Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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Physics 3313 – Review 1 3/8/2010 1 3313 Andrew Brandt Monday March 8 , 2010 Dr. Andrew Brandt

Transcript of Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

Page 1: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

Physics 3313 – Review 1

3/8/2010 13313 Andrew Brandt

Monday March 8 , 2010Dr. Andrew Brandt

Page 2: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. 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

116

1vc

Page 3: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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

sec

mvt x x x km

0 6 /16 0.38L

km

Page 4: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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 .80.962

.7 .81

x

c cv c

c cc

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'

'

21

xx

x

v vv

vv

c

Page 5: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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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) 0.4v

ac 1.09 8 81.09 1 0.4 3 10 1.31 10

sec sec

m kg mp mv kg

) 0.8v

bc 1.67 8 81.67 1 0.8 3 10 4.01 10

sec

kg mp

8) 1.2 10sec

kg mc p mv

8) 2.4 10sec

kg md

Page 6: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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

22 / 0.8 2.5

1

m cMc M kg kg

vc

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

Page 7: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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:

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slope=E/ =h (planck’s constant)

Page 8: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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 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 )

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346.626 10 J sech

maxKE h 0h h

Page 9: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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

1.24 104.5

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

Page 10: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

Photon Energy Loss

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e e

22 .511 2em c MeV

(1 cos )h

mc

Page 11: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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 behavior

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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

Page 12: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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 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

2ppc m c

Page 13: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

Wave Velocities

• Definition of phase velocity:

• Definition of group velocity:

• For light waves :

• For de Broglie waves

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2 / 2pv k

g

dv

dk

g

dv

k dk

g pv v c

Page 14: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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

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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

Page 15: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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 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

Page 16: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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

Page 17: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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

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Page 18: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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

Page 19: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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

3/8/2010 193313 Andrew Brandt

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

Page 20: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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

Page 21: Physics 3313 – Review 1 3/8/20101 3313 Andrew Brandt Monday March 8, 2010 Dr. Andrew Brandt.

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

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