Physics 9HE-Modern Physics Sample Final Exam (100 points total) ----------------------------------------------------------------------------------------------

Physics 9HE-Modern Physics Sample Final Exam (100 points total) ----------------------------------------------------------------------------------------------

1
Physics 9HE-Modern Physics
Sample Final Exam
(100 points total)
---------------------------------------------------------------------------------------------Miscellaneous data:
c = 3.00 x 108 m/s
e = 1.60 x 10-19 C
MSun = 2 x 1030 kg
1 eV = 1.60 x 10-19 J
1 Å = 10-10 m
MEarth = 5.98 x 1024 kg rEarth = 6.38 x 106 m
me = 9.1094 x 10-31 kg = 0.5110 MeV/c2
mp = 1.6726 x 10-27 kg = 938.27 MeV/c2
mn = 1.6749 x 10-27 kg = 939.57 MeV/c2
1 u = 1.6605 x 10-27 kg = 931.49 MeV/c2
m(1H) = 1.0078 u G = 6.67 x 10-11 Nt-m2/kg2
g = 9.81 m/s2
σ = 5.67 x 10-8 W-m-2-K-4
h = h/2π = 1.05 x 10-34 J-s
h = 6.63 x 10-34 J-s
kB = 1.38 x 10-23 J-K-1 a0 = 0.529 Å
N(t) = Noexp(-t/τO) = Noexp(-0.693t/τ1/2) kC ≡ 1/(4πεo) = 8.98 x 109 N-m2-C-2 RH = 1.09678 x 107 m-1
1
1
γ =
=
≅ 1 + 0.5 β 2 if v c
β = v/c = [(γ2-1)/γ2]1/2
2
2 1/ 2
(1 − v c )
(1 − β 2 )1/ 2
T= γT
o
L = Lo/γ
x ' = γ ( x − vt )
y'= y
u x' =
ux − v
⎛v ⎞
1 − ⎜ 2 ⎟ ux
⎝c ⎠
ν =ν0
z' = z
u y' =
(1 ± β )1/ 2
≅ ν 0 [1 ± β ] for β << 1
(1 m β )1/ 2
⎛v ⎞
⎝ ⎠
uz' =
⎤
γ ⎢1 − ⎜ 2 ⎟ u x ⎥
c
⎣
ν (lect.) = f(book)
⎡ ⎛v ⎞ ⎤
t ' = γ ⎢t − ⎜ 2 ⎟ x ⎥
⎣ ⎝c ⎠ ⎦
uy
⎡
ν = c/λ
⎦
uz
⎡ ⎛v ⎞ ⎤
γ ⎢1 − ⎜ 2 ⎟ u x ⎥
⎣ ⎝c ⎠ ⎦
x2 + y2 + z2 –c2t2 = invariant
r
r
p = γ mu
E = γ mc 2 = K + mc 2
Δν
E = pc = hν
ν
e
V tanθ
=
me
dB 2 l
=
E 2 = p 2 c 2 + m 2c 4
ΔT
GM ⎡ 1 1 ⎤
gH
=− 2 ⎢ − ⎥≈− 2
T
c ⎣ r1 r2 ⎦
c
λmaxT = 2.898 x 10-3 m-K
2
Δν
ν
I (λ ,T ) =
≈−
Δλ
λ
2π c h
2
λ
5
(if
Δν
ν
<< 1)
1
⋅
e
hc
λ kT
RSch =
R(T) = εσT
2GM
c2
4
−1
Ni nt ⎛ e ⎞
Z Z2
ZZe
Z Z e2
f (≥ θ ) = ntσ
b = 1 2 cot(θ / 2)
rmin = 1 2
⎜
⎟ 2 2
4
16 ⎝ 4πε o ⎠ r K sin (θ / 2)
4πε o K
8πε o K
h
hc
Δλ = λ ' − λ =
(1 − cos θ ) = 2.43 × 10 −12 (1 − cos θ ) ( in m ) hν = Kmax + φ λmin =
eV
mc
qq
mv 2
4πε 0 h 2 2
Z 2 e2
13.6Z 2
Fcoul = kC 1 2 2 Fradial =
=−
En = −
( eV ) rn =
n = a0
2
2
8πε 0 a0 n
n
me Ze 2
r
r
N (θ ) =
a0 =
2
2
1
4πε 0 h 2
= 0.529 Å
me e 2
λ = h/p
p = hk
e ± ix = cos x ± i sin x
sin 2t = 2 sin t cos t
vn =
2
nh
me rn
ΔxΔpx ≥ h/2
2
⎡ 1
1 ⎤
= Z 2 RH ⎢ 2 − 2 ⎥
λ
nu ⎦
⎣ nl
1
ΔEΔt ≥ h/2
⎡
M
v ph = ω /k
v gr = d ω /dk
1 ix
1 ix
⎡⎣e + e − ix ⎤⎦ sin x =
⎡e − e − ix ⎤⎦
2
2i ⎣
cos 2t = cos2 t – sin2 t = 2 cos2 t – 1 = 1 – 2 sin2 t
cos x =
⎤
μe = me ⎢
⎥
⎣ M + me ⎦
= ntπ b 2
nλ = 2dsinθ
n2
Z
2
∞
∞
a0
2π n
2π n
2π n
+ ∑ an cos(
x ) + ∑ bn sin(
x ), with k n =
, and
2 n =1
l
l
l
n =1
2 l
2π n
2 l
2π n
a0 = aver . of ψ over l , an = ∫ ψ ( x')cos(
x')dx', bn = ∫ ψ ( x') sin(
x')dx'
ψ(x ) =
l
1
ψ(x ) =
2π
∫
∞
−∞
c( k )e ikx dk , with c( k ) =
∂Ψ
Hˆ Ψ = i h
∂t
Hˆ = Kˆ + V
∂
pˆ x = −i h
∂x
h2 ∂ 2
ˆ
Kx = −
2m ∂ x 2
k=
2m( E − V )
h2
1/ 2
⎛2⎞
⎝ ⎠
ψn = ⎜ ⎟
L
ψ ∝ e ± ikx
⎛ nπ x ⎞
sin ⎜
⎟
⎝ L ⎠
ψ n = Hn ( x )e −α x
2
/2
⎡
V 2 sin 2 ( kL ) ⎤
T = ⎢1 + 0
⎥
4 E ( E − V0 ) ⎦
⎣
TFE
mκ
−1
2
h
1
2π
∫ ψ ( x')e
y = r sin θ sin φ
l
− ikx '
l
0
dx'
−∞
Ψ =ψe
−
iE t
ΔA =
h
Hˆ ψ = Eψ
= ψ e − iω t
2
A2 − A ∞
κ =α =
ˆ ψ dx
< A >= ∫ ψ * A
ˆ ψ = aψ
A
a
a
2m(V − E )
h2
π 2 h 2 n2
ψ ∝ e ±κ x
2mL2
ω=
κ
m
1⎞
⎛
En = ⎜ n + ⎟ h ω ;
2
⎝
⎠
⎡
V 2 sinh 2 ( κ L ) ⎤
T = ⎢1 + 0
⎥
4 E (V 0 − E ) ⎦
⎣
⎡
⎤
⎢ 4φ 3 / 2 2me ⎥
≈ exp ⎢ −
⎥
dV ⎥
⎢
3e h
dx ⎥⎦
⎣⎢
x = r sin θ cos φ
∞
ψ ∝ sin kx,cos kx
En =
α=
l
0
ISTM ∝ e −2 KL
z = r cos θ
λ = fcollTα
−1
≈ 16
Erot =
E
V0
h 2 l ( l + 1)
2I
I=
mR1
2
⎡
E ⎤ −2 κ L
( when κ L >> 1 )
⎢1 −
⎥e
V
0 ⎦
⎣
⎡
ZR nuc ( m ) ⎤
0.0993 MeV
+8
Tα = exp ⎢ − 4 π Z
⎥
E α ( MeV )
7.3 × 10 − 15 ⎦
⎣
dV = r 2 dr sin θ dθ dφ
ψ nl ml ( r ,θ , φ ) = Rnl ( r )Θ l ml ( θ )Φ ml ( φ ) = Rnl ( r )Yl ml ( θ ,φ ) Pnl ( r ) = r 2 Rn2l ( r )
r
r
r r
r
r
e r
L
e r
s
μ = iA E = − μ B
μl = −
L = − μB
μ s = − s = −2 μ B
2m
m
h
h
μB ≡
eh
= 9.274 × 10 −24 J / Tesla
2me
r
ψ MO
(r ) =
j
∑
r
μN ≡
eh
= 5.058 × 10 −27 J / Tesla
2mp
c Ai , j ϕ Ai ( r )
Atoms A
Orbitals i
B( ZA X ) = [ Nmn + Zm( 1H ) − M( ZA X )]c 2 ≈ aV A − aA A2 / 3 −
Rnuc ≈ (1.2 × 10 −15 m )A1 / 3
3 Z( Z − 1)e 2
( N − Z )2
− aS
+δ
5 4πε 0 Rnuc
A
Q = [ Minitial − Mfinal ] c 2
---Tear off this sheet and begin exam---
1
Physics 9HE-Modern Physics
Sample Final Examination
(100 points total)
Name (printed)_____SAMPLE ANSWERS_____
Name (signature)_____________________________________
Student ID No.____________________________________
[1.1] (10 points) The free neutron at rest decays into a proton and an electron with
an exponential lifetime of τ0 = 920 s. Now consider a beam of neutrons with a kinetic
energy of 100 MeV travelling along the +x direction relative to an observer in the
laboratory.
(a) At what speed will the neutrons in the beam be moving?
(b) What would the observer in the laboratory measure for the lifetime of the neutrons
in the beam?
2
[1.2] (25 points) Answer the following questions with brief statements or
calculations
(a) State three types of experimental observations that challenged classical physics
around 1900.
Any three of:
Blackbody radiation
Electromagnetism and the origin of magnetic fields, which did not transform
from one coordinate system to another
Line spectra of hydrogen and other atoms
The photoelectric effect
Compton scattering
(b) State three experimentally verified consequences of Special Relativity.
Any three of:
Time dilation, e.g. in planes circling earth
Length contraction, e.g. in muon viewing mountain
Relativistic momentum: effective mass increase of moving object as viewed in
fixed frame
Relativistic energy: E = mc2
Special version of Doppler effect
B fields as a relativistic manifestation of E fields
(c) So-called L x-rays are emitted from a copper atom in which an initial 2p vacancy is
created. What transitions from the n = 3 shell are permitted in generating these x-rays
and why?
Dipole selection rules are: Δl = ±1 and Δml = 0, ±1, so the allowed transitions are
from 3d with Δl = 1-2 = -1 and 3s with Δl = 1-0 = +1, and this would get full credit. In
more detail using the other selection rule, we would have:
-2
-1
0
+1
+2
3d: l = 2, ml = _______
_______
_______
_______
_______
3s: l = 0, ml = _______
0
2p: l = 1, ml =
-1
0
+1
_______ _______ _______
3
(d) A red laser is used to produce a perfect sinusoidal traveling light wave of frequency
νo = 1014 Hz. However, by special means, the wave is abruptly cut off at each end so
that it has a finite length in time of 10-13 s. If this finite wave is now described in a
Fourier representation, what would be the approximate range of frequencies Δν
involved?
z
(d) Define degeneracy and give one example of it from the systems we have studied.
(e) The position of an electron along the x direction is measured with an uncertainty
of 1 Å. What can you say about the uncertainty in its x momentum? What, if anything,
can you say about the uncertainty in its y momentum?
(f) In the oxygen molecule pointing along the x direction, one of the bonding wave
functions can be made up of oxygen 2px functions on the two atoms. Indicate the
4
equation for this wave function, and sketch it, using a 3D contour of equal probability
density.
ψ Molecular ∝ φO2 p
Orbital
x
on atom 1
− φO2 px on atom 2
( Simple + s ign OK too )
-
+
+
-
x
5
[1.3] (15 points) Consider a hydrogenic atom consisting of a single electron
around a uranium nucleus (Z = 92).
(a) What will be the energy in eV and the radius in Å of the first Bohr orbit for this
atom?
(b) Balance coulombic force and acceleration and use the result of part (a) to derive
the speed of the electron in the first Bohr orbit around this atom, using a nonrelativistic approach. Based on your answer, is a non-relativistic model adequate?
c
(c) Now consider this hydrogenic atom initially in an n = 10 excited state, from which it
deexcites to the n = 1 state by emitting electromagnetic radiation. What would be the
weight change of the atom due to emitting this radiation, including its sign (increase or
decrease)?
Δm =
ΔE
c2
1
1
{ 2 − 2}
E10 − E1
2
1
10
(1.602 x10 −19 J / eV )
=
= (13.6eV )( 92 )
c2
( 3 x10 8 )2
= 20,284 x10 −35 kg = 2.03 x10 −31 kg, about 1 / 5 the mass of an electron !
Energy lost , so mass decreases : Δ m = negative
6
[1.4] (10 points) Consider an electron trapped inside the potential well shown below,
at an energy of 200 eV, at the position indicated:
∞
Higher amplitude and
Longer wavelength in 3
•
•
e-
…∞
Classical turning points
Zero here
Exponential decay:
exp(-κ4x)
sin k2x
sin k3x, cos k3x
1
3
2
0Å
50 Å
4
100Å
x
(a) Qualitatively sketch on the diagram the form of the wavefunction in each of the
regions 1-4 indicated, being careful to show the relative wavelengths and the relative
amplitudes in each region. (That is, in which region of the box will the particle be most
likely to be found?)
From (c) below, the wavelength is pretty small compared to this well’s
dimensions, so there are something like 50/0.868 = 57 cycles in each of regions 2 and
3. Too hard to draw this, but I have indicated relative heights of ψ such that |ψ|2 is
inversely proportional to the particle velocity in the region (a correspondence
principle type of argument), which is in turn proportional to the square root of kinetic
energy. So |ψ2|2/|ψ3|2 ≈ [KE3/KE2]1/2 = [100/200]1/2 = 0.707 and finally ψ2/ψ3 ≈ 0.840.
Particle is more likely to be found in region 3.
Full credit here was simply showing a smaller wavelength and amplitude in
region 2.
(b) Indicate the classical turning points appropriate to this state on the diagram as well.
See diagram. Two of them.
7
(c) For the energy values shown on the diagram, what is the wavelength of the particle
in region 2?
In general, λ = h/p, and in non-relativistic limit energy of particle in region 2 = E2 =
KE = kinetic energy = p2/2m, so p = sqrt[2mE] and finally λ = h/sqrt[2mE].
With E = 200 eV = 200(1.60 x 10-19 J/eV) = 3.20 x 10-17J, we thus have finally, in SI
units:
λ = h/sqrt[2mE] = [6.63 x 10-34J-s]/sqrt[2(9.11 x 10-31kg)(3.20 x 10-17J)
= [6.63 x 10-34]/7.63 x 10-24 = 0.868 x 10-10 m = 0.868 Å
[1.5] (15 Points) Consider an electron incident from the left side of the infinitely
long potential step at x = 0 shown below. The particle energy E = 10 eV and the step
has a height of Vo = 7 eV.
e-
E = 10 eV
I
V0 = 7 eV
II
x
(a) What is the relevant time-independent Schroedinger equation to the left of the step
(region I) and to the right of the step (region II)? You need not solve these equations.
8
(b) What would be the time-dependent form of the wave function for this problem in
both regions, including both transmitted and reflected components and taking
advantage of the known general solutions to this type of problem?
9
(d) Now set up the boundary conditions needed to solve this problem, but you need
not go beyond this.
OK to this line
[1.6] (15 Points)
Two of the 3d hydrogen-atom wavefunctions are given by:
ψ 3 ,2 ,±2 ( r ,θ , φ ) = ψ 3d = Cr e
2
±2
−
r
3ao
sin 2 θ e ±2 iφ
(a) Indicate how the probability density associated with these wavefunctions would be
calculated, and how the normalization constant C would be determined from this
density, going as far as you can without evaluating any non-trivial integrals.
(b) Sketch and describe in words in an unambiguous way the three-dimensional
probability density of these two hydrogenic wave functions, using either a 3D contour
of equal probability or a plot in which greater darkness means a higher probability of
finding the electron at a given position.
10
(c) Write down three eigenfunction relations satisfied by these wave functions,
including a specification of the eigenvalues involved and the physical meaning of the
eigenvalue.
(d) Write down the time-dependent form of these two wave functions.
Just add the usual complex exponential as:
ψ 3 ,2 ,±2 ( r , θ , φ ) = ψ 3d = Cr e
2
−
r
3 ao
±2
6
where ω = E3 / h = −(13.6eV )
sin 2 θ e ±2 iφ e − iωt
12
1.51 eV
=−
2
3 h
h
[1.7] (10 Points) Consider the nuclide 2449Crx of chromium, with atomic mass of
48.951341 u.
(a) What is x here, and what is the binding energy of this nuclide per nucleon?
(b) This nuclide decays by positron emission to form a nuclide of vanadium =
V. Write down the overall reaction, including the new nuclide that would be formed.
Positron emission reduces the atomic no. Z by one, and overall reaction is:
+
Cr25 → 49
23V26 + e + ν e
49
24
11
OK without neutrino stated.
(c) Which one of the four fundamental interactions is responsible the force between
quarks and what is the particle mediating this force?
The strong interaction is responsible for quark interaction and the mediating
particle is the gluon.
---End of examination---