Physics 3313 – Review 1

3/8/2010 13313 Andrew Brandt

Monday March 8 , 2010Dr. Andrew Brandt

Time Dilation Example

• Muons are essentially heavy electrons (~200 times heavier)• Muons are typically generated in collisions of cosmic rays in upper

atmosphere and, unlike electrons, decay ( sec)• For a muon incident on Earth with v=0.998c, an observer on Earth would

see what lifetime of the muon?• 2.2 sec?

• t=35 sec • Moving clocks run slow so when an outside observer measures, they see a

longer time than the muon itself sees. (It’s who is the observer that matters not who is actually moving)

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0 2.2t

2

2

1 161 vc

More about Muons• They are typically produced in atmosphere about 6 km above surface of Earth and

frequently have velocities that are a substantial fraction of speed of light, v=.998 c for example

• How do they reach the Earth?• Standing on the Earth, we see the muon as moving so it has a longer life (moving clocks

run slow) and we see it living t=35 sec not 2.2 sec , so it can travel 16 times further than 0.66 km that it “thinks” it can travel, or about 10 km, so it can easily cover the 6km necessary to reach the ground.

• But riding on a muon, the trip takes only 2.2 sec, so how do they reach the ground???• Muon-rider sees the ground moving, so the length contracts and is only • so muon can go .66km, but reaches the ground in only .38km

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8 60 2.994 10 2.2 10 sec 0.66

secmvt x x x km

0 6 /16 0.38L

km

Velocity Addition Example• Lance is riding his bike at 0.8c relative to observer. He throws a ball at

0.7c in the direction of his motion. What speed does the observer see?

2

.7 .8 0.962.7 .81

xc cv cc cc

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'

'

21

xx

x

v vvvvc

Relativistic Momentum Example• A meteor with mass of 1 kg travels 0.4c• Find its momentum, what if it were going twice as fast? compare with

classical case

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) 0.4vac 1.09 8 81.09 1 0.4 3 10 1.31 10

sec secm kg mp mv kg

) 0.8vbc 1.67 8 81.67 1 0.8 3 10 4.01 10

seckg mp

8) 1.2 10seckg mc p mv

8) 2.4 10seckg md

Relativistic Energy

• Ex. 1.6: A stationary bomb (at rest) explodes into two fragments each with 1.0 kg mass that move apart at speeds of 0.6 c relative to original bomb. Find the original mass M.

• What if the bomb were not stationary?

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2 2E mc KE mc

2 2 2 21 1 1 20i i fE Mc KE Mc E m c m c

22 1

2

2

2 2 / 0.8 2.51

m cMc M kg kgvc

2( 1)KE mc 2 2 2 2( )E mc p c

Properties of Photoelectric Effect• Existence of photoelectric effect was not a surprise, but the details were

surprising1) Very little time (nanoseconds) between arrival of light pulse and emission

of electron2) Electron energy independent of intensity of light3) At higher frequency get higher energy electrons

Minimum frequency (0) required for photoelectric effect depends on material:

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slope=E/ =h (planck’s constant)

Einstein Explains P.E. Effect• Einstein explained P.E. effect: energy of light not distributed evenly over

classical wave but in discrete regions called quanta and later photons 1) EM wave concentrated in photon so no time delay between incident

photon and p.e. emission 2) All photons of same frequency have same energy E=h, so changing

intensity changes number (I=Nh, where N is rate/area) but not energy3) Higher frequency gives higher energy

• Electrons have maximum KE when all energy of photon given to electron.• is work function or minimum energy required to liberate electron from

material ( =h 0 )

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346.626 10 J sech

maxKE h 0h h

Example of PE for Iron• a) Find given

• b) If p.e.’s are produced by light with a wavelength of 250 nm, what is stopping potential?

• =4.96-4.5 =0.46 eV (is this your final answer?)• NO! it is V0=0.46V (not eV)• x-ray is inverse photo-electric effect, can neglect binding energy, since x-ray is very

energetic

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0h

6

9

1.24 10 4.5250 10

eV m eVm

150 1.1 10 Hz

15 15 14.14 10 sec 1.1 10eV s 4.5eV

h max 0KE eV

Photon Energy Loss

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e e

22 .511 2em c MeV

(1 cos )hmc

De Broglie• Tiger example• He hits a 46 g golf ball with a velocity of 30 m/s (swoosh) .• Find the ; what do you expect?•

• The wavelength is so small relevant to the dimensions of the golf ball that it has no wave like properties

• What about an electron with ?• This large velocity is still not relativistic so

• Now, the radius of a hydrogen atom (proton + electron) is • Thus, the wave character of the electron is the key to understanding atomic

structure and behavior3/8/2010 3313 Andrew Brandt 11

dB

1v c hmv

346.6 10 sec

.046 30m / secx Jkg

344.8 10 m

710 m / secv

3411

31 7

6.63 10 7.3 10 m9.1 10 10 m/s

J skg

115 10 m

hmv

hv hpc

E pc h hp

De Broglie Example 3.2

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• What is the kinetic energy of a proton with a 1fm wavelength• Rule of thumb, need relativistic calculation unless

• Is pc = energy? Units are right, but pc .ne. energy

• Since need relativistic calculation

• This implies a relativistic calculation is necessary so can’t use KE=p2/2m

2 .938ppc m c GeV15 8

15

4.136 10 3 10 /1 10ev s m s

m

2 2 2E mc p c 2 2(.938) (1.24) 2KE E mc

hcpc

1.24GeV

1.55GeV

1.55 0.938 0.617 617GeV MeV

2ppc m c

Wave Velocities• Definition of phase velocity:• Definition of group velocity: • For light waves :• For de Broglie waves

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2 / 2pv k

gdvdk

gdv

k dk

g pv v c

Particle in a Box Still• General expression for non-rel Kinetic Energy:• With no potential energy in this model, and applying constraint on wavelength gives:

• Each permitted E is an energy level, and n is the quantum number• General Conclusions: 1) Trapped particle cannot have arbitrary energy like a free particle—only

specific energies allowed depending on mass and size of box2) Zero energy not allowed! v=0 implies infinite wavelength, which means particle is not trapped

3) h is very small so quantization only noticeable when m and L are also very small

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2 22

2

12 2 2

p hKE mvm m

22 2 2

2 / 82 (2 / )n

hE n h mLm L n

Example 3.5• 10 g marble in 10 cm box, find energy levels

• For n=1 and• Looks suspiciously like a stationary marble!!• At reasonable speeds n=1030

• Quantum effects not noticeable for classical phenomena

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2 34 264 2

2 1 2

(6.63 10 ) 5.5 108 10 (10 )n J s n J

kg m

2 2 2/ 8nE n h mL

64min 5.5 10E J 313.3 10 /v m s

Uncertainty Principlea) For a narrow wave group the position is

accurately measured, but wavelength and thus momentum cannot be precisely determined

b) Conversely for extended wave group it is easy to measure wavelength, but position uncertainty is large

• Werner Heisenberg 1927, it is impossible to know exact momentum and position of an object at the same time

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2x p

2E t

Rutherford Scattering

• The actual result was very different—although most events had small angle scattering, many wide angle scatters were observed

• “It was almost as incredible as if you fired a 15 inch shell at a piece of tissue paper and it came back at you”

• Implied the existence of the nucleus. • We perform similar experiments at

Fermilab and CERN to look for fundamental structure

4 2( )

sin2

KNKE

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Spectral Lines• For Hydrogen Atom (experimental observation):

• where nf and ni are final and initial quantum states• R=Rydberg Constant

• Balmer Series nf = 2 and ni=3,4,5 visible wavelengths in Hydrogen spectrum 656.3, 486.3,…364.6 (limit as n)

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2 2

1 1 1( )f i

Rn n

7 1 11.097 10 0.01097m nm

212

08nm

EeEr n

1 112 2 2

1 1 8( ) 13.6( ) 12.13 1 9f i

f i

E EE E E E eVn n

5

110

1 10 4355.3 10

nrna

512 7.19 10nEE eVn

12 2

1 1 1( )f i

Ech n n

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Bohr Energy Levels

How much energy is required to raise an electron from the ground state of a Hydrogen atom to the n=3 state?

Rydberg atom has r=0.01 mm what is n? E?

21nr n r

1 12 2 /f if i

E EE E E h hcn n

Bohr atom explains energy levels

Correspondence principle: For large n Quantum Mechanics Classical Mechanics

Expectation Values• Complicated QM way of saying average• After solving Schrodinger Eq. for particle under particular condition contains all info on a particle permitted by uncertainty principle in the

form of probabilities. • This is simply the value of x weighted by its probabilities and summed

over all possible values of x, we generally write this as

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*( , ) ( , )x x t x x t dx

Eigenvalue Examples• If operator is and wave function is find eigenvalue

•

• So eigenvalue is 4• How about sin(kx)? • eigenvalue is –k2

• How about x4? • Not an eigenvector since

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2xe 2

2

dGdx

2

22

xdG edx

2xd d edx dx

22 xd edx

24 xe 4G

2

4 2 42 12dG x x cx

dx