Solucionario Capitulo 39 - Paul E. Tippens

Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition

Chapter 39. Nuclear Physics and the Nucleus

The Elements

39-1. How many neutrons are in the nucleus of 82208 Pb ? How many protons? What is the ratio

N/Z? (N is the number of neutrons and Z is the number of protons.)

A = N + Z; N = A – Z = 126 neutrons; Z = 82 protons; 1.54A

Z

39-2. The nucleus of a certain isotope contains 143 neutrons and 92 protons. Write the symbol

for this nucleus.

A = N + Z = 143 + 92 = 235; Z = 92: 23592 U

39-3. From a stability curve it is determined that the ratio of neutrons to protons for a cesium

nucleus is 1.49. What is the mass number for this isotope of cesium?

Z = 55; 1.49; 1.49(55) 81.95; ;N N A N Z

Z

A = 137

39-4. Most nuclei are nearly spherical in shape and have a radius that may be approximated by

1153

0 0 1.2 x 10 mr r A r

What is the approximate radius of the nucleus of a gold atom 19779 Au ?

15 -153(1.2 10 m) 197 6.98 x 10 mr x ; r = 6.98 x 10-15 m

538


39-5. Study Table 39-4 for information on the several nuclides. Determine the ratio of N/Z for

the following nuclides: Beryllium-9, Copper-64, and Radium 224.

Beryllium: A = 9; Z = 4; N = 9 – 4 = 5; 1.25N

Z

39-5. (Cont.) Copper: A = 64; Z = 29; N = 64 – 29 = 35; 1.21N

Z

Radium: A = 224; Z = 88; N = 224 – 88 = 136; 1.55N

Z

The Atomic Mass Unit

39-6. Find the mass in grams of a gold particle containing two million atomic mass units?

-276 1.66 x 10 kg2 x 10 u ;

1.00 um

m = 3.32 x 10-21 kg

39-7. Find the mass of a 2-kg copper cylinder in atomic mass units? In Mev? In joules?

-27

1 u2 kg ;1.6606 x 10 kg

m

m = 1.20 x 1027 u

27 931 MeVm = 1.204 x 10 u ;1 u

m = 1.12 x 1030 MeV

539


-1330 1.6 x 10 Jm = 1.12 x 10 MeV ;

1 MeV m = 1.79 x 1017 J

39-8. A certain nuclear reaction releases an energy of 5.5 MeV. How much mass (in atomic

mass units) is required to produce this energy?

1 u5.5 MeV ;931 MeV

m

m = 0.00591 u

39-9. The periodic table gives the average mass of a silver atom as 107.842 u. What is the

average mass of the silver nucleus? (Recall that me = 0.00055 u )

For silver, Z = 47. Thus, the nuclear mass is reduced by the mass of 47 electrons.

m = 107.842 u – 47(0.00055 u); m = 107.816 u

*39-10. Consider the mass spectrometer as illustrated by Fig. 39-3. A uniform magnetic field of

0.6 T is placed across both upper and lower sections of the spectrometer., and the electric

field in the velocity selector is 120 V/m. A singly charged neon atom (+1.6 x 10-19 C) of

mass 19.992 u passes through the velocity selector and into the spectrometer. What is the

velocity of the neon atom as it emerges from the velocity selector?

120,000 V/m ;0.6 T

EvB

v = 2 x 105 m/s

*39-11. What is the radius of the circular path followed by the

neon atom of Problem 19-10?

540

B = 0.6 T into paper

R


-27261.66 x 10 kg19.992 u 3.32 10 kg

1 um x

2

; mv mvqvB RR qB

;

-27 5

-19

(3.32 x 10 kg)(2 x 10 m/s) ;(1.6 x 10 C)(0.600 T)

R R = 6.92 cm

Mass Defects and Binding Energy (Refer to Table 39-4 for Nuclidic Masses.)

*39-12. Calculate the mass defect and binding energy for the neon-20 atom 2010 Ne .

[( )] 10(1.007825 u) 10(1.008665 u) 19.99244 uD H nm Zm Nm M

mD = 20.016490 u –19.99244 u ; mD = 0.17246 u

2 931 MeV(0.17246 u) 160.6 MeV1 uDE m c

; E = 161 MeV

*39-13. Calculate the binding energy and the binding energy per nucleon for tritium 31 H . How

much energy in joules is required to tear the nucleus apart into its constituent nucleons?

[( )] 1(1.007825 u) 2(1.008665 u) 3.016049 uD H nm Zm Nm M

39-13. (Cont.) mD = 0.009106 u;

2 931 MeV(0.009106 u) 8.48 MeV1 uDE m c

541


E = 8.48 MeV;

8.48 MeV3

BEA

;

2.83 MeV/nucleonBEA

Energy to tear apart: EB = 8.48 MeV(1.6 x 10-13 J/MeV) = 1.36 x 10-12 J

*39-14. Calculate the mass defect of 37 Li . What is the binding energy per nucleon?

[( )] 3(1.007825 u) 4(1.008665 u) 7.016930 uD H nm Zm Nm M

mD = 7.058135 u –7.016930 u ; mD = 0.041205 u

2 931 MeV(0.041205 u) 38.4 MeV1 uDE m c

; E = 38.4 MeV

38.4 MeV ;7

BEA

5.48 MeV/nucleonBEA

*39-15. Determine the binding energy per nucleon for carbon-12 ( 126 C ).

6(1.007825 u) 6(1.008665 u) 12.0000 u 0.09894 uDm

2 931 MeV(0.09894 u) ;1 uDE m c

92.1 MeV12

BEA

= 7.68 MeV/nucleon

*39-16. What is the mass defect and the binding energy for a gold atom 19779 Au ?

542


79(1.007825 u) 118(1.008665 u) 196.966541 u 1.674104 uDm

2 931 MeV(1.674104 u) 1.56 GeV;1 uDE m c

1.56 GeV197

BEA

= 7.91 MeV/nucleon

*39-17. Determine the binding energy per nucleon for tin-120 ( 12050 Sn ).

50(1.007825 u) 70(1.008665 u) 119.902108 u 1.09569 uDm

(1.09569 u)(931 MeV/u ;120

BEA

BEA

8.50 MeV/nucleon

Radioactivity and Nuclear Decay

39-18. The activity of a certain sample is rated as 2.8 Ci. How many nuclei will have

disintegrated in a time of one minute?

10 -13.7 x 10 s2.8 Ci (60 s);1 Ci

Nuclei

nuclei = 6.22 x 1012 nuclei

543


39-19. The cobalt nucleus 6027 Co emits gamma rays of approximately 1.2 MeV. How much mass

is lost by the nucleus when it emits a gamma ray of this energy?

1 u1.2 MeV ;931 MeV

E m = 0.00129 u

39-20. The half-life of the radioactive isotope indium-109 is 4.30 h. If the activity of a sample is

1 mCi at the start, how much activity remains after 4.30, 8.60, and 12.9 h?

½/ 1

0½

1 4.3 h 1; 1; (1 mCi) ;2 4.3 h 2

t T tR R RT

R = 0.5 mCi

½/ 2

0½

1 8.6 h 1; 2; (1 mCi) ;2 4.3 h 2

t T tR R RT

R = 0.25 mCi

½/ 3

0½

1 12.9 h 1; 3; (1 mCi) ;2 4.3 h 2

t T tR R RT

R = 0.125 mCi

39-21. The initial activity of a sample containing 7.7 x 1011 bismuth-212 nuclei is 4.0 mCi. The

half-life of this isotope is 60 min. How many bismuth-212 nuclei remain after 30 min?

What is the activity at the end of that time?

½/ ½11

0½

1 30 min 1; 0.5; (7.7 x 10 )2 60 min 2

t T tN N NT

= 5.44 x 1011 nucl.

544


½/ ½

01 1; (4 mCi) ;2 2

t T

R R R R = 2.83 mCi

*39-22. Strontium-90 is produced in appreciable quantities in the atmosphere during a nuclear

explosion. If this isotope has a half-life of 28 years, how long will it take for the initial

activity to drop to one-fourth of its original activity?

00

1 1 1; ; 22 2 4

n nRR R nR

½

2; 2(28 yr); 28 yr

t tn tT

t = 56 yr

*39-23. Consider a pure, 4.0-g sample of radioactive Gallium-67. If the half-life is 78 h, how

much time is required for 2.8 g of this sample to decay?

When 2.8 g decay, that leaves 4 g – 2.8 g or 1.20 g remaining.

00

1 1.2 g 1; 0.300; 0.3002 4 g 2

n nmm mm

The solution is accomplished by taking the common log of both sides:

0.523log(0.5) log(0.300); 0.301 0.523; 1.740.301

n n n

545


½

1.74;78 h

t tnT

t = 1.74 (78 h) t = 135 h

*39-24. If one-fifth of a pure radioactive sample remains after 10 h, what is the half-life?

1 1 ; 0.200 = (0.5) ; log(0.5) log(0.2); 0.301 0.6995 2

nn n n

½½ ½

10 h 10 h2.32; ;2.32

tn TT T

T½ = 4.31 h

Nuclear Reactions

*39-25. Determine the minimum energy released in the nuclear reaction

19 1 4 169 1 2 8F H He + O + energy

The atomic mass of 919 F is 18.998403 u,

4 12 1He 4.002603 u, H = 1.007824 u.

19 1 4 169 1 2 8 F + H - He - O ;E (Energy comes from mass defect)

E = 18.998403 u + 1.007825 u – 4.002603 – 15.994915 u = 0.0087 u

931 MeV(0.00871 u) ;1 u

E E = 8.11 MeV

546


*39-26. Determine the approximate kinetic energy imparted to the alpha particle when Radium-

226 decays to form Radon-222. Neglect the energy imparted to the radon nucleus.

226 222 4 22688 86 2 88Ra Rn + He + Energy; Ra = 226.02536

931 MeV226.02536 u - 222.017531 u - 4.002603 u = 0.00523 u1 u

E = 4.87 MeV

*39-27. Find the energy involved in the production of two alpha particles in the reaction

7 1 4 43 1 2 2Li H He He + energy

931 MeV7.016003 u + 1.007825 u - 2(4.002603 u) = 0.018622 u1 u

E = 17.3 MeV

*39-28. Compute the kinetic energy released in the beta minus decay of thorium-233.

233 233 0 233 23390 91 +1 90 91Th Pa + + energy; Th = 233.041469 u; Pa = 233.040130

233.041469 u - 233.040130 u - 0.00055 u = 0.000789 uE

931 MeV0.000789 u ;1 u

E E = 0.735 MeV

*39-29. What must be the energy of an alpha particle as it bombards a Nitrogen-14 nucleus

producing 17 1 178 1 8O and H? ( O = 16.999130 u) .

547


4 14 17 12 7 8 1He + N + Energy O + H

931 MeV16.999130 u + 1.007825 u - 14.003074 - 4.002603 u = 0.001278 u1 u

E

E = 1.19 MeV This is the Threshold Energy for the reaction

Challenge Problems

*39-30. What is the average mass in kilograms of the nucleus of a boron-11 atom?

-271.66 x 10 kg11.009305 u ;1.00 u

m

m = 1.83 x 10-26 kg

*39-31. What are the mass defect and the binding energy per nucleon for boron-11?

[( )] 5(1.007825 u) 6(1.008665 u) 11.009305 uD H nm Zm Nm M

mD = 11.09112 u –11.009305 u ; mD = 0.08181 u

2 931 MeV(0.08181 u) 76.2 MeV1 uDE m c

; E = 76.2 MeV

76.2 MeV ;11

BEA

6.92 MeV/nucleonBEA

*39-32. Find the binding energy per nucleon for Thallium-206.

548


81(1.007825 u) 125(1.008665 u) 205.976104 u 1.740846 uDm

2 931 MeV(1.740846 u) ;1 uDE m c

1621 MeV206

BEA

= 7.87 MeV/nucleon

*39-33. Calculate the energy required to separate the nucleons in mercury-204.

80(1.007825 u) 124(1.008665 u) 203.973865 u 1.7266 uDm

2 931 MeV(1.7266 u) ;1 uDE m c

EB = 1610 MeV

*39-34. The half-life of a radioactive sample is 6.8 h. How much time passes before the activity

drops to one-fifth of its initial value?

0

1 1 ; 0.200 = (0.5) ; log(0.5) log(0.2); 0.301 0.6995 2

nnR n n

R

½½ ½

0.699 6.8 h 6.8 h2.32; 2.32; ;0.301 2.32

tn n TT T

T½ = 2.93 h

*39-35. How much energy is required to tear apart a deuterium atom? 21( H 2.014102 u)

549


1(1.007825 u) 1(1.008665 u) 2.014102 u 0.002388 uDm

2 931 MeV(0.002388 u) ;1 uDE m c

EB = 2.22 MeV

*39-36. Plutonium-232 decays by alpha emission with a half-life of 30 min. How much of this

substance remains after 4 h if the original sample had a mass of 4.0 g? Write the equation

for the decay?

½/ 8

0½

1 4 h 1; 8.00; (4 g) ;2 0.5 h 2

t T tm m mT

m = 15.6 mg

232 228 494 92 2Pu U + He + energy

*39-37. If 32 x 109 atoms of a radioactive isotope are reduced to only 2 x 109 atoms in a time of

48 h, what is the half-life of this material?

9

90

2 x 10 1 ; 0.0625 = (0.5) ; log(0.5) log(0.0625); 0.301 1.20432 x 10 2

nnN n n

N

550


½½ ½

1.204 48 h 48 h4.00; 4.00; ;0.301 4.00

tn n TT T

T½ = 12.0 h

*39-38. A certain radioactive isotope retains only 10 percent of its original activity after a time of

4 h. What is the half-life?

0

1 1 ; 0.100 = (0.5) ; log(0.5) log(0.100); 0.301 1.0010 2

nnN n n

N

½½ ½

1.00 4 h 4 h3.32; 3.32; ;0.301 3.32

tn n TT T

T½ = 72.3 min

*39-39. When a 36 Li nucleus is struck by a proton, an alpha particle and a produce nucleus are

released. Write the equation for this reaction. What is the net energy transfer in this case?

6 1 4 33 1 2 2Li H He He + energy

931 MeV6.015126 u + 1.007825 u - 4.002603 u - 3.016030 = 0.00432 u1 u

E = 4.02 MeV

*39-40. Uranium-238 undergoes alpha decay. Write the equation for the reaction and calculate

the disintegration energy.

238 224 492 90 2U Th + He + energy

551


238.05079 u - 234.04363 u - 4.002603 u = 0.00456 uE

931 MeV0.00456 u ;1 u

E E = 4.24 MeV

*39-41. A 9-g sample of radioactive material has an initial activity of 5.0 Ci. Forty minutes later,

the activity is only 3.0 Ci. What is the half-life? How much of the pure sample remains?

0

3 Ci 1 ; 0.600 = (0.5) ; log(0.5) log(0.6); 0.301 0.2225 Ci 2

nnR n n

R

½½ ½

0.222 40 min 40 min0.737; 0.737; ;0.301 0.737

tn n TT T

T½ = 54.3 min

0.737

01 1(9 g)2 2

n

m m m = 5.40 g

Critical Thinking Problems

*39-42. Nuclear fusion is a process that can produce enormous energy without the harmful

byproducts of nuclear fission. Calculate the energy released in the following nuclear

fusion reaction:

3 3 4 1 12 2 2 1 1H He He H+ H energy

552


931 MeV2(3.016030 u) - 4.002603 u - 2(1.007825 u) = 0.013807 u1 u

E

E = 12.9 MeV

*39-43. Carbon-14 decays very slowly with a half-life of 5740 years. Carbon dating can be

accomplished by seeing what fraction of carbon-14 remains, assuming the decay process

began with the death of a living organism. What would be the age of a chunk of charcoal

if it was determined that the radioactive C-14 remaining was only 40 percent of what

would be expected in a living organism?

0

10.40 ; 0.400 = (0.5) ; log(0.5) log(0.4); 0.301 0.3992

nnR n n

R

½

0.399 0.737; 1.32; t 1.32(5740 yr);0.301 5740 yr

t tn nT

t = 7590 yr

*39-44. The velocity selector in a mass spectrometer has a magnetic field of 0.2 T perpendicular

to an electric field of 50 kV/m. The same magnetic field is across the lower region. What

is the velocity of singly charged Lithium-7 atoms as they leave the selector? If the radius

of the path in the spectrometer is 9.10 cm, find the atomic mass of the lithium atom?

50,000 V/m ;0.2 T

EvB

v = 2.5 x 105 m/s

553B = 0.2 T into paper

R


2

; mv qBRqvB mR v

;

-19

5

(1.6 x 10 C)(0.2 T)(0.091 m) ;(2.5 x 10 m/s)

m

-26-27

1 um = 1.165 x 10 kg ;1.66 x 10 kg

m = 7.014 u

*39-45. A nuclear reactor operates at a power level of 2.0 MW. Assuming that approximately

200 MeV of energy is released for a single fission of U-235, how many fission processes

are occurring each second in the reactor?

6 1919

-19

2 x 10 J/s 1.25 x 10 MeV/s1.25 x 10 MeV/s; 1.6 x 10 J/MeV 200 MeV/fission

P 6.25 x 10216 fissions/s

*39-46. Consider an experiment which bombards 714 N with an alpha particle. One of the two

product nuclides is 11H . The reaction is

2 He N X H4714

11 Z

A

What is the product nuclide indicated by the symbol X? How much kinetic energy must

the alpha particle have in order to produce the reaction?

Conservation of nucleons means: 4 14 17 12 7 8 1He N + O Henery

16.99913 u + 1.007825 u - 4.002603 u - 14.003074 u = 0.001282 uE

554


931 MeV0.001282 u ;1 u

E E = 1.19 MeV

*39-47. When passing a stream of ionized lithium atoms through a mass spectrometer, the radius

of the path followed by 37 Li (7.0169 u) is 14.00 cm. A lighter line is formed by the 3

6 Li

(6.0151 u). What is the radius of the path followed by the 36 Li isotopes?

2

; ;mv mvqvB RR qB

(See Problems 39-10 and 39-11)

1 1 2 12

2 2 1

(6.0151 u)(14 cm); ;7.0169 u

m R m RRm R m

R2 = 6.92 cm

555

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Solucionario Capitulo 39 - Paul E. Tippens

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