Physics 3313 – Review 1 3/9/20091 3313 Andrew Brandt Monday March 9, 2009 Dr. Andrew Brandt.

3313 Andrew Brandt 1

Physics 3313 – Review 1

3/9/2009

Monday March 9, 2009Dr. 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.

3/9/2009

0 2.2t

2

2

116

1vc


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?• We see the muon moving so t=35 sec not 2.2 sec , so they can travel 16 time

further, or about 10 km, so they easily 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

3/9/2009

8 60 2.994 10 2.2 10 sec 0.66

sec

mvt 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

2

.7 .80.962

.7 .81

x

c cv c

cc

3/9/2009

'

'

21

xx

x

v vv

vv

c


Relativistic Momentum Example

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

classical case

1/26/2009

) 0.4v

ac

1.09 8 81.09 1 0.4 3 10 1.31 10sec sec

m kg mp mv kg

) 0.8v

bc

1.67 8 81.67 1 0.8 3 10 4.01 10sec

kg mp

8) 1.2 10sec

kg mc p mv

8) 2.4 10sec

kg md


Relativistic Energy

• Ex. 1.6: A stationary high-tech bomb 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.

•

1/26/2009

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

22 / 0.8 2.5

1

m cMc M kg kg

vc

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 surprising

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

2) Electron energy independent of intensity of light

3) At higher frequency get higher energy electrons

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

2/9/2009

slope=E/ =h (planck’s constant)


Einstein Explains P.E. Effect

• Einstein 1905 (I had a good year) explained P.E. effect: energy of light not distributed evenly over classical wave but in discrete regions called quanta and later photons (draw)

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 energy

3) 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 )

2/9/2009

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 the answer?)

• NO! it is V0=0.46V (not eV)

2/9/2009

0h

6

9

1.24 104.5

250 10

eV meV

m

150 1.1 10 Hz

15 15 14.14 10 sec 1.1 10eV s 4.5eV

h max 0KE eV


Photon Energy Loss

2/11/2009

e e

22 .511 2em c MeV

(1 cos )h

mc

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 behavior2/16/2009 3313 Andrew Brandt 11

dB

1v c h

mv

346.6 10 sec

.046 30m / sec

x J

kg

344.8 10 m

710 m / secv

3411

31 7

6.63 107.3 10 m

9.1 10 10 m/s

J s

kg

115 10 m

h

mv

hv hp

c E pc h h

p

De Broglie Example 3.2

2/16/2009 3313 Andrew Brandt 12

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

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

2 .938ppc m c GeV

15 8

15

4.136 10 3 10 /

1 10

ev 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


More about Wave Velocities

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

2/18/2009

2 / 2pv k

g

dv

dk

g

dv

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 box

2) 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

2/18/2009

2 22

2

1

2 2 2

p hKE mv

m m

22 2 2

2/ 8

2 (2 / )n

hE n h mL

m 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

2/18/2009

2 34 264 2

2 1 2

(6.63 10 )5.5 10

8 10 (10 )

n J sn 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

2/18/2009

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

( )sin

2

KN

KE


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)

2/23/2009

2 2

1 1 1( )

f i

Rn n

7 1 11.097 10 0.01097m nm


212

08n

m

EeE

r n

1 112 2 2

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

f i

E EE E E E eV

n n

5

110

1 10435

5.3 10nrna

512

7.19 10n

EE eV

n

12 2

1 1 1( )

f i

E

ch n n

3/2/2009

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 hc

n 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

3/4/2009

*( , ) ( , )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

3/4/2009

2xe 2

2

dG

dx

2

22

xdG e

dx 2xd d

edx dx

22 xde

dx 24 xe 4G

2

4 2 42

12d

G x x cxdx

Physics 3313 – Review 1 3/9/20091 3313 Andrew Brandt Monday March 9, 2009 Dr. Andrew Brandt.

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Transcript of Physics 3313 – Review 1 3/9/20091 3313 Andrew Brandt Monday March 9, 2009 Dr. Andrew Brandt.