PHYSICS 2D
PROF. HIRSCH
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FINAL EXAM
Formulas:
Time dilation; Length contraction : Δt = γΔt'≡ γ Δt p ;
L = Lp /γ
; c = 3 ×10 8 m /s
Lorentz transformation : x'= γ (x − vt) ; y' = y ; z' = z ; t'= γ (t − vx /c 2 ) ; inverse : v → -v
uy
ux − v
Velocity transformation : ux '=
; uy '=
; inverse : v → -v
2
γ (1− ux v /c 2 )
1− ux v /c
Spacetime interval : (Δs) 2 = (cΔt) 2 - [Δx 2 + Δy 2 + Δz 2 ]
Relativistic Doppler shift : f obs = f source 1+ v /c / 1− v /c
r
r
Momentum : p = γ mu ; Energy : E = γ mc 2 ; Kinetic energy : K = (γ −1)mc 2
Rest energy : E 0 = mc 2
;
Electron : me = 0.511 MeV /c 2
E=
p 2c 2 + m 2c 4
Proton : mp = 938.26 MeV /c 2
Neutron : mn = 939.55 MeV /c 2
Atomic mass unit : 1 u = 931.5 MeV /c 2
; electron volt : 1eV = 1.6 ×10 -19 J
4
Stefan's law : etot = σT , etot = power/unit area ; σ = 5.67 ×10−8 W /m 2K 4
∞
hc
etot = cU /4 , U = energy density = ∫ u( λ,T)dλ ;
Wien's law : λm T =
4.96kB
0
-E/(kB T )
Boltzmann distribution : P(E) = Ce
8π
hc / λ
8πf 2
Planck's law : uλ ( λ,T) = N λ ( λ) × E ( λ,T) = 4 × hc / λkB T
;
N( f ) = 3
λ
e
−1
c
Photons : E = hf = pc ; f = c / λ ; hc = 12,400 eV A ; k B = (1/11,600)eV /K
Photoelectric effect : eVs = K max = hf − φ , φ ≡ work function; Bragg equation : nλ = 2d sin ϑ
Compton scattering : λ'- λ =
h
(1 − cos θ ) ;
mec
h
= 0.0243A
mec
kq q
kq
kq q
Coulomb force : F = 12 2 ; Coulomb energy : U = 1 2 ; Coulomb potential : V =
r
r
rr
r
r r
Force in electric and magnetic fields (Lorentz force): F = qE + qv × B
1
Z2
ke 2 = 14.4 eV A
Rutherford scattering : Δn = C 2
4
Kα sin (φ /2)
1
1
1
1
Hydrogen spectrum :
= R( 2 − 2 )
;
R = 1.097 ×10 7 m−1 =
λmn
m
n
911.3A
2
2
2
2
Z
ke Z
ke
me (ke )
mev 2
ke 2 Z
Bohr atom : E n = −
= −E 0 2 ; E 0 =
=
= 13.6eV ; K =
; U =−
n
2a0
2
2rn
2h 2
r
hf = E i − E f ; rn = r0 n 2 ; r0 =
a0
Z
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WINTER QUARTER 2016
MARCH 15, 2016
de Broglie : λ =
h
E
;f =
p
h
; a0 =
h2
= 0.529A ; L = me vr = nh angular momentum
me ke 2
; ω = 2πf ; k =
2π
;
λ
Wave packets : y(x,t) = ∑ a j cos(k j x − ω j t), or y(x,t) =
E = hω ; p = hk ;
∫ dk a(k) e
i(kx -ω (k )t )
E=
p2
2m
; ΔkΔx ~ 1 ; ΔωΔt ~ 1
j
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group and phase velocity : v g =
dω
ω
; vp =
;
dk
k
Heisenberg : ΔxΔp ~ h ; ΔtΔE ~ h
PHYSICS 2D
PROF. HIRSCH
FINAL EXAM
WINTER QUARTER 2016
MARCH 15, 2016
b
Probability: P(x)dx =| Ψ(x) |2 dx
;
P(a ≤ x ≤ b) =
!c = 1973 eVA
∫ dxP(x)
a
E
-i t
h2 ∂ 2Ψ
∂Ψ
+ U(x)Ψ(x,t) = ih
;
Ψ(x,t) = ψ (x)e h
2
2m ∂x
∂t
∞
h 2 ∂ 2ψ
Time − independent Schrodinger equation : +
U(x)
ψ
(x)
=
E
ψ
(x)
;
∫ dx ψ *ψ = 1
2m ∂x 2
-∞
Schrodinger equation : -
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∞ square well : ψ n (x) =
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π 2h2n 2
2
nπx
sin(
) ; En =
2mL2
L
L
Harmonic oscillator : Ψn (x) = H n (x)e
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−
mω 2
x
2h
;
h2
= 3.81eVA 2 (electron)
2me
1
p2 1
1
; E n = (n + )hω ; E =
+ mω 2 x 2 = mω 2 A 2 ; Δn = ±1
2m 2
2
2
Expectation value of[Q] :< Q >= ∫ ψ * (x)[Q]ψ (x) dx ; Momentum operator : p =
Eigenvalues and eigenfunctions : [Q] Ψ = q Ψ (q is a constant) ; uncertainty :
(k1 − k 2 ) 2
Step potential : reflection coef : R =
, T = 1− R ;
(k1 + k 2 ) 2
k=
h ∂
i ∂x
ΔQ = < Q2 > − < Q > 2
2m
(E − U)
h2
x2
∫
-2 α (x )dx
Tunneling :
ψ (x) ~ e -α x
;
T = e -2αΔx ;
T =e
;
x1
α (x) =
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2m[U(x) - E]
h2
r
r
r
r -i E t
h2 2
∂Ψ
∇ Ψ + U( r )Ψ( r ,t) = ih
;
Ψ( r ,t) = ψ ( r )e h
2m
∂t
2 2
π h n12 n 22 n 32
3D square well : Ψ(x,y,z) = Ψ1 (x)Ψ2 (y)Ψ3 (z) ; E =
( +
+ )
2m L12 L22 L23
Spherically symmetric potential: Ψn,l,m (r,θ, φ ) = Rnl (r)Ylm (θ, φ ) ; Ylm (θ, φ ) = Plm (θ )e im φ
Schrodinger equation in 3D : -
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l
l
l
l
l
r r r
h ∂
; [L2 ]Ylm l = l(l +1)h 2Ylm l ; [L z ]Ylm l = ml hYlm l
Angular momentum : L = r × p ; [Lz ] =
i ∂φ
ke 2 Z 2
Radial probability density : P(r) = r 2 | Rnl (r) |2 ;
Energy : E n = −
2a0 n 2
1 Z
Ground state of hydrogen and hydrogen - like ions : Ψ1,0,0 = 1/ 2 ( ) 3 / 2 e−Zr / a 0
π
a0
→
→
€ Orbital magnetic moment : µ = −e L ; µz = −µB ml ; µB = eh = 5.79 ×10−5 eV /T
2me
2me
r
1
−e r
Spin 1/2 : s = , | S |= s(s + 1)h ; Sz = msh ; ms = ±1/2 ;
µs =
gS
2
2me
g=2
r
−e r
r r
r
µ=
( L + gS ) ; Energy in mag. field : U = −µ⋅ B
Orbital + spin mag moment :
2me
r r
r r
Two particles : Ψ(r1, r2 ) = + /− Ψ( r2 , r1 ) ; symmetric/antisymmetric
Screening in multielectron atoms : Z → Z eff , 1 < Z eff < Z
Orbital ordering: 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<6d~5f
PHYSICS 2D
PROF. HIRSCH
FINAL EXAM
WINTER QUARTER 2016
MARCH 15, 2016
Problem 1
The moving walkway at the airport moves at speed 0.5c. You are about to miss your
plane, so you run on the walkway at speed 0.5c with respect to the walkway to get to your
gate. According to your watch, you were 60 seconds on the walkway and got to the gate
right on time, unfortunately according to the agent at the gate you missed the plane by:
A: 10 s ; B: 20 s; C: 30 s; D: 40 s; E: not sure
(E always counts 0.31 pts)
Problem 2
A mass m moving at speed 0.8c collides inelastically with a mass m at rest, the two
masses stick together forming a mass M that is equal to
A: 2m; B: 2.14m; C: 2.31m; D: 2.66m; E: not sure
Problem 3
A 60W light bulb emits maximum power at wavelength 5700A. Its surface area is
A: 1.58mm2; B: 2.44mm2; C: 3.81mm2; D: 5.32mm2; E: not sure
Problem 4
In a photoelectric experiment with a certain metal it is found that the stopping voltage
(voltage that will stop the photocurrent) is 1.5V when the incident light has wavelength
4000A. What is the maximum wavelength of light that will give rise to a photocurrent for
this metal?
A: 8650A; B: 7750A; C: 6850A ; D:5950A; E: not sure
Problem 5
In a Compton scattering experiment with incident photons of wavelength 1A, the
scattered electrons at a certain angle θ have kinetic energy 150eV. The angle θ is
approximately
A: 30o; B: 45o; C: 60o; D: 90o; E: not sure
Problem 6
In a Rutherford scattering experiment with atoms of atomic number Z=35 and α particles
of incident kinetic energy 7 Mev, it is found that for every 800 α particles scattered at a
90o angle there are approximately 100 α particles scattered at angle 180o. What can you
say about the radius R of this nucleus? (1fm=10-5A).
A: R<14.4 fm; B: R>14.4 fm; C: R<7.2 fm; D: R>7.2 fm; E: not sure
Problem 7
The ratio of speeds of an electron in Bohr orbits of angular momentum 3! and angular
momentum 6! is:
A: 1/4; B: 4; C 1/2; D: 2; E: not sure
Problem 8
A proton and an electron both have de Broglie wavelength λ=0.001A. The ratio of the
speed of the proton to the speed of the electron is (Hint: consider whether you need to use
relativity)
A: 0.00054; B: 0.0028; C: 0.013; D: 0.11; E: not sure
PHYSICS 2D
PROF. HIRSCH
FINAL EXAM
WINTER QUARTER 2016
MARCH 15, 2016
Problem 9
For an electron wavepacket with peak at k=0.2A-1, the ratio of phase velocity to group
velocity is
A: 1; B: 1/2; C: 2; D: >10; E: not sure
Problem 10
An electron moving in a one-dimensional potential U(x) is in a stationary state with
energy E. Its wavefunction in the interval 0<x<5A is
ψ (x) = Cx 4
with C a constant. What is the difference in the values of its potential energy U(x) at
positions x1=0.5A and x2=1A, i.e. what is U(x1)-U(x2)?
A: 137eV; B: 84eV; C: 3.81eV; D: 7.62eV; E: not sure
Hint: use the Schrodinger equation
Problem 11
An electron in a one-dimensional box of length L can emit photons of wavelength exactly
1000A and 2000A when it makes a transition from an excited state to the ground state.
What is the length of the box?
A: 3A; B: 6A; C: 9A; D: 12A; E: not sure
Problem 12
The classical amplitude for an electron in the first excited state of a one-dimensional
harmonic oscillator potential is 3A. What is the ground state energy?
A: 1.27eV; B: 2.54eV; C: 3.81eV; D: 5.08eV; E: not sure
Problem 13
Which of the following functions is an eigenfunction of the operator x 2
2
∂
?
∂x
2
A: e− λ x ; B: e− λ x ; C: e− λ /x ; D: e− λ /x ; E: not sure
Problem 14
3eV
2eV
For electrons incident with kinetic energy 1 eV, the transmission probability through the
two barriers shown above, of height 3eV and 2eV, is the same. If the 3eV barrier has
width 1A, the width of the 2eV barrier is
A: 0.71A; B: 1.41A C: 2A; D: 0.5A; E: not sure
PHYSICS 2D
PROF. HIRSCH
FINAL EXAM
WINTER QUARTER 2016
MARCH 15, 2016
Problem 15
For an electron in the ground state of a two-dimensional box of side length L=4A, the
probability of finding the electron in a small region around (x=2A, y=2A) is how many
times larger than the probability of finding the electron in a small region (of the same
area) around (x=1A, y=1A)?
A: 2; B: 4 C: 8; D: 16; E: not sure
Problem 16
The wavefunction for an electron in a hydrogen-like ion with nuclear charge Ze is
ψ (r, θ , φ ) = C(1− Dr + Fr 2 )e−r/a0
where C, D, and F are constants
The values of the quantum numbers n, l, ml are
A: 3, 2, 1; B: 4, 0, 0; C: 3, 0, 0; D: 2, 0, 0; E: not sure
Problem 17
€ when the electron in the wavefunction of problem 16
The energy of the photon emitted
makes a transition to the ground state is, in terms of E0=13.6eV:
A: 2E0 B: 4E0; C: 8E0; D: E0; E: not sure
Problem 18
The radial wavefunction for an electron in a hydrogen-like ion is
R(r) = Cre−r/(2a0 )
The average radius, i.e. < r > , is
A: 2a0 ; B: 2.5a0; C: 3.5a0; D: 5a0; E: not sure
∞
n!
Hint: use ∫ dx x n e − λx = n +1
λ
€
0
Problem 19
For an electron in hydrogen with quantum numbers n=4, ℓ =3, s=1/2, in a magnetic field
B=10T, the difference in energy between the highest and lowest states, assuming the
€
spin-orbit interaction can be ignored, is
A: -4.05x10-3eV; B: 4.63x10-3eV; C: 2.32x10-3eV ; D: 2.03x10-3eV; E: not sure
Problem 20
The ground state energy for an electron in a cubic box is Eg. Taking into account electron
spin and the Pauli exclusion principle and assuming no interaction energy between
electrons, if there are 14 electrons in this box their total energy is:
A: 32Eg B: 36Eg; C: 24Eg; D: 28Eg; E: not sure
Problem 21
Estimate the kinetic energy of an electron confined inside a nucleus of diameter 10-5A.
A: 4x106eV B: 2x108eV; C: 4x1010eV; D: 2x1012eV; E: not sure