x) = sin(x)

cos(x) , both can blow up to infinity if the bottom trig

function goes to 0.Considering the second limiting case, when M → ∞ then µ/M → 0. It is also the case that

when mass goes to infinity, the ring won’t move from any amount of force placed on the string sowhat we have is a fixed end on the ring side. This means that the only wavelengths possible are

407


lengths of L = n λ2 . This is the case because the fixed end on each side acts as a node and you have

a bound standing wave. Checking this requirement with both (A) and (B), gives

(A) µ/M = 0 = cot(2πLλ ) =⇒ cot−1(0) = π/2 + n π = 2πL

λ =⇒ L = n2 + 1

4

(B) µ/M = 0 = tan(2πLλ ) =⇒ tan−1(0) = 0 + n π = 2πL

λ =⇒ L = nλ2

Of which, only (B) meets the necessary criteria.

Correct Answer

(B)

408


4.86 PGRE9677 #86

409



As a general rule, particles moving in an orthogonal direction to a magnetic field ( ~B) will exhibitcyclotron (helix shaped) motion with a direction of spin in agreement with the right hand rule.Of the choices, only (B) and (E) exhibit this phenomena and only (B) demonstrates actual helicalmotion.

Correct Answer

(B)

410


4.87 PGRE9677 #87


Immediately eliminate (D) and (E) because they both suggest that charges in a magnetic fieldwon’t cause the pith balls to move which is incorrect. Eliminate (B) because it doesn’t have adependence on R and if R went to 0, the magnetic field would as well. Finally, consider that if wehad a dependence on d as in (C) and let d → ∞, then angular momentum would also need to beinfinite which would violate conservation of the momentum built up from the magnetic field.

Correct Answer

(A)

411


4.88 PGRE9677 #88


Immediately eliminate all solutions that suggest that the magnetic field is anything other than0 at the origin, specifically (E) and (D). There are a number of ways you can convince yourself

412


this criteria must be true. For example, consider integrating the magnetic field generated over aninfinitely small surface area which would give you no magnetic field at all. Alternatively, recall thata larger magnetic field vector is indicative of a larger magnetic field and that the magnetic fieldvectors get smaller as R gets smaller, until it approaches R = 0 when ~B = 0 (Figure 4.13).

Figure 4.13: Magnetic field vectors decrease in magnitude as R→ 0

Next, eliminate all solutions which don’t suggest that the magnetic field at point c and all radiilarger than c is 0., i.e. (C) and (A). You can convince yourself that this must be the cases becausethe two cables have equivalent magnetic fields moving in opposite directions which will cancel eachother out at and past point c.

Correct Answer

(B)

413


4.89 PGRE9677 #89


As soon as you see a charged particle in a magnetic field, think Lorentz force, F = q(~v × ~B). Inour particular problem the velocity vector and magnetic field vectors are orthogonal, so the crossproduct of ~v × ~B = vB. Since the object is rotating the net force on the particle should also beequal to the centripetal force, F = m v2

R . Set these two equations equal to one another and solvefor momentum, p = mv.

qvB =m v2

R(4.283)

m v = qBR (4.284)

Using Pythagorean theorem, solve for the hypotenuse of the triangle drawn in the diagram,which also happens to be our radius R. You should get

414


R2 = l2 + (R− s)2 (4.285)

expand the (R− s)2 term to get

R2 = l2 +R2 − 2sR+ s2 (4.286)

l2 − 2sR+ s2 = 0 (4.287)

Now, recall that s << l so we can let s = 0. Solve for R, and you get

R = l2/2s (4.288)

Substitute that back into our previous equation and you have

p = qBl2/2s (4.289)

Correct Answer

(D)

Alternate Solution

First, we would expect the momentum of the particle to increase as the arc of the particles motionbecomes more linear (i.e. as s → 0). From this, you can immediately get rid of (A) and (B)because both predict that momentum goes to 0 as the sagitta goes to 0. Next, we know that thefinal result should have units of N ·s or kg m

s . Recall that the unit for the magnetic field, the Tesla,

is kgA·s2 and units for electrical charge, the Coulomb, is A · s. From this, you can see only (D) and

(E) can have correct units, so get rid of (C). Finally, you will need to find some way to decidewhether the denominator of the solution is 2s or 8s. You can do this by recognizing, as we did inthe ”Recommended Solution”, that the force on the particle will be a centripetal force, m v2

R . Sincem and v aren’t lengths, we only care about the relationship between s and R. Using Pythagoreantheorem, R2 = (R−s)2 + l2. Expand everything and you will see that you get a 2s term. Eliminatechoice (E) based on this

Correct Answer

(D)

415


4.90 PGRE9677 #90


Pick your favorite letter between A, B, C, and D.

Alternate Solution

In the middle of the quiet exam room, with many other test takers concentrating on what couldvery well change their futures forever, stand up and walk to the nearest wall. Using a piece oftape, gum, other adhesive compound, attach your test booklet to the wall so that it is open andthe possible solutions are facing you. Locate a long sharp object, perhaps a small flag pole or awooden pointer and from 10 paces away, charge the test booklet while making horse noises (yourmotivation is a medieval knight in a jousting competition). Stab through your test booklet andcheck to see which of the 4 choices was punctured. If you managed to spear a bunch of white spaceor some non-related part of the problem, return to your starting point and charge the test again.Repeat this process until you’ve managed to skewer one of the 4 possible solutions. While thismethod is preferable to the Recommended Solution in both style and form, it can be more timeconsuming and so it should be your second method of attack. Note that the word “attac” in theprevious sentence is used both figuratively and literally.

416


4.91 PGRE9677 #91


There are a number of different descriptions of the second law of thermodynamics. One of those,the one relevant to this problem, is known as the Clausius statement, which says

It is impossible to move heat from a high temperature source to a low temperaturesource unless external work is done

This tells us that the oven will transfer heat to the sample as long as the temperature of the ovenis higher than the sample. However, if the temperature of the sample were able to get higher thanthe oven, then temperature flow would switch to moving from the sample back to the oven, whichhas become the lower temperature source. In other words, the temperature of the sample will neverbe able to exceed that of the oven because temperature flow will simply reverse any time it does.The sample can never achieve 900 K.

Correct Answer

(E)

417


4.92 PGRE9677 #92


First of all, we know that there will be a dependence on the mass of the particle so eliminate (A).Based on the statement about small oscillations, you can conclude that bx4 term is not particularlyinfluential so it likely doesn’t depend on b and we can eliminate (C). For (B), ETS is trying to playoff of your possibly memorized equation for angular frequency as

ω =2π

T(4.290)

But don’t be fooled because we shouldn’t expect to get a π out of this oscillator. Finally,between (D) and (E), recall that the angular frequency for a SHO is

ω =

√k

m(4.291)

Since angular frequency requires us to take the derivative of the potential and we are treating thepotential as a 2nd order polynomial, you can bet we are going to get a 2 down from differentiation.

Correct Answer

(D)

418


4.93 PGRE9677 #93


If we let k → 0 then V = 0. This tells us, at the very least, that the period of motion is dependenton k. (C) doesn’t account for this, get rid of it. If g → 0, again the period should be influenced.(A) and (B) don’t appropriately account for this so eliminate them. Finally, recall that the periodfor a simple harmonic oscillator is

T = 2π

√m

k(4.292)

But this only applies to one side of the equation so we want half of the term. One of thecomponents should be (1

2)2π√m/k = π

√m/k. This requirement matches the first term of (D).

Correct Answer

(D)

419


4.94 PGRE9677 #94


Recall from Statistical Mechanics the fact that with infinite energy available to a system, thetotal possible energy states will be populated equally in order to minimize the total number of“alternative” microstates the system could occupy. From this, we know that as T →∞ we wouldexpect a system with only two energy levels to each contain half of the total particles. Let T →∞to see which solution fulfills this requirement.

(A) Nε: No Temperature dependence so this isn’t a possible solution.

(B) 32NK(∞) =∞

(C) Nεe−ε/k(∞) =∞

(D) Nε(eε/kT+1)

= Nε2

(E) Nε(1+e−ε/kT )

= Nε2

Both (D) and (E) fulfill the requirement, so now you can figure that (D) is correct because wewould expect energy to approach 0 as T → 0 which isn’t true of (E).

Correct Answer

(D)

420


4.95 PGRE9677 #95

421



The key to solving this problem comes in the description, which tells us that we are concernedwith the, “region where it BECOMES superconducting”. When an object hits that temperaturethat enables it to exhibit superconductivity (a jump in resistance), the specific heat also changesinstantly, meaning we would expect an instantaneous jump at some temperature. Of the 5 possiblesolutions, only (E) exhibits this jump.

Correct Answer

(E)

422


4.96 PGRE9677 #96


One of the unique differences between the photon and the electron (well the photon and otherparticles) is its ability to maintain the same speed in all conceivable reference frames. From this,it should be clear that kinetic energy of a photon can never be 0 in any frame. However, it is alsotrue of a particle with mass that a frame can be chosen in which kinetic energy and momentumcan be 0.

Correct Answer

(A)

423


4.97 PGRE9677 #97


The probability current equation is

~J(x, t) =h

2mi

(ψ∗∂ψ

∂x− ∂ψ∗

∂xψ

)(4.293)

Take the derivative of both wave functions

ψ′ = keiωt [−αsin(kx) + βcos(kx)] (4.294)

ψ∗ ′ = −keiωt [−αsin(kx) + βcos(kx)] (4.295)

Plug everything in to get

h

2miγ (4.296)

where γ is

γ =[(e−iωt [α cos kx+ β sin kx]

) (keiωt [−α sin kx+ β cos kx]

)]−

[(−keiωt [−α sin kx+ β cos kx]

) (eiωt [α cos kx+ β sin kx]

)]

424


At this point, you could simplify the equation and get an exact solution. However, it is muchquicker to recognize that all terms with α2 and β2 cancel out and so you only get left with αβ andβα terms. From this, you can get that the solution must be (E).

Correct Answer

(B)

425


4.98 PGRE9677 #98


First off, eliminate solution that suggests the first energy level is ever 0. Having an energy of 0 isa problem because this would imply that we have a completely stationary particle which also thenmeans that we have precisely defined the particles position and momentum (oops!). Next, recallthe equation for the energy levels of the one-dimensional harmonic oscillator from QM is

Vn = hω

(n+

1

2

)=hω

2,3hω

2

5hω

2

7hω

2, · · · (4.297)

However, note that the problem tells us that there is an infinite potential wall at the center.This won’t effect all of our odd termed energies because they always have a node at the center buteven valued energies have a peak at this point and so they will be disrupted.

Taking only the odd valued energies,

Vn =3hω

2,7hω

2,11hω

2,15hω

2, · · · (4.298)

Correct Answer

(D)

426


4.99 PGRE9677 #99


A metastable state in an n-level laser is any state that acts as an “energy trap” or an intermediateenergy level. You can think of a metastable energy level as helping to support transitions from lowerto higher levels by providing a stepping stone and temporary energy location for the transition.Metastable states are, by definition, only intermediate states so our solution shouldn’t include n = 1or n = 3. The only energy level between n = 1 and n = 3 is n = 2.

Correct Answer

(B)

427


4.100 PGRE9677 #100


By definition, a Hermitian operator is an observable operator and an observable operator commutesso If I is true, II is also true. Without looking at anything else, we know that the only possiblecombination of choices will be (I and II), (I, II and III) or (III only). We could go through theexplanation for why I and II being true implies III is false and vice versa, but since all 3 beingtrue isn’t even an option, we won’t bother. The only two options are

(C) III only

(D) I and II only

Consider that for an operator to be adjoint, it must satisfy the condition

〈ax, y〉 = 〈x, a†y〉 (4.299)

a† isn’t technically a complex conjugate but it is analogous to one (at least it is in a Hilbertspace in which you treat operators as complex numbers, which is true of a). If we were to applya complex conjugate to a then we would get a sign change at which point it would be clear thata† 6= a. Since III is true, I and II can’t be true.

Correct Answer

(C)

428

Chapter 5

PGRE0177 Solutions

429


5.1 PGRE0177 #1


For a moving pendulum in which we can neglect gravity, the only two forces on the the pendulumbob is centripetal acceleration due to its rotation and acceleration due to gravity, g. At point c,the centripetal acceleration should be pointing upwards and acceleration due to gravity must bepointing downwards. Sum these two forces to see that the net force on the pendulum bob can onlypoint up or down, and we can eliminate (B) and (D). Next, check (A) to see that the net forces atpoint a and b should only have a horizontal acceleration in the left direction and, alternatively, ahorizontal acceleration in the right direction for points d and e. Since this is not true in (A), wecan eliminate it. Finally, recall that the centripetal acceleration for a pendulum will be minimized

430


at its peaks, so the net acceleration should be pointing downward more than left or right, whichmore closely matches (C) than (E).

Correct Answer

(C)

431


5.2 PGRE0177 #2


Start with the equation for friction,

f = µFN (5.1)

the normal force will be equal and opposite to the force due to the gravity, making Equation5.1

f = µmg (5.2)

but since the net force in the horizontal will be rotational, f in 5.2 is then

f = µmg (5.3)

mv2

r= µmg (5.4)

v2

r= µg (5.5)

r =v2

µg(5.6)

Now, convert velocity into rotational units, by

v = rω (5.7)

= r

(33.3

rev

min

)(2π

rad

rev

)(1 min

60 s

)(5.8)

≈ πrrad

s(5.9)

Finally, substitute Equation 5.9 into Equation 5.6 and solve to get

432


r =π2r2

µg(5.10)

=µg

π2(5.11)

≈ 3

π2(5.12)

≈ 1

3(5.13)

which is closest to (D).

Correct Answer

(D)

433


5.3 PGRE0177 #3

Figure 5.1: Electric potential at point P in relation to ring of radius R


Recall Kepler’s third law,”The square of the orbital period of a planet is directly proportional to the cube of the semi-

major axis of its orbit”or, equivalently

T 2 ∝ R3 (5.14)

T ∝ R3/2 (5.15)

which is option (D).

Correct Answer

(D)

434


5.4 PGRE0177 #4


Despite the change in energy, we can utilize the conservation of momentum to get the final andinitial momentums

pi = 2mvi (5.16)

pf = 3mvf (5.17)

by conservation, pi = pf and we get

2mvi = 3mvf (5.18)

vf =2

3vi (5.19)

Now, we need to find the initial and final energy of the system, as

Ei =1

2(2m)v2

i = mv2i (5.20)

Ef =1

2(3m)v2

f =3

2m

(2

3

)2

v2i =

2

3mv2

i (5.21)

At which it is clear that the difference between final and initial kinetic energy is 1/3.

Correct Answer

(C)

435


5.5 PGRE0177 #5


From the equipartion theorem, we know that the average energy for an ”ideal gas” with onlytranslational degrees of freedom is

〈H〉 =3

2kBT (5.22)

where the 3 in our 3/2 comes from the 3 degrees of freedom for translational motion. For aharmonic oscillator, we must add to this 2 rotational degrees of freedom and 1 degree of freedomfor its single dimension of oscillations. This brings the grand total to 6 degrees of freedom, makingthe average total energy

〈H〉 =6

2kBT = 3kBT (5.23)

Correct Answer

(D)

436


5.6 PGRE0177 #6


The quickest solution to this problem is simply to be familiar with the P-V curves for adiabaticand isothermal processes. You will likely have seen these in a thermodynamics lab course,

Figure 5.2: P-V curve comparison of an Isothermal and Adiabtic process

which is clearly (E).

Correct Answer

(E)

437


5.7 PGRE0177 #7


From the diagram, we know that the two magnets have similar poles next to one another so weshouldnt have magnetic field lines from one to another, like in (A) (C) and (D). Next, we shouldget some repelling between the field lines which doesnt show up in (E), so we must select (B).

Correct Answer

(B)

438


5.8 PGRE0177 #8


This specific problem is commonly used as one of the more simplistic, but by no means trivial,problems to introduce the method of image charges. Going through the entire proof during thisexam would be a much more lengthy process than we can finish in a reasonable amount of time sowe can really only solve this problem by knowing that a point charge will induce an exactly equaland opposite charge in the grounded plate, in fact choice (D).

Correct Answer

(D)

439


5.9 PGRE0177 #9


With 5 charges set symmetrically and equidistant about the center, we should have 5 equal electricfields canceling out fields pointing in the opposite direction. Because of symmetry, the sum of allelectric fields will equal 0.

Correct Answer

(A)

440


5.10 PGRE0177 #10


Given two capacitors, C1 = 3 microfarad and C2 = 6 microfarad and potential difference V =300 volt, the total energy can be found with

U =1

2CeqV

2 (5.24)

for capacitors in series, the equivalent capacitance is

Ceq =C1C2

C1 + C2(5.25)

=

((3 µF)(6 µF)

(3 µF) + (6 µF)

)(5.26)

=18 µF2

9 µF(5.27)

= 2 µF (5.28)

plug Equation 5.28 into Equation 5.24 and solve

U =1

2(2 µF)(300 volt)2 (5.29)

= 0.09 (5.30)

Correct Answer

(A)

441


5.11 PGRE0177 #11


We start with the thin lens equation

1

do+

1

di=

1

f(5.31)

plug in the given values for do and f1 to solve for di

1

40 cm+

1

di=

1

20 cm(5.32)

1 +40 cm

di= 2 (5.33)

di = 40 cm (5.34)

since the first image is to the right of the second lens by 10 cm, for our next calculation wemust use an object distance of d0 = −10 cm. Again, using the same equation as before, we get

1

−10 cm+

1

di=

1

10 cm(5.35)

1 +−10 cm

di= −1 (5.36)

di = 5 cm (5.37)

which predicts a final image 5.0 cm to the right of the final lens.

Correct Answer

(A)

442


5.12 PGRE0177 #12


Start with the mirror equation

1

d0+

1

di=

1

f(5.38)

1

di=

1

f− 1

d0(5.39)

Now, since we know from the image that the focal distance is larger than the object distance,we know that the RHS of Equation 5.39 must be negative and, therefore, the image distance isvirtual and behind the mirror.

Correct Answer

(E)

Alternate Solution

If we add an object at point O and do some ray tracing, we get Figure 5.3

443


Figure 5.3: Ray tracing diagram of a concave mirror with a longer focal length than object distance

Correct Answer

(E)

444


5.13 PGRE0177 #13


The Rayleigh Criterion can be used to give us the minimum resolution detail of a telescope,

θ = 1.22λ

d(5.40)

so we can solve for d in Equation 1 and plug in our known values to get

d = 1.22λ

θ(5.41)

= 1.22600 nm

3× 10−5 rad(5.42)

= 2.5 cm (5.43)

Correct Answer

(B)

445


5.14 PGRE0177 #14


From the description, we know that the detector is 100% efficient because exactly 50% of thesamples are being detected when pressed up exactly on one of its two sides. Since this is the case,we just need to find out how much fewer gamma rays will hit the detector at a distance of 1 maway. Since the rays will disperse spherically, we want to take the ratio of the samples as it passesthrough a circular surface area (AC = πr2) to that of the surface area of a sphere (AS = 4πr2) ata distance of 1 m = 100 cm. The radius of the circular surface area is 4, so we let AC = 16π cm2.Next, we can find the spherical surface area as

AS = 4πr2 (5.44)

= 4π(100 cm)2 (5.45)

= 40, 000 cm2 (5.46)

Finally, take the ratio of the two areas to get

ACAS

=16π cm2

40, 000 cm2(5.47)

= 4× 10−4 (5.48)

Correct Answer

(C)

446


5.15 PGRE0177 #15

447



Recall that precision is distinct from accuracy in that accuracy describes how close to the corrector true value a measurement is, while precision is a measurement of how closely grouped or howwell a result can be reproduced. In other words, a group of measurements can be entirely incorrectbut still be “precise” if they are all extremely similar to one another. Of the plots given, (A)demonstrates the closest grouping of data points.

Correct Answer

(A)

448


5.16 PGRE0177 #16


From a laboratory methods course, you probably did some problems on measurement uncertaintyand came up with the uncertainty equation

u =σ√N

(5.49)

where σ is the standard deviation and N is the number of samples. Since our data set is discreteand completed over time, we can find the standard deviation via the Poisson distribution

σ =√x (5.50)

where the average of our current data set appears to be, x = 2. This makes our standarddeviation, σ =

√2. Since we want to be within 1% of the average, we take 1% of 2 to get u = 0.02.

N =

√σ

2

u2(5.51)

=2

0.022(5.52)

= 5000 (5.53)

Correct Answer

(D)

449


5.17 PGRE0177 #17


Your first approach to this problem should be to ensure that each of the possible solutions containsall 15 electrons for phosophorous. Checking this, by summing the subscripts, you’ll find 15 foreach. This means that every single solution, except for the correct one, should have an incorrectprogression from our standard energy level diagram.

From the energy level diagram above, it should be clear that the progression is

1s22s22p63s23p3

Correct Answer

(B)

450


5.18 PGRE0177 #18


In the problem, 79.0 eV is given as the ionization energy for both electrons which also tells usthat the ionization for each electron individually must sum to this number. We also know that thefirst electron will be easier to pull from the atom because although it feels the same pull from thepositively charged nucleus, it also has a negatively charged electron trying to push it away. Thus,we know that the first electron will have a lower ionization energy than half of the total. Since24.6 eV is the only choice that meets this criteria, we choose (A).

Correct Answer

(A)

451


5.19 PGRE0177 #19


The sun is powered by nuclear fusion of hydrogen atoms into helium atoms. Because the atomicmass of hydrogen is approximately 1 and the atomic mass of helium is roughly 4, it must be true,by conservation of energy, that 4 hydrogen atoms combine to make 1 helium atom.

Correct Answer

(B)

452


5.20 PGRE0177 #20


Bremsstrahlung radiation refers to E&M radiation generated when a charged particle gets acceler-ated as the result of a collision with another charged particle, i.e choice (E). This is one of thoseproblems that you either know, or you don’t. Unless you speak German, in which case you couldbreak the Bremsstrahlung into its components, bremsen ”to brake” and Strahlung ”radiation”.

Correct Answer

(E)

453


5.21 PGRE0177 #21


The Lyman and Balmer series both refer to different types of transitions of an electron in a hydrogenatom from one radial quantum level (n) to another. The Lyman series is a description of all suchtransitions from n=r to n=1, such that r ≥ 2 and is an integer. The first Lyman transition(commonly called Lyman-α) is n=2 going to n=1, the second (Lyman-β) involves a transition ofn=3 to n=1, etc. The Balmer series, on the other hand, involves transitions from some n=s ton=2, such that s ≥ 3 and is an integer. The longest wavelength for both series involves the smallesttransition, i.e. n=2 going to n=1 for the Lyman Series and n=3 going to n=2 for the Balmer. TheRydberg formula can then be used to find the wavelength for each of the two transitions

1

λ= R

(1

n2f

− 1

n2i

)(5.54)

For this problem we won’t need to compute anything, just compare λL and λB. Doing this forthe shortest Lyman transition gives

1

λL= R

(1

12− 1

22

)(5.55)

1

λL=

3

4R (5.56)

λL = 4/(3R) (5.57)

and for the Balmer transition

1

λB= R

(1

22− 1

32

)(5.58)

1

λB=

5

36R (5.59)

λB = 36/(5R) (5.60)

454


making the ratio

λL/λB =4/(3R)

36/(5R)= 5/27 (5.61)

Correct Answer

(B)

455


5.22 PGRE0177 #22


Recall our good friend Galileo, who demonstrated that an object in free fall in a vacuum will fall atthe same acceleration regardless of its mass. In the problem given, the moon is our object, space isour vacuum and we conclude that the mass of the moon is unknowable with the given information.

Correct Answer

(B)

456


5.23 PGRE0177 #23


In this problem, we have three vectors to consider. The first is the centripetal acceleration vectorgenerated by the rotation of the particle. The second two are both tangential to the circle but oneis a velocity vector, v = 10 m/s, and the other is a tangential acceleration, aT = 10 m/s2. First,get the net acceleration by adding aT and aC

Figure 5.4: Tangential and centripetal acceleration of a particle constrained to a circle

Since aT and aC are both 10 m/s2, the angle between them must be 45. This tells us that thenet acceleration is always 45 from any tangential vectors and since the velocity vector is also atangential vector, the angle between v and anet is 45.

Correct Answer

(C)

457


5.24 PGRE0177 #24


Start by realizing that without air resistance, the horizontal velocity will always be constant andpositive, so vx vs. t must be plot II and we can eliminate (A) and (E). Next, in the y-direction weknow that velocity must not be constant as it reaches some peak and its velocity becomes zero,before then changing direction and increasing its velocity in a negative direction. This descriptionperfectly describes plot III and so we choose (C).

Correct Answer

(C)

458


5.25 PGRE0177 #25


Start with the moment of inertia for a disk, I = 1/2mr2, which is given in the front of your testbooklet. Next, we can find the moment of inertia for the rest of the pennies using the parallel axistheorem,

I = ICOM +ml2 (5.62)

again, the moment of inertia for a single penny is I = 1/2mr2 but they have a radius from thecenter of mass of l = 2r, so we get

Iop =1

2mr2 +ml2 (5.63)

=1

2mr2 +m(2r)2 (5.64)

=9

2mr2 (5.65)

however, since we have 6 pennies, the sum of all 6 is just

Iop,6 =54

2mr2 (5.66)

then add the moment of inertia of all 6 pennies with the inner penny to get,

459


Iip + Iop,6 =54

2mr2 +

1

2mr2 =

55

2mr2 (5.67)

Correct Answer

(E)

460


5.26 PGRE0177 #26


Before the rod has moved, it only has potential energy of

U =mgL

2(5.68)

where we use L/2 because that is the location of the center of mass. Next, recall the equationfor rotational kinetic energy, which all of the energy is converted to at the bottom of the rods fall,

mgL

2=

1

2Iω2 (5.69)

The moment of inertia for a rod rotating at its tip is

I =1

3mL2 (5.70)

substitute Equation 5.70 into Equation 5.69 and substitute ω = v/r to get

mgL

2=

Iω2

2(5.71)

461


=mL2v2

6L2(5.72)

gL =v2

3(5.73)

v =√

3gL (5.74)

Correct Answer

(C)

462


5.27 PGRE0177 #27


The Hermitian operator is defined as

〈A〉 =

∫ψ∗(r)Aψ(r)dr (5.75)

where A is the expectation value and is an observable. Therefore, A is, like all other observables,is real valued and we choose (A).

Correct Answer

(A)

463


5.28 PGRE0177 #28


Recall the condition for orthogonality,

〈x|x〉 = 1 (5.76)

from this, we can solve for x by taking

〈ψ1|ψ2〉 = 0 (5.77)

(5 · 1) + (−3 · −5) + (2 · x) = 0 (5.78)

20 + 2x = 0 (5.79)

x = −10 (5.80)

Correct Answer

(E)

464


5.29 PGRE0177 #29


Recall that the expectation value of any state is,

〈A〉ψ =∑j

aj |〈ψ|φj〉|2 (5.81)

where aj is the eigen value. Thus, we square each term, multiply it by its eigenvalue and sumeverything up to get

〈O〉 = −1

6+

1

2+

2

3(5.82)

= 1 (5.83)

Correct Answer

(C)

465


5.30 PGRE0177 #30


Start by realizing that the wave function should decrease as the radius goes off to infinity, whichis not true of option II and so we eliminate (B), (C) and (E). Next, as the radius goes to zero, thewave function should go to A, which is only true of I and we can choose (A).

Note that even if you don’t recognize that the wave function should go to A when r → 0, youcan at least be sure that it shouldn’t blow up to infinity like it does in option III.

Correct Answer

(A)

466


5.31 PGRE0177 #31


One exceedingly useful piece of information to memorize is that the ground state energy of positro-nium is half that of the hydrogen atom,

E0,pos =E0,H

2=−13.6 eV

2= −6.8 eV (5.84)

Then, using the Bohr model energy equation, substitute −13.6 eV with −6.8 eV and solve

E = −6.8 eV

(1

n21

− 1

n22

)(5.85)

= −6.8 eV

(1− 1

9

)(5.86)

= 6.0 eV (5.87)

Correct Answer

(A)

467


5.32 PGRE0177 #32


Start by equating the total relativistic energy, E = γm0c2, to the rest energy, E = m0c

2, in theproportions given

Enet = γm0c2 = 2m0c

2 (5.88)

which clearly tells us that the lorentz factor is γ = 2. Next, solve for velocity in the Lorentzfactor

2 =1√

1− v2/c2(5.89)

2√

1− v2/c2 = 1 (5.90)

v2/c2 =3

4(5.91)

v =

√3

2c (5.92)

Finally, using our relativistic momentum and Lorentz factor, we can solve for p

p = γmov (5.93)

= 2m0

√3

2c (5.94)

=√

3m0c (5.95)

Correct Answer

(D)

468


5.33 PGRE0177 #33


First, immediately eliminate (E) because a pion can’t achieve light speed. Next, notice that if youtry to apply the classical concept of velocity (i.e. ∆x/∆T ), you get a ridiculously large number sorelativity must be in full effect and (A) isn’t nearly fast enough. Now, start with our equation forthe space time interval

∆s2 = ∆r2 − c2∆t2 (5.96)

in the pion’s frame, its change in position is 0 so we have

∆s2π = −c2∆t2 (5.97)

= −(3× 10−8)2(1× 108)2 (5.98)

= −9 m/s (5.99)

Now, in the lab frame, we get the space time interval,

∆s2lab = (30 m)2 − c2∆t2lab (5.100)

∆t2labc2 = 909 (5.101)

∆lab =

√909√c2

(5.102)

=√

101× 10−8 (5.103)

Finally, using our standard velocity equation, plug in our values to get

v =∆Xlab

∆tlab(5.104)

=30 m√

101× 10−8(5.105)

≈ 2.98× 108 (5.106)

Correct Answer

(D)

469


5.34 PGRE0177 #34


From the the displacement 4-vector, we get 3 general types of space-time intervals: time-like,space-like and light-like. These are typically drawn out on a space-time diagram as such

Figure 5.5: Relativistic space time diagram for time-like, space-like and light-like reference frames

each interval is characterized by the following criteria

time-like: if |∆x/∆t| < c, two events occur at the same location

space-like: if |∆x/∆t| > c, two events occur at the same time

light-like: if |∆x/∆t| = 0, two events are connected by a signal that moves at the speed of light.

Of the choices, the problem is describing a space-like interval which is (C).

Correct Answer

(C)

470


5.35 PGRE0177 #35


According to our models for simple blackbody radiation, a blackbody’s power is proportional tothe 4th power of its temperature

P = κT 4 (5.107)

From this, the increase in power from tripling the temperature is

P = κ(3T )4 = 81κT 4 (5.108)

which is (E).

Correct Answer

(E)

471


5.36 PGRE0177 #36


An adiabatic expansion is not the same thing as an isothermal expansion and an isothermal ex-pansion is the only type which maintains a constant temperature. Additionally, from the ideal gasequation we know that a change in volume should be accompanied by a change in temperature

P∆V = nR∆t (5.109)

Correct Answer

(E)

472


5.37 PGRE0177 #37


The quickest method to determine the thermodynamic work is to estimate the area under the P-Vcurve. The curve is roughly triangular shaped with base of 2 and height of 300. Calculate the areaunder the triangle as

1

2PV =

1

2(2)(300) ≈ 300 (5.110)

this eliminates all but (B) and (D). We can then determine whether the work is negative orpositive by the direction of the process. Recall that a clockwise process represents a heat engineand the work is positive. Alternatively, if the process is counter-clockwise, it’s a heat pump andthe work is negative. This process is counter-clockwise so we choose (D).

Correct Answer

(D)

473


5.38 PGRE0177 #38


Amplitude is maximized in an RLC circuit when we’ve reached the resonant frequency. This canbe calculated by

ω =1√LC

(5.111)

plug in your values to get

ω =1√LC

(5.112)

C =1

Lω2(5.113)

=1

(25 millihenries)(1× 103 rad)2(5.114)

=1

2.5× 107 millihenries-rad2 (5.115)

= 40 µF (5.116)

Correct Answer

(D)

474


5.39 PGRE0177 #39


At low frequencies and low energies, a capacitor and a resistor look like an impassible gap so thesewill act as a high pass filter (i.e. only high frequencies can pass) but an inductor will appear likejust another bit of wire. At high frequencies and high energies, capacitors and resistors can bepassed quite easily but passing through an inductor will generate a strong magnetic field and willimpede the flow, making this a low pass filter. Looking through the choices,

I. The inductor impedes high frequencies before it reaches the terminals so this can’t be a highpass filter

II. The resistor allows the high energy to pass and the low frequencies will drop via the resistor.We know this is a high pass filter.

III. The capacitor will allow high energies to pass through and low frequency voltage will drop.We know this is a high pass filter.

IV. The high energy passes the resistor and passes with the capacitor without a drop in voltage.Without a drop in voltage, this can’t be a high pass filter.

Correct Answer

(D)

475


5.40 PGRE0177 #40


After the switch is closed, we will start off with the maximum voltage and continually decrease asthe resistor and inductor eat away at the initial voltage. From this, we can eliminate all optionsbut (D) and (E). Next, compare the amount of time between the two remaining options to seethat only half of the voltage dropping after 200 seconds is a vastly unrealistic voltage drop for anycircuit, so we choose (D).

Correct Answer

(D)

476


5.41 PGRE0177 #41


The existence of magnetic charge would be, in essence, the same thing as saying that magneticmonopoles exist. If this were the case, then II would not have to be changed to allow magnetic fieldlines to diverge completely and we can eliminate (A), (C) and (D). Next, consider that a magneticmonopole will also allow magnetic field lines to exist without curling back to its opposite pole, soIII must change.

Correct Answer

(E)

477


5.42 PGRE0177 #42


From Lenz’s law, we know that anytime a current is generated by a change in the magnetic or electricfield, the generated emf will be such that it opposes this change in direction and magnitude. As thecenter ring moves towards ring A, this increases the field present and ring A will oppose this changewith a current of opposite current, i.e. clockwise. Ring B, on the other hand, will have its fielddecreased so a current will be induced to oppose this decrease, which requires an anti-clockwisemotion.

Correct Answer

(C)

478


5.43 PGRE0177 #43


Start with the commutator relation

[AB,C] = A[B,C] + [A,C]B (5.117)

and apply it to the given commutator

[LxLy, Lz] = Lx[Ly, Lz] + [Lx, Lz]Ly (5.118)

and replace the bracketed portions on the RHS with the commutation relations given in theproblem (while recalling that the have the commutation relation in the opposite order from left toright simply changes the sign), to get

[LxLy, Lz] = Lx[Ly, Lz] + [Lx, Lz]Ly (5.119)

= Lx(ihLx) + (−ihLy)Ly (5.120)

= ih(L2x − L2

y) (5.121)

Correct Answer

(D)

479


5.44 PGRE0177 #44


The problem gives us the energy equation as

En =n2π2h2

2mL2(5.122)

this tells us that the only difference between, say E1 and E2 will be the change in Energy fromthe squared quantum number, n = 1, 2, 3, . . .. So, we know that all allowed energies will be somesquared integer multiple of E1. Of those given, only (D) has a coefficient with a perfect squarevalue (i.e. 32 = 9).

Correct Answer

(D)

480


5.45 PGRE0177 #45


First, recall that the expectation value for any operator is the sum of its terms, each one individuallysquared and multiplied by it’s respective eigenvalue. First, find the ”ket” values for |1〉, |2〉 and 3〉,

|1〉 =3

2hω (5.123)

|2〉 =5

2hω (5.124)

|3〉 =7

2hω (5.125)

then multiply these by their respective coefficients, having been squared, to get

|ψ〉 =1√14|1〉 − 2√

14|2〉+

3√14|3〉 (5.126)

=

(1

14

)(3

2hω

)+

(4

14

)(5

2hω

)+

(9

14

)(7

2hω

)(5.127)

=86

28hω (5.128)

=43

14hω (5.129)

Correct Answer

(B)

481


5.46 PGRE0177 #46


Start with the de Broglie wavelength equation

λ =h

p(5.130)

and then recall the Schrdinger 1-D energy equation for a particle in a potential

E =p2

2m+ V (x) (5.131)

without doing much work, you can see that combining Equation 5.130 with Equation 5.131 willforce you to have an inverse square root to substitute the p2 in Equation 2 into the p in Equation1. This only matches option (E) so it must be the correct choice.

Correct Answer

(E)

482


5.47 PGRE0177 #47


First, eliminate choices (D) and (E) as both predict a negative change in entropy, which should notbe true of a gas expanding outward to occupy a new open space. Next, recall from thermodynamicsthat entropy is defined as the natural log of the total number of states of a system

S = kb ln(Ω) (5.132)

where Ω is the number of states. For a volume divided into two parts, for n particles the numberof states will be

Ω = 2n (5.133)

Plugging Equation 5.133 into Equation 5.132 and applying the rules of logarithms gives

S = kb ln(Ω) (5.134)

= kb ln (2n) (5.135)

= nkb ln(2) (5.136)

which is given by choice (B).

Correct Answer

(B)

483


5.48 PGRE0177 #48


You can initially eliminate all solutions that are less than 1, as we would expect the smaller andlighter N2 molecules to move faster than the larger and heavier O2 molecules. This step eliminates(A), (B), and (E). Between (C) and (D), you can either take a guess, or recall Graham’s law ofeffusion, which states

Rate1

Rate2=

√M2

M1(5.137)

at which point, (C) becomes the obvious choice.

Correct Answer

(C)

484


5.49 PGRE0177 #49


A Maxwell-Boltzmann system has the canonical partition function

Z =∑s

g e−Es/kbT (5.138)

Where g is the degeneracy. Since we have two energies, we will need at least two terms in ourenergy equation so we can eliminate (A), (B) and (C). Next, plug everything in to see our coefficientof 2 appear, as in (E).

Z =∑s

g e−Es/kbT (5.139)

= 2e−ε/kT + 2e−2ε/kT (5.140)

= 2[e−ε/kT + e−2ε/kT

](5.141)

Correct Answer

(E)

485


5.50 PGRE0177 #50


Recall our equation relating frequency to wavelength,

ν =v

λ(5.142)

if our velocity is 3

vf = 0.97vo (5.143)

so the initial velocity is

νo =ν0

λ(5.144)

440 hz =v0

λ(5.145)

v0 = (440 hz)(λ) (5.146)

then, compare this to the final velocity

νf =vfλ

(5.147)

=(0.97v0)

λ(5.148)

=(0.97)(440 hz)

λ(5.149)

=427 hz

λ(5.150)

Correct Answer

(B)

486


Note: if you have any experience with wind instruments you know that a cold instrument alwaysstarts off a bit flat and the pitch of your instrument continually increases as it warms up. Fromthis little fact, you should at least be able to determine that (D) and (E) can’t be correct.

487


5.51 PGRE0177 #51


Consider a beam of unpolarized light headed towards you and you have three ideal polarizers inhand.

In the figure above, grey vectors indicate individual directions of oscillation, black vectorsindicate the net polarization, greyed out areas indicate areas of absorbed polarization and whiteareas indicate unaffected polarization. If we wanted to completely eliminate all light using twoof the polarizers, we could place them in series with a rotation of exactly π/2 between them, forexample using the horizontal and vertical polarization. In this arrangement, anything that survivedthrough the horizontal polarizer would be caught by the vertical polarizer and no light would betransmitted on the other side. However, imagine we were to place another polarizer between thehorizontal and vertical polarizers such that this third polarizer is rotated π/4 or 45 with respect tothe other two. As the light first passes through the horizontal polarizer, only half of the photonsthat hit the polarizer will pass through. These photons will continue to the 45 polarizer where,

488


again, half of the remaining photons get absorbed and the other half pass through. After thephotons pass through the 45 polarizer, the remaining photons will spread out from -22.5 to +67.5.This occurs because linearly polarized light will always completely fill a full 90 of angular spread,which I didn’t mention previously because the previous instances came out with a 90 spread.Finally, the remaining photons will pass through the vertical polarizer giving a final total of 1/8the original number of photons

Correct Answer

(B)

489


5.52 PGRE0177 #52


Recall the three generic types of cubic crystals: Simple Cubic, Body Centered Cubic and FaceCentered Cubic.

Sum up the total area in each cubic to find

Simple Cubic 1 Atom

Body Centered Cubic 2 Atoms

Face Centered Cubic 4 Atoms

490


(a) Simple Cubic (b) Body Centered Cubic (c) Face Centered Cubic

Figure 5.6: 3 generic forms of cubic crystals

Thus the primitive unit cell (simple cubic) contains half as much area, a3/2

Correct Answer

(C)

491


5.53 PGRE0177 #53


Recall that semi-conductors, unlike regular conductors, conduct well at high temperatures andeffectively not at all at extremely low temperatures. From this, we know that the resistivity at0 should be very high and it should decrease as temperature increases. Only (B) matches thisdescription.

Correct Answer

(B)

492


5.54 PGRE0177 #54


Recall that impulse is

I =

∫ t2

t1F dt (5.151)

So we just want to find the area under this plot. If you don’t recall this equation, take notethat we are given a y-axis in newtons and an x-axis in seconds. The only way to get these twounits to end with kg ·m/s is to take the area. Using the equation for the area under a triangle,you can quickly get I = 2 kg ·m/s

Correct Answer

(C)

493


5.55 PGRE0177 #55


We can quickly eliminate choices (B) and (D) based on the fact that they both suggest an asym-metrical set of values for an exceptionally symmetric problem. We also know that (A) can’t beright because it would violate conservation of momentum for a particle of mass m and velocityv0 to generate two particles with mass m and velocity v0. Finally, since we know that horizontalmomentum must be conserved and each piece gets half the horizontal momentum of the initial,then the addition of a vertical component of velocity requires, by the Pythagorean theorem, thatthe net velocity be greater than just the horizontal component. In other words

v2net = v2

x + v2y (5.152)

Correct Answer

(E)

494


5.56 PGRE0177 #56


Start by finding the difference between the air density and the helium density to get the averagedensity the balloon will experience

ρavg = ρair − ρHe = 1.29 kg/m3 − 0.18 kg/m3 = 1.11 kg/m3 (5.153)

Now, note that the only way to get units of m3 from kg and kg/m3 is to do the following

V =m

ρavg(5.154)

=300 kg

1.11 kg/m3 (5.155)

= 270 m3 (5.156)

Correct Answer

(D)

495


5.57 PGRE0177 #57


Consider that if the density of the water were to increase, so to would the force and we don’t seethis feature in (D) and (E). Next, eliminate (C) because the acceleration due to gravity,g, shouldnot be a part of these calculations. Finally, compare units on (A) and (B) to get

(A) ρv2A = (kg/m3)(m2/s2)(m2) = kg m/s2 = N

(B) ρvA/2 = (kg/m3)(m/s)(m2) = kg/s

and we choose (A).

Correct Answer

(A)

496


5.58 PGRE0177 #58


The problem tells us that the electric field is pushing the proton in the +x-direction and, fromFleming’s left hand rule, we can quickly deduce that the magnetic field is pushing the proton inthe -x-direction. Since we are told the proton isn’t deflected with a potential difference of V , weknow that the forces must have balanced out exactly. On the second pass, however, the potentialdifference is doubled and, from the Lorentz force F = qvB, this will increase the force from themagnetic field but not for the electric field. In the second pass, the magnetic field force will begreater and will exceed the electric field force and the overall direction of deflection will be in the-x-direction.

Correct Answer

(B)

497


5.59 PGRE0177 #59


The equation for a simple harmonic oscillator is

mx+ kx = 0 (5.157)

which we get when L = m, C = 1/k and Q = x.

Correct Answer

(B)

Alternate Solution

Start by eliminating any choices that suggest that Q = m, i.e. (C) and (E), as a change in massover time is not a necessary condition for a simple harmonic oscillator. Next, since Q has to bex, from our previous elimination, we can reasonably conclude that 1/C should be 1/k so that thesecond term gives us Hooke’s law, i.e. kx, and you can eliminate (A) and (D).

Correct Answer

(B)

498


5.60 PGRE0177 #60


Start by recognizing that the flux through the Gaussian surface should have some dependence onx, because if x blows up to infinity, so to should the flux. With (A) and (B) eliminated, nexteliminate (C) because the area under consideration should be the area cut out by the entire circleminus the area cut out by x, not the area of the difference of the two. Finally, note that if x = 0,the area under consideration should just be that of a circle, which is πR2 as opposed to 2πR2.

Correct Answer

(D)

499


5.61 PGRE0177 #61


Considering the electric field first, when an electric field orthogonal to a conductor interacts withthat conductor, the field (in a sense) disperses over it and the E field just to the left and to theright will be 0, from which we can eliminate (B), (D) and (E). Next, as the ~E interacts with theconductor, charges are moved around and this induces a magnetic field, so ~B 6= 0 and we choose(C).

Correct Answer

(C)

500


5.62 PGRE0177 #62


Start with our cyclotron frequency equation

f =Bq

2πm(5.158)

and then re-write Equation 5.158 to solve for m

m =Bq

2πf(5.159)

Now, plugging everything in gives us

m =(π/4 teslas)(2e−)

2π(1600 hz)(5.160)

=e−

(4)(1600 hz)(5.161)

=1.6× 10−19 kg/s

(4)(1600 hz)(5.162)

=1× 10−22 kg

4(5.163)

= 2.5 × 10−23 kg (5.164)

Correct Answer

(A)

501


5.63 PGRE0177 #63


Ideally, we would be able to recall Wien’s displacement law

λ =b

T(5.165)

in which b is the Wien displacement law constant. Alternatively, you could quickly get exactlythe same relationship with a bit of dimensional analysis by noticing that the only way to get afinal unit of Kelvins would be to take the quotient of Wien’s displacement law constant with themaximum wavelength. Either way you do it, convert the peak wavelength of 2 µm to 2× 10−6 mand solve

T =2.9× 10−3 m ·K

2× 10−6 m(5.166)

= 1, 500 K (5.167)

Correct Answer

(D)

502


5.64 PGRE0177 #64


Options (C) and (D) are very much true and are frequently mentioned in a lot of pop physicsbooks, television and other forms of media, so these two should be easily dismissed. (E) is also truebecause band spectrum, i.e. spectral lines so close together as to form a band, can’t appear whena single atom produces a single spectral line. Lastly, (B) is true because absorption spectroscopyand emission spectroscopy are essentially different ways of measuring the same thing, specificallywavelengths. Only (A) is left so that is our answer.

Correct Answer

(A)

503


5.65 PGRE0177 #65


Start with the Taylor Series expansion of ex to get an approximation

ex ≈ 1 + x (5.168)

ehν/kT = 1 + hν/kT (5.169)

Plug this approximation into the denominator of Einstein’s formula

C = 3kNA

(hν

kT

)2 ehν/kT

(1 + hν/kT − 1)2(5.170)

= 3kNA

(hν

kT

)2 ehν/kT

(hν/kT )2(5.171)

= 3kNAehν/kT (5.172)

Finally, as temperature blows up to infinity, ehν/kT = 1 and so we are left with

C = 3kNA (5.173)

Correct Answer

(D)

504


5.66 PGRE0177 #66


Ignore the fact that this problem mentions anything about physics and treat it like a standard rateproblem. If Jimmy can eat a whole apple pie in 24 minutes and Susie can eat a whole apple pie in36 minutes, how long will it take to eat that same pie if both Jimmy and Susie are sharing a singlepie?

24 ∗ 36

24 + 36= 14.4 minutes (5.174)

Correct Answer

(D)

505


5.67 PGRE0177 #67


First, eliminate (A) because we know that Uranium-238 can fission spontaneously. Next, eliminate(B) because, aside from it being a very vague claim that is not typical of a GRE answer, it is astrange conclusion to try and draw from the splitting of a Uranium atom that barely loses anymass/energy in the process. We can then eliminate (C) simply because it isn’t true that a nucleiwith roughly half the particles of Uranium-238 would be equally massive. Finally, between (D) and(E), recall that binding energy generally increases as we move towards the most stable element,iron, and decreases as we move away. A change from Uranium-238 to a nuclei with A=120 wouldbe a movement closer to iron, so we would likely so a binding energy of 8.5 MeV/nucleon ratherthan 6.7 MeV/nucleon.

Correct Answer

(E)

506


5.68 PGRE0177 #68


Since an alpha particle is a helium atom, specifically He2+ with two protons and two neutrons,if lithium lost an alpha particle it would have lost one to many protons to become beryllium and(A) is eliminated. Losing an electron, positron or neutron wouldn’t alter the actual type of atom,so emitting these wouldn’t change anything and we can eliminate (B), (C) and (D). This leaves us(E).

Correct Answer

(E)

507


5.69 PGRE0177 #69


From the description, we know that the light first interacts with the oil surface and the phase ofthe blue light is shifted 180. Next, the light interacts with the glass surface and, because of theprevious phase shift, we use the thin film interference equation

2nd = mλ (5.175)

where m = 1, 2, 3, . . .. The thinnest oil film will be when m = 1, so we solve for d in Equation5.175 and plug in our values

d =mλ

2n(5.176)

=480 nm

2(1.2)(5.177)

= 200 nm (5.178)

Correct Answer

(B)

508


5.70 PGRE0177 #70


Starting with the double slit equation

ω sin(θ) = mλ (5.179)

re-write Equation 5.179 to solve for the angle between fringes, θ, to get

sin(θ) =mλ

ω(5.180)

from Equation 5.180, we can easily see that doubling the frequency, ω, will result in a separationangle half as large as the initial 1.0 millimeter and we pick (B).

Correct Answer

(B)

509


5.71 PGRE0177 #71


We can eliminate (E) right away because the proposed velocity is faster than light. Next, eliminate(A) and (B) because we know, from the Doppler effect, that an increase in wavelength correspondsto an object moving away from the observer. Lastly, we can choose between (C) and (D) by utilizingthe doppler redshift equation

λ

λ0=

√1 + v/c

1− v/c(5.181)

607.5 nm

121.5 nm=

√1 + v/c

1− v/c(5.182)

25 =1 + v/c

1− v/c(5.183)

24− 26(v/c) = 0 (5.184)v

c= 12/13 (5.185)

v = (12/13)c (5.186)

v = 2.8× 108 m/s (5.187)

Correct Answer

(D)

510


5.72 PGRE0177 #72


Right as the string breaks the block will be accelerating due to gravity and it will also be acceleratingfrom the spring pulling down on the top block from the bottom block. Based on this, we knowthat the net acceleration of the top block must be larger than just that from acceleration, and weeliminate (A), (B), and (C). Lastly, summing the vertical forces on the top block, we get

−ma = −mg − k∆x (5.188)

2mg = k∆x (5.189)

which tells us that the acceleration should be 2g.

Correct Answer

(E)

511


5.73 PGRE0177 #73


In order for block B to not fall, the gravitational force downward must be in equilibrium with thefrictional force upwards. Adding up the forces and setting the net force to 0, we get

Fnet = f − FG (5.190)

0 = µFN −mg (5.191)

FN =mg

µ(5.192)

=(20 kg)(10 m/s2)

0.50(5.193)

= 400 N (5.194)

Correct Answer

(D)

512


5.74 PGRE0177 #74


The Lagrangian equation of motion can be calculated with

d

dt

(∂L

∂q

)− ∂L

∂q= 0 (5.195)

perform both derivatives to get

d

dt

(∂L

∂q

)= 2aq (5.196)

∂L

∂q= 4bq3 (5.197)

plugging our results from Equation 5.197 into Equation 5.195 gives

d

dt

(∂L

∂q

)− ∂L

∂q= 0 (5.198)

2aq − 4bq3 = 0 (5.199)

2aq = 4bq3 (5.200)

q =2b

aq3 (5.201)

Correct Answer

(D)

513


5.75 PGRE0177 #75


We could approach this problem theoretically, but it will be quicker to just give the matrix a vectorand see where it goes. to simplify things, use the vector

~v = 〈1, 0, 0〉 (5.202)

multiplying everything out, we get a transformed vector

~v′ = 〈1/2,√

3/2, 0〉 (5.203)

which represents a 60 rotation clockwise about the z-axis.

Correct Answer

(E)

514


5.76 PGRE0177 #76


The first of the five choices we can eliminate is (D), because electrons travel quite slowly in metals,making them exceptionally non-relativistic. We can also quickly eliminate (B) because there is nocondition specified in the problem that would take the metal out of equilibrium. Next, eliminate(A) because electrons in a metal move freely but are generally more constrained in their degrees offreedom than free atoms. Finally, between (C) and (E), choose (C) because interactions betweenphonons and electrons are more technical than we would likely ever see on the GRE.

Correct Answer

(C)

515


5.77 PGRE0177 #77


By Maxwell-Boltzmann statistics, we get the equation

Ni = Ngie−εi/kT

Z(5.204)

kT and the energy are both given so plug them both into Equation 5.204 and solve to get

Ni = Ngie−εi/kT

Z(5.205)

= Ngie

(−0.1 eV)/(0.025 eV)

Z(5.206)

= Ngie−4

Z(5.207)

which results in our ratio of e−4.

Correct Answer

(E)

516


5.78 PGRE0177 #78


Since we are talking about changing the arrangement of neutrinos and anti-neutrinos, both of whichare leptons, we would be wise to check lepton number first. Doing so, we find the typical leptonicdecay as

µ− → e− + νe + νeL : 1 = 1 − 1 + 1Le : 0 = 1 − 1 +0Lµ : 1 = 0 + 0 + 1

removing the anti-neutrino from our typical lepton decays results in

µ− → e− + νeL : 1 = 1 + 1Le : 0 = 1 + 0Lµ : 1 = 0 + 1

so lepton number is not conserved.

Correct Answer

(E)

517


5.79 PGRE0177 #79


Start with the relativistic energy equation

E2 − (pc2) = (mc2)2 (5.208)

we are given E = 10 GeV and p = 8 GeV/c so we plug these in and solve for m

(10 GeV)2 − (8 GeV/c)2c2 = m2c4 (5.209)

(100 GeV2)− (64 GeV2) = m2c4 (5.210)

36 GeV2 = m2c4 (5.211)

m2 = 36 GeV2/c4 (5.212)

m = 6 Gev/c2 (5.213)

Correct Answer

(D)

518


5.80 PGRE0177 #80


Start by eliminating option (E) because light will move slower than c when interacting with somemedium, relativistic or not. Next, consider that when the tube and water are moving at 1/2c andmoving in the same direction as the light, we would expect the light to pass through more quicklythan if the tube was stationary. Recalling our equation for light speed in a medium,

v =c

n=

3

4c (5.214)

so our answer should be larger than this and we eliminate (A) and (B). Finally, we can decidebetween (C) and (D) by considering the sum of our two relativistic velocities using

v′ =u+ v

1 + vuc2

(5.215)

=1/2c+ 3/4c

1 + (1/2c)(3/4c)c2

(5.216)

=5/4

11/8c (5.217)

=10

11c (5.218)

Correct Answer

(D)

519


5.81 PGRE0177 #81


Start with the equations for L2 and Lz

L2Ylm(θ, φ) = l(l + 1)h2Ylm(θ, φ) (5.219)

LzYlm(θ, φ) = mhYlm(θ, φ) (5.220)

applying both eigenvalues to their respective operators, we get

l(l + 1)h2 = 6h2 (5.221)

l = 2 (5.222)

and

mh = −h (5.223)

m = −1 (5.224)

at which point we plug in our values for m and l into the original momentum eigenfunction andget

Y ml (θ, φ) = Y −1

2 (θ, φ) (5.225)

Correct Answer

(B)

520


5.82 PGRE0177 #82


For a two electron system, there are three triplet state configurations and one singlet state config-uration. The single singlet configuration occurs when the total spin, S = S1 + S2 equals 0, whichis only true when S1 = −S2 and can be written as

1√2

(|α〉1|β〉2 − |α〉2|β〉1) (5.226)

we can then eliminate all choices with II as an option and, more to the point, all other config-urations must be one of the three possible triplet configurations.

Correct Answer

(D)

521


5.83 PGRE0177 #83


Depending on how well you recall your linear algebra, you might quickly recognize that the Paulispin matrix given in this problem is a relatively typical transformation/rotation matrix. To demon-strate this, you can pick any arbitrary up and down spin vectors (I’ll be using 1,2,3 & 4 just for thepurposes of distinguishing separate elements of the vectors). Start with spin up and down vectors

| ↑ 〉 = (1, 2) (5.227)

| ↓ 〉 = (3, 4) (5.228)

Now multiply each of those to the transformation matrix given in the problem to get thetransformed vectors

| ↑ 〉σx = (2, 1) (5.229)

| ↓ 〉σx = (4, 3) (5.230)

522


which shows that this Pauli matrix performs an orthogonal transformation on the vectors. Ofthe potential solutions, only (C) gives us the necessary result that swapping the order of the vectorswill also swap the sign.

Correct Answer

(C)

523


5.84 PGRE0177 #84


For a single electron transition, utilize our selection rules for the orbital and total angular quantumnumbers

∆l = ±1 (5.231)

∆j = 0,±1 (5.232)

we can eliminate transition A because ∆l = 0. For B and C, however, we have transitions

B: ∆l = −1 and ∆j = −1

C: ∆l = −1 and ∆j = 0

both of which are allowed and so we choose (D).

Correct Answer

(D)

524


5.85 PGRE0177 #85


The resistance of the wire is related to its length and area by

R =ρl

a(5.233)

this gives us two resistances

R1 =ρ2L

A(5.234)

R2 =ρL

2A(5.235)

which gives us a ratio for the resistances of R1/R2 = 4/1. This ratio also represents, fromV = IR, the ratio of voltages

V1

V2=

4

1(5.236)

Finally, find the net voltage as

V1 + V2 = 7 volts (5.237)

and use Equation 5.236 and Equation 5.237 to get

V = 1 + IR (5.238)

= 1 V + 1.4 V (5.239)

= 2.4 V (5.240)

Correct Answer

(A)

525


5.86 PGRE0177 #86


Start with Ohm’s law and solve for current

I =V

R(5.241)

Then, we can find the induced emf from a changing magnetic field by

ε = N |dφBdt| (5.242)

φB =

∫~B · d ~A (5.243)

526


The area and, therefore, the magnetic flux changes during rotation so we find the magnetic fluxas

φB = Bπr2 sin(ωt) (5.244)

and, therefore, the induced EMF is

ε = NBπr2ω cos(ωt) (5.245)

Finally, substitute the EMF from Equation 5.245 into Equation 5.241 to get the final solution

I =NBπr2ω cos(ωt)

R(5.246)

=(15 turns)(0.5 tesla)(0.01 m)2(300 rad/sec) cos(ωt)

9 Ω(5.247)

= 25π cos(ωt) (5.248)

Correct Answer

(E)

527


5.87 PGRE0177 #87


For the left side of the diagram, the test charge falls inside the sphere meaning that the potentialis constant and the field is 0. This means that the only force on the test charge will be the resultof sphere Q pushing q to the left. Using Gauss’s law with a distance of 10d− d/2 = 19/2d, the netforce is

F =1

4πε0

qQ

r2(5.249)

=4qQ

4πε0(361d2)(5.250)

=qQ

361πε0d2(5.251)

Correct Answer

(A)

528


5.88 PGRE0177 #88


Eliminate (A) since a changing field must generate a magnetic field. Next, compare the units foreach of the potential solutions to get

(B): kgA·s2 = tesla

(C): kgA·s2 = tesla

(D): kgA·s2m 6= tesla

(E): kgA·s2m 6= tesla

We are left with (B) and (C) at which point you can guess or try to recall that the Biot-Savartlaw is proportional to 1/4π rather than 1/4π2

B =

∫µ0

4π

I dl × r|r|2

(5.252)

Correct Answer

(C)

529


5.89 PGRE0177 #89


From conservation of angular momentum,

Lf = L0 (5.253)

When the child is in the middle of the merry-go-round, he/she/it won’t contribute to the angularmomentum, so the final momentum will be

Lf = Iωf =

(MR2

2

)ωf (5.254)

however, for the initial momentum we must use both the child at a distance R and the merry-go-round,

530


Li = Iω0 =

(MR2

2+mR2

)ω0 (5.255)

Plug Equations 5.254 and 5.255 into Equation 5.253 to get

(MR2

2

)ωf =

(MR2

2+mR2

)ω0 (5.256)

ωf = ω0 +2m

Mω0 (5.257)

= 2.0 rad/sec +2

5(2.0 rad/sec) (5.258)

= 2.8 rad/sec (5.259)

Correct Answer

(E)

531


5.90 PGRE0177 #90


From simple common sense, with a shorter length in which oscillations can occur and a strongerspring constant in Figure 1, the period should be less than that of Figure 2. From this, we caneliminate all options that are 1 or greater, i.e. (C), (D) or (E). Between (A) and (B), recall thatthe period of a SHO can be written as

T = 2π

√m

k(5.260)

and

T = 2π

√l

g(5.261)

which tells us that doubling k and doubling l will both scale the period by√

2. However, sinceeach of these changes are happening in such a way as to give Figure 1 a longer period, the twoscalings combine to give a total change in period of twice as much for figure 1 or, equivalently,

T1

T2=

1

2(5.262)

Correct Answer

(A)

532


5.91 PGRE0177 #91


At the top of the wedge, net energy is just the gravitational energy,

Enet = mgh (5.263)

all of this energy is converted to translational and rotational at the bottom of the wedge,

mgh =1

2mv2 +

1

2Iω2 (5.264)

solving for I in Equation 5.264 gives,

I =2mghR2

v2−R2m (5.265)

and substituting the value given for v2

533


I =2R2mgh

v2−R2m (5.266)

=7

4R2m−R2m (5.267)

=3

4R2m (5.268)

Correct Answer

(B)

534


5.92 PGRE0177 #92


The Hamiltonian is the sum of kinetic and potential energy terms, as opposed to the Lagrangianwhich is the difference. The energy of a harmonic oscillator is

535


V =1

2k(∆l)2 (5.269)

and the kinetic energy term is

T =1

2mv2 =

p2

2m(5.270)

adding the two of these gives us

H = T + V (5.271)

=p2

1

2m+

p22

2m+

1

2k(l − l0)2 (5.272)

=1

2

[p2

1

m+p2

2

m+ k(l − l0)2

](5.273)

Correct Answer

(E)

536


5.93 PGRE0177 #93


If you do well with your physics history, you may recall that Bohr radius is defined as the smallestpossible orbital distance for the hydrogen atom in its ground state, which is sufficient to choose a0.

Correct Answer

(C)

Alternate Solution

If you don’t recall the fact in the recommended solution, start by taking the squared wavefunctionand multiplying it by a spherical shell

dP =

[1

√πa

3/20

e−r/a0

]2

4πr2 dr (5.274)

=4

a30

r2e−2r/a0 dr (5.275)

differentiating dP in Equation 5.275 with respect to r and setting it equal to 0 allows us tosolve for the minimum radius

2re−2r/a0 − 2

a0r2e−2r/a0 = 0 (5.276)

2re−2r/a0

[1− r

a0

]= 0 (5.277)

537


at which point, it is easy to solve for the radius as

r = a0 (5.278)

Correct Answer

(C)

538


5.94 PGRE0177 #94


From quantum mechanics, the first order energy shift for our specific hamiltonian is

E(1)n = 〈n0|V (a+ a†)|n0〉 (5.279)

multiply the Hamiltonian out to get(a+ a†

)2= a2 + aa† + a†a+ a†2 (5.280)

at which point we can eliminate a2 and a†2 by orthogonality. Finally, solve for the energy withour remaining terms

E = 〈n|(aa† + a†a|n〉 (5.281)

= (2n+ 1)V (5.282)

= 5V (5.283)

Correct Answer

(E)

539


5.95 PGRE0177 #95


Start by recalling the relative permitivity equation

κ =ε(ω)

ε0(5.284)

where ε(ω) and ε0 are the absolute permitivity and electric constant, respectively. Next, re-calling that the Electric field is inversely proportional to ε(ω), which you could potentially realizefrom Coulomb’s law, we get

E ∝ 1

ε(ω)(5.285)

540


∝ 1

ε0κ(5.286)

Then applying E0 ∝ 1/ε0, we get

E =E0

κ(5.287)

Correct Answer

(A)

541


5.96 PGRE0177 #96


Recall the Larmor formula, which gives the power radiated by a charged object,

P =e2a2

6πε0c3(5.288)

since the sphere is not moving, just expanding at a stationary location, we get an acceleration,a = 0. Plugging this into Equation 1 clearly results in a total radiated power of zero as well.

Correct Answer

(E)

542


5.97 PGRE0177 #97

543



Consider the limiting case of θ = 0. In this case, there will be no difference between δθ′ and θ′,which tells us that the solution should have some dependence on θ and, furthermore, that δθ′ = 0when θ = 0. This eliminates (A), (B) and (C) due to their lack of theta dependence and (D) basedon the its failure to have δθ′ = 0 when θ = 0. Note that taking the limit in (D) requires you toutilize L’hospitals rule to deal with the division of 0/0, at which point you get a pair of cosineswhich go to 1 at θ′ = 0 and θ = 0 and keeps δθ′ from going to 0.

Correct Answer

(E)

Alternate Solution

Rather than applying a limiting case to the angle, θ, we can apply the limiting case n = 1. Whenn = 1, there is no transition to a new index of refraction and δθ′ should be equivalent to θ′. We canimmediately eliminate (D) because it lacks the dependence on n and we can next eliminate (A),(B) and (C) because their lack of θ′ dependence means we could never get δθ′ = θ′ when n = 1.

Correct Answer

(E)

544


5.98 PGRE0177 #98


Examining both sums in the expression, it should become quite clear that whatever it represents,it will have units of energy. Looking at the units for the potential solutions,

(A) Average energy is an energy!

(B) The denominator of this particular expression is the partition function. You don’t really needto know that, however, provided you recognize that it is not an energy.

(C) Absolutely not an energy.

(D) Probability of a certain energy value is not, in itself, an energy value.

(E) Entropy has units of J/K so it is not an energy.

Correct Answer

(A)

545


5.99 PGRE0177 #99


Considering the problem qualitatively, after the collision occurs, we have three particles each witha mass m and some non-zero velocity. This means that our final energy, which by conservationof momentum must equal our initial energy, must be the sum of three particles with rest energyof mc2 (i.e. net rest energy will be 3mc2) and some kinetic energy. This tells us that the initialphoton energy must be greater than just the final rest energies so we can eliminate (A), (B) and(C). Finally, equating the coefficients 4 and 5 in choices (D) and (E), respectively, to the Lorentzfactor, γ, gives us a strong hint that higher values of γ will give us less realistic particle velocities(i.e. far too high) and we can choose (D).

Correct Answer

(D)

546


5.100 PGRE0177 #100


Recalling the acronym for the visible light spectrum, ROYGBIV, it should be clear that red lighthas a longer wavelength and lower energy than green light. Red light is given to us in this instanceas λred = 632.82 nm so we know that the wavelength should be less than this and we can eliminate(D) and (E). Next, recall width of the visible spectrum is roughly between 380 nm and 750 nmand that green light is exactly in the middle of the spectrum. Thus, averaging the two wavelengthsshould give us a good approximation for the wavelength of green light

380 nm + 750 nm

2=

1130 nm

2= 565 nm (5.289)

and this average is closest to (B).

Correct Answer

(B)

547

Chapter 6

Appendix

6.1 Equations

It would be very convenient if every question on these tests could be solved with nothing more thanour intuitions and common sense. Sense this isn’t the case, we will need to do some calculationsand, on this test, it isn’t feasible in the given time to re-derive most of the classical results. Beloware the equations that had to be used at least once in my solutions.

6.1.1 Classical Mechanics

Kinematics:

∆x = v0t+1

2at2

v2 = v20 + 2ad

v = +at

∆x =v0 + vf

2t

Newton’s Second Law:

F = mx

Momentum:

P = mx

Angular Momentum:

L = Iω

Torque:

τ = ~r × ~F

τ = Iα

548

6.1. EQUATIONS CHAPTER 6. APPENDIX

Centripetal Acceleration:

ac = ω2r

F = mrω2

ω =2π

T

Gravitational Potential Energy:

U = mgh

U = −Gm1 m2

RU = −W

Kinetic Energy:

Uk =1

2mv2

Uk =p2

2m

Uk−rotational =1

2Iω2

Impulse:

I =

∫ t2

t1F dt

Parallel Axis Theorem:

I = Icom +Mh2

Work:

W =

∫F · s

Moment of Inertia (Point Mass):

I = MR2

Schwarzschild Radius:

Re =2GM

c2

Lagrangian:

L = T − V

Lagrangian Equation of Motion:

d

dt

(∂L

∂q

)− ∂L

∂q= 0

549


Hamiltonian:

H = T + V

Angular Frequency:

ω =2π

T

ωSHO =

√k

m

Period of a SHO:

T = 2π

√m

k

Rocket Mechanics:

mdv

dt+ u

dm

dt= 0

Kepler’s third law:

T 2 ∝ βr3

Center of Mass:

COM =m1x1 +m2x2

mtot

6.1.2 Electricity & Magnetism

Maxwell’s Equations:

∇ · ~E =ρ

ε0

∇ · ~B = 0

∇× ~E = −∂~B

∂t

∇× ~B = µ0~J + µ0ε0

∂ ~E

∂t

Coulombs Law:

U =1

4πε0

dQ

R

U =1

4πε0

q1 q2

R

F =1

4πε0

q1 q2

R2

Hooke’s Law:

F = −kx

550


Faraday’s Law of Induction

|ε| = N

∣∣∣∣dφBdt∣∣∣∣

Magnetic Flux:

φB =

∫ ∫S

~B(~r, t) · dA

Malus’ Law:

I =1

2cε0E

20cos2(θ)

Gauss’ Law:

~E · d ~A =q

ε0

Ampere’s Law

~B =µ0i

2πR

Larmor Formula:

P =q2a2

6πε0c3

Lorentz Force:

F = q(~v × ~B

)Hall Voltage:

VH = − IBdne

Magnetic Dipole Moment:

µ =

∫i dA

Relative Permitivity:

κ =ε(ω)

ε0

551


6.1.3 Optics

Photoelectric Equation:

Uk = hν − φ

Snell’s Law:

sin(θ1)

sin(θ2)=n2

n1

n-slit constructive interference:

d sin(θn) = nλ

n-slit destructive interference:

d sin(θn) =

(n+

1

2

)λ

Photon Energy:

E = hν

E =hc

λ

Thin Film Interference:

2nd =

(m+

1

2

)λ

Light speed through a medium:

v =1

εµ

Traveling wave:

y(x, t) = A sin

(2π

λ

)(x± vt)

Wave Propagation Velocity:

v = fλ

Drift velocity:

vd =i

nqA

Compton Equation:

λ′ − λ =h

mpc(1− cos(θ))

552


Bragg’s Law:

2d sin(θ) = nλ

Optical Magnification (Convex Lens):

M =fobjfeye

Rayleigh Criterion:

sin(θ) = 1.22λ

d

Thin Lens Equation/Mirror Equation:

1

d0+

1

di=

1

f

6.1.4 Thermodynamics & Stat Mech

Rydberg Formula:

1

λ= R∞

(1

n21

− 1

n22

)Rydberg Formula for Hydrogen like atoms:

1

λH−like= RZ2

(1

n21

− 1

n22

)Moseley’s Law:

1

λK−α= R (Z − β)2

(1

n21

− 1

n22

)Rydberg Energy:

E = −mee4

2h2

Z2

n2

E = E0

(1

λ21

− 1

λ2

)E = n(13.6 eV )

(1

λ21

− 1

λ22

)Heat Capacity:

C =∆Q

∆T

C =∂Q

∂T

C = T∂S

∂T

CV =

(∂Q

∂T

)V

=

(∂U

∂T

)V

Cp =

(∂Q

∂T

)p

=

(∂H

∂T

)p

553


Heat:

Q = cm∆T

First Law of Thermodynamics:

dU = dQ+ dW

dU = dQ− PdV

Ideal Gas Law:

pV = nRT

Thermodynamic Work:

W =

∫ Vf

Vi

PdV

Entropy:

∆S =

∫ T2

T1

dq

T

Fourier’s Law of Heat Conduction:

∂Q

∂t= −k

∮s∇T · d ~A

Mean free path:

P (x) = nσdx

Stefan-Boltzmann’s Law:

j∗ = σT 4

Bohr Energy:

Etot =−13.6 eV

n2

Carnot Engine Efficiency:

η =W

QH= 1− TC

TH

Fermi Velocity:

vf =

√2Efm

Graham’s Law of Effusion

Rate1

Rate2=

√M2

M1

Wien’s Displacement Law

λ =b

T

554


6.1.5 Quantum Mechanics

Particle Location (Probability):

Pab =

∫ b

a|ψ(x)|2 dx

Infinite Square Well/Particle in a box:

ψn(x, t) = Asin(knx)e−iωnt

Planck Length:

lp =

√Gh

c3

Expectation Value:

〈A〉 =

∫〈ψ |A|ψ〉

Heisenberg Uncertainty Principle:

∆x∆p ≥ h

2

∆E∆t ≥ h

2

Spin Operator:

S21ψ1 = S1(S1 + 1)ψ1

Probability Current:

~J(x, t) =h

2mi

(ψ∗∂ψ

∂x− ∂ψ∗

∂xψ

)Quantum Harmonic Oscillator:

En = hω

(n+

1

2

)

Wave Speed (de Broglie relations):

vp =E

p

vp =c2

v

Time-Independent Schrodinger Equation:

E ψ(x) = − h2

2m∇2 ψ(x) + V (x) ψ(x)

555


Quantum Momentum Operator:

P =h

i∇ψ

Russel-Saunders Term Symbols:

2s+1LJ

Total Angular Momentum Quantum Number:

j = l + s

Hermitian Operator:

〈A〉 =

∫ψ∗(r)Aψ(r) dr

Canonical Partition Function:

Z =∑s

ge−Es/kbT

6.1.6 Special Relativity

Rest Energy:

E = m0c2

Lorentz Factor:

γ =1√

1− v2

c2

Relativistic Energy:

ER = γmc2

Relativistic Momentum:

prel = γm0v =m0v√1− v2

c2

Relativistic Energy-Momentum:

E2 =(mc2

)2+ (pc)2

Relativistic sum of velocities:

u′ =u+ v

1 + vuc2

u =u′ + v′

1 + v′u′

c2

556


Proper Time:

∆τ2 = ∆t2 −∆x2

∆t2 = ∆τ2 + ∆x2

Space-Time Interval:

∆S2 = −(C∆t)2 + ∆x2

Length Contraction:

L′ =L

γ= L

√1− v2

c2

Time Dilation:

∆t′ = γ∆t =∆t√

1− v2/c2

6.1.7 ElectronicsOhm’s Law:

V = IR

Kirchoff’s First Law:

n∑k=1

I = 0

Kirchoff’s Second Law:

n∑k=1

V = 0

Current:

i =dq

dt

Faraday’s law of induction:

ε =

∣∣∣∣dφBdt∣∣∣∣

Capacitance:

C =Q

V

Frequency of an RLC Circuit:

f =1

4π√LC

557


Impedance Matching:

ZS = Z∗L

Rg + jXg = Rg + jXl

Capacitor Energy:

U =1

2CV 2

Capacitor Discharge:

V = V0e−t/RC

Resistance of a Wire:

R =ρl

A

Biot-Savart Law:

B =

∫µ0

4π

I dl × r|r|2

6.1.8 Special Topics

Acoustic Beats:

beats = |f2 − f1|

Doppler Effect:

f =

[1

1± vsourcevwave

]f0

Cyclotron Resonance Frequency:

ωc =eB

m

Frequency of a Pendulum:

ω =

√mgrcom

I

Poisson Distribution Uncertainty Equation:

u =σ√N

558

6.2. UNITS AND CONVERSIONS CHAPTER 6. APPENDIX

6.2 Units and Conversions

6.2.1 UnitsCurrent:

A =C

s

Pressure:

P =N

m2

Energy:

J = N ·m =kg ·m2

s2= Pa ·m3 = W · s

Power:

W =J

s=

N ·ms

=kg ·m2

s3

Electric Charge:

C = A · s = F ·V

Electric Potential:

V =W

A=

J

A · s=

J

C

Capacitance:

F =C

V=

J

V2

Resistance:

Ω =V

A=

J

s ·A2

Conductance:

S = Ω−1 =A

V

Magnetic Flux Density:

T =V · sm2

=N

A ·m=

Wb

m2

Inductance:

H =m2 · kg

s2 ·A2 =Wb

A=

V · sA

= Ω · s

Entropy/Heat Capacity:

J

K

559

6.2. UNITS AND CONVERSIONS CHAPTER 6. APPENDIX

6.2.2 Conversions

Volume:

1 cm3 = 1 mL

1 dm3 = 1 L

Angstrom:

1 A = 0.1 nm = 1× 10−9 m

Micron:

1 µ = 1× 10−6 m

Electronvolt

1 eV = 1.7826× 10−36 kg

Kilogram per Cubic Meter:

1, 000kg

m3= 1

g

mL= 1

kg

L

Revolutions:

1 rpm = 0.105rad

s

6.2.3 SI Prefixes

Prefix Symbol 10n

peta P 1015

tera T 1012

giga G 109

mega M 106

kilo k 103

hecto h 102

deca da 101

BASE 101 = 1

deci d 10−1

centi c 10−2

milli m 10−3

micro µ 10−6

nano n 10−9

pico p 10−12

femto f 10−15

atto a 10−18

560

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