Download Common Exam - 2009 Department of Physics University of Utah August 22, 2009

Common Exam - 2009
Department of Physics
University of Utah
August 22, 2009
Examination booklets have been provided for recording your work and your solutions.
Please note that there is a separate booklet for each numbered question (i.e., use
booklet #1 for problem #1, etc.).
To receive full credit, not only should the correct solutions be given, but a sufficient
number of steps should be given so that a faculty grader can follow your reasoning.
Define all algebraic symbols that you introduce. If you are short of time it may be helpful
to give a clear outline of the steps you intended to complete to reach a solution. In some
of the questions with multiple parts you will need the answer to an earlier part in order to
work a later part. If you fail to solve the earlier part you may represent its answer with an
algebraic symbol and proceed to give an algebraic answer to the later part. This is a
closed book exam: No notes, books, or other records should be consulted. YOU MAY
ONLY USE THE CALCULATORS PROVIDED. The total of 250 points is divided
equally among the ten questions of the examination.
All work done on scratch paper should be NEATLY transferred to answer booklets.
SESSION 1
COMMON EXAM DATA SHEET
e = - 1.60 × 10-19 C = - 4.80 × 10 -10 esu
c = 3.00 × 108 m/s = 3.00 × 10 10 cm/s
h = 6.64 × 10 -34 JAs = 6.64 × 10 -27 ergAs = 4.14 × 10 -21 MeVAs
S = 1.06 × 10 -34 JAs = 1.06 × 10 -27 ergAs = 6.59 × 10 -22 MeVAs
k = 1.38 × 10 -23 J/K = 1.38 × 10 -16 erg/K
g = 9.80 m/s2 = 980 cm/s 2
G = 6.67 × 10-11 NAm2/kg2 = 6.67 × 10 -8 dyneAcm 2/g 2
NA = 6.02 × 1023 particles/gmAmole = 6.02 × 10 26 particles/kgAmole
go(SI units) = 8.85 × 10 -12 F/m
:o(SI units) = 4B × 10 -7 H/m
m(electron) = 9.11 × 10 -31 kg = 9.11 × 10 -28 g= 5.4859 × 10 -4 AMU = 511 keV
M(proton) 1.673 × 10 -27 kg = 1.673 × 10 -24 g = 1.0072766 AMU = 938.2 MeV
M(neutron) 1.675 × 10 -27 kg = 1.675 × 10 -24 g = 1.0086652 AMU = 939.5 MeV
M(muon) = 1.88 × 10 -28 kg = 1.88 × 10 -25 g
1 mile = 1609 m
1 m = 3.28 ft
1 eV = 1.6 × 10 -19 J = 1.6 × 10 -12 ergs
hc = 12,400 eVAD
Table of Integrals and Other Formulas
Spherical Harmonics
Conic Section in Polar Coordinates
(origin at the focus; e = 1 for parabola)
Normal Distribution
Cylindrical Coordinates (orthonormal bases)
Spherical Coordinates (orthonormal bases)
Maxwell’s Equations (Rationalized MKS)
Maxwell’s Equations (Gaussian Units)
Problem 1 – Electromagnetism
An insulating sphere of radius R carries an electric charge Q uniformly distributed in its volume. At time
t = 0, the sphere is made conductive with a constant and uniform conductivity σ.
(a) [9 pts.] Using the continuity equation, Ohm’s law and Gauss’s law, write a differential equation for the
charge density ρ(r, t) as a function of time and find its solution in the form ρ(r, t) = ρ0 f (t).
(b) [8 pts.] Write the expression of the magnitude of the electric field E(r, t) and current density j(r, t)
inside the sphere.
(c) [8 pts.] What is the total amount of energy dissipated by Joule effect between t = 0 and t = ∞?
(Hint: the Joule dissipation power density is
dP
dV
= j(r, t)E(r, t).)
Problem 1 – Electromagnetism – sample solution
An insulating sphere of radius R carries an electric charge Q uniformly distributed in its volume. At time
t = 0, the sphere is made conductive with a constant and uniform conductivity σ.
(a) [9 pts.] Using the continuity equation, Ohm’s law and Gauss’s law, write a differential equation for the
charge density ρ(r, t) as a function of time and find its solution in the form ρ(r, t) = ρ0 f (t).
∂ρ
+ ∇j
∂t
get ∂ρ
+ σ0 ρ
∂t
Charge conservation writes
Gauss’ law ∇E =
− σ t
0
ρ(r, t) = ρ0 e
ρ
,
0
we
= 0. Then, Ohm’s law writes j = σE so
∂ρ
∂t
+ σ∇E = 0 and with
= 0. This differential equation is satisfied by a solution of the form
where ρ0 is the initial charge density: ρ0 =
3Q
.
4πR3
(b) [8 pts.] Write the expression of the magnitude of the electric field E(r, t) and current density j(r, t)
inside the sphere.
From Gauss’ law, with r < R, E(r, t) =
that j(r, t) =
4πr 3 ρ(t)
3×4πr2 0
rρ(t)
30
=
=
− σ t
Q·r
0 .
e
3
4π0 R
From Ohm’s law it then follows
σ
σ·Q·r − 0 t
e
3
4π0 R
(c) [8 pts.] What is the total amount of energy dissipated by Joule effect between t = 0 and t = ∞? (Hint:
the Joule dissipation power density is
The dissipation power density is
sphere is P =
σ·Q2 ·
(4π0 R3 )2
−2 σ t
0
e
RR
0
dP
dV
dP
dV
= j(r, t)E(r, t).)
r2 · 4πr2 dr =
obtained by time integration:∆U =
σ
σ·Q2 ·r 2 −2 0 t
e
(4π0 R3 )2
−2 σ t
= j(r, t)E(r, t) =
R∞
0
P dt
σ·Q2 ·
e
20π20 R
Q2
= 40π
.
0R
0
so the total power dissipated in the
. The total amount of energy dissipated is then
Problem 2 – Quantum Mechanics
An initial state of a particle of mass m in an infinite one-dimensional potential well of width a is described
by the following wave function
s
f (x) =


12 

a 
x
,
a
0≤x≤
a−x a
, 2
a
a
2
≤x≤a
(a) [10 pts.] Write down explicit expression for coefficients cn that determine f (x) in terms of the eigenstates
ψn (x) =
(b) [2 pts.]
q
2
a
sin[πnx/a] of the problem: f (x) =
P∞
n=1 cn ψn (x).
Express probability Pn to find the particle in the n-th eigenstate (with energy En =
h̄2 π 2 n2 /(2ma2 )) in terms of cn .
(c) [6 pts.] Calculate probability P2 to find the particle in n = 2 state.
(d) [6 pts.] Calculate the ratio P3 /P1 .
(e) [1 pts.] What is the most probable outcome of the energy measurement for this problem?
f(x)
0
a
FIG. 1: The potential well and the initial wave function.
Problem 2 – Quantum Mechanics – sample solution
An initial state of a particle of mass m in an infinite one-dimensional potential well of width a is described
by the following wave function
s
f (x) =


12 

a 
x
,
a
0≤x≤
a−x a
, 2
a
a
2
≤x≤a
(a) [10 pts.] Write down explicit expression for coefficients cn that determine f (x) in terms of the eigenstates
ψn (x) =
q
2
a
P∞
n=1 cn ψn (x).
sin[πnx/a] of the problem: f (x) =
Z
a
cn = hψn |f i =
dxψ(x)f (x) =
√ Z 0
πnx x Z a
πnx a − x
24 a/2
=
dx sin[
dx sin[
(
] +
]
) = (z = x/a)
a
a a
a
a
0
a/2
Z 1
√ Z 1/2
dz sin[πnz](1 − z)).
= 24(
dz sin[πnz]z +
(1)
1/2
0
Introduce substitution y = 1 − z in the second integral: it now reads
Z
1/2
dy sin[πn − πny]y = − cos[πn]
Z
1/2
dy sin[πny]y = (−1)n+1
Hence
√
1/2
dy sin[πny]y.
0
0
0
Z
√
1/2
24 Z πn/2
cn = [1 + (−1) ] 24
dx sin[πnx]x = [1 + (−1) ] 2 2
dx sin[x]x.
π n 0
0
Thus cn = 0 for all even values of quantum number n. For odd values of n,
√
√
πn/2
2 24
πn
2 24 = 2 2 sin
.
cn = 2 2 sin[x] − x cos[x] 0
π n
π n
2
n+1
(b) [2 pts.]
Z
n+1
(2)
Express probability Pn to find the particle in the n-th eigenstate (with energy En =
h̄2 π 2 n2 /(2ma2 )) in terms of cn .
Pn = |cn |2 = c2n .
(c) [6 pts.] Calculate probability P2 to find the particle in n = 2 state. c2 = 0 since f (x) is even while ψ2 (x)
is odd. This also follows from the general result in Eq.(2). Thus P2 = 0.
(d) [6 pts.] Calculate the ratio P3 /P1 .
P3 /P1 = (c3 /c1 )2 .
Now,
√ Z
c1 = 2 24
0
1/2
√
2 24
dx sin[πx]x =
,
π2
and
√ Z
c3 = 2 24
0
1/2
√
−2 24
dx sin[3πx]x =
.
9π 2
Thus c3 /c1 = −1/9 and P3 /P1 = 1/81.
(e) [1 pts.] What is the most probable outcome of the energy measurement for this problem? With probability P1 =
96
π4
= 98.6% the energy measurement will result in E1 = h̄2 π 2 /(2ma2 ).
Problem 3 – Classical Mechanics
In this problem we will study some mechanical properties of the CO2 molecule. The CO2 molecule is an
axial molecule (see figure below). The distance between oxygen and carbon atoms is r, the mass of the carbon
atom is 12m and the mass of the oxygen atom is 16m, where m is the mass of the hydrogen atom.
(a) [3 pts.] Find a moment of inertia of the CO2 molecule with respect to the axis that is perpendicular to
the axis of the molecule and goes through the center of the carbon atom.
(b) [7 pts.] A hydrogen atom with mass m moving with speed v0 collides with one of the carbon atoms
of the CO2 molecule at the right angle to the molecule axis (see figure below). The CO2 molecule is
initially at rest. The hydrogen atom bounces straight back and CO2 molecule acquires translational and
rotation motion. We assume that CO2 is rigid. Write down the laws of momentum, angular momentum
and energy conservation for the described collision process.
(c) [6 pts.] From the system of equations found in (b) compute the angular frequency of rotation of the
CO2 molecule after the collision.
(d) [9 pts.] Let us consider an inelastic collision process that has the same geometry as in (b) and leads to
the formation of a chemical bond between the hydrogen and one oxygen atom. The kinetic energy in
this collision is not conserved. Compute the angular frequency of rotation of the HO-C-O molecule after
the collision. For simplicity assume that the molecule retains an axial symmetry, the length of both C-O
bonds is unchanged, and that the H-O group can be considered as a point mass with mOH = 17m.
Problem 3 – Classical Mechanics – sample solution
Let us define angular frequency of rotation of the CO2 molecule as (ω) and velocities of hydrogen atom and
the CO2 molecule after collision (v1 ) and (v2 ), respectively. (a)
J=
X
mi ri2 = 2 × 16m × r2 = 32mr2
(3)
(b) The mass of the CO2 molecule is 44m.
¿From the laws of momentum, angular momentum and energy conservation we have the system of equations
mv0 = −mv1 + 44mv2
mrv0 = −mrv1 + 32mr2 ω
1
1 2 1 2 1
mv0 = mv1 + 44mv12 + 32mr2 ω 2
2
2
2
2
(c) By canceling common factors the system of equations obtained in (b) can be simplified as
(4)
(5)
(6)
v0 = −v1 + 44v2
(7)
v0 = −v1 + 32rω
(8)
v02 = v12 + 44v12 + 32r2 ω 2
(9)
¿From equation (5) and (6) we find that v2 =
8
rω
11
ω=
and v1 = 32rω − v0 . Substituting these in eq. (7) gives
8 v0
v0
= 0.0216
371 r
r
(10)
(d)After the formation of the chemical bond the center of mass is displaced by a distance x from the carbon
atom. From the equation
17m(r − x) = 12mx + 16m(r + x),
we find that x =
1
r.
45
(11)
The moment inertia of the HO-C-O molecule with respect to the new center of mass is
I = 17m
442 2
1 2
462 2
r
+
12m
r
+
16m
r = 32.98mr2 ≈ 33mr2 .
452
452
452
(12)
¿From the law of the angular momentum conservation
mv0 r
44
= 33mr2 ω,
45
(13)
we find
ω = 0.0297
v0
r
(14)
Problem 4 – Thermodynamics
There are two bodies with temperatures T1 and T2 (T1 > T2 ). The heat capacities of these bodies, C1 and
C2 respectively, do not depend on temperature.
(a) [3 pts.] The two bodies are brought in thermal contact and after exchanging heat come to the thermal
equilibrium. What is the corresponding equilibrium temperature, Te ?
(b) [5 pts.] What is the total change in the entropy of the two body system after the thermal equilibrium
is established?
(c) [8 pts.] The first body is used as a heater and the second body as a refrigerator in an ideal heat engine.
(The efficiency of an ideal heat engine is 1 − (Tr /Th ), where Tr is the temperature of the refrigerator
and Th is the temperature of the heater.) The operation of the engine stops when the temperatures two
bodies are the same. Compute this terminal temperature, Tf .
(d) [6 pts.] What is the total work produced by the engine in the process described in question (c)?
(e) [3 pts.] What is the total change in the entropy of the universe after the process described in the question
(c) is completed?
Problem 4 – Thermodynamics – sample solution
a) In the simple heat exchange process the heat released by the first body is equal to the heat taken by the
second body. From ∆Q = C1 (T1 − Te ) = C2 (Te − T2 ) we find that
Te = (C1 T1 + C2 T2 )/(C1 + C2 )
(15)
b)The change in the entropy can be computed as
∆S =
Z
Te
T1
C1 dT Z Te C2 dT
T1
Te
+
= C1 ln − C2 ln ,
T
T
Te
T2
T2
(16)
where Te is given by the equation (1).
c)The ideal heat engine has maximum efficiency because no entropy is produced in its operation. That
means
dQh
C1 dTh
dQr
C2 dTr
=−
=
=
Th
Th
Tr
Tr
(17)
where dQh is positive heat released by the heater, dQr is positive heat taken by the refrigerator and Th and
Tr are instantaneous temperatures of the heater and the refrigerator, respectively. The minus sign appears in
front of C1 because the temperature of the heater decreases and dTh is negative. The process stops when the
terminal temperature Tf is reached. Then because of the zero net entropy production
−
Z
Tf
T1
C1 dTh Z Tf C2 dTr
=
.
Th
Tr
T2
(18)
After the integration we find that
C1
C2
Tf = T1C1 +C2 T2C1 +C2
(19)
d) From the energy balance total work produced in the process is
W = C1 (T1 − Tf ) − C2 (Tf − T2 )
e) Zero. No entropy is produced in the operation of an ideal heat engine.
(20)
Problem 5 – General Physics
A capacitive fluid level measurement system (see figure) is based on a capacitor C with two rectangular
plates of distance s, length d (which equals the container depth) and width w. The plates are immersed in the
dielectric fluid on a length h. The fluid’s dielectric constant = 0 r differs from the dielectric constant of air
As
) by a relative factor r . The fluid level therefore determines the capacity C. Thus, using
(0 = 8.85 × 10−12 Vm
an electric circuit which transforms C(h) into a voltage Vout (h), one can electrically measure the fluid level h.
(a) [5 pts.] Derive an expression for the capacity C as a function of w, s, d, h, r and 0 .
(b) [9 pts.] An electric circuit shown in the figure is used for the capacity to voltage conversion. It consists
of a series of the capacity C and resistor R which are connected in series to a sinusoidal input voltage
vin (root mean square) and angular frequency ω. Derive an expression for the root mean square of the
output voltage vout as a function of R, C and ω. (Hint: Note that the impedance of a capacitor is:
Z=
1
)
iωC
(c) [5 pts.] By inspection of the result of (b), determine the condition under which the output voltage
becomes a linear function of the capacity and write down this linear function vout (C) ≈ kC (This means
find the proportionality factor k).
(d) [6 pts.] When r ≈ 80 (water), w = 10cm, s = 5mm, d = 2m, vin = 100V, ω = 104 s−1 and R = 100Ω,
the measurement system shows a linear fluid level dependency vout (h) = v0 + ξh. By using the results of
(c) find both both v0 and ξ with correct units. Give your answer up to three significant digits.
Problem 5 – General Physics – sample solution
(a) The capacity C can be viewed as a net capacity of a parallel circuit of two capacitors with equal spacing
s and width w but different dielectric fillings (air with dielectric constant 0 and fluid with dielectric
constant 0 r ) and also with different areas wh and w(d − h), respectively. Thus:
C(h) = Cfull + Cempty
0 r hw 0 (d − h) w
+
=
s "
s
#
0 wd
h
=
1 + (r − 1)
s
d
(b) The capacity to voltage conversion circuit is a voltage divider consisting of the resistance R and the
capacitors impedance Z =
1
.
iωC
Thus:
R
R+Z
R
= vin
1
R + iωC
iωRC
= vin
1 + iωRC
ω 2 R2 C 2 + iωRC
= vin
.
1 + ω 2 R2 C 2
ṽout (C) = vin
The expression ṽout represents the rms value times the phase factor of the output voltage. The magnitude
of the output voltage is
vout (C) = |ṽout (h)|
√
ω 2 R2 C 2 + 1
= vin ωRC
1 + ω 2 R2 C 2
vin ωRC
= √ 2 2 2
.
ω R C +1
(c) Since vout (C) =
√ vin ωRC ,
ω 2 R2 C 2 +1
it is clear that when ωRC 1, then vout (C) ≈ vin ωRC. Thus, the condition
for the linearity of the output voltage as function of the capacity C is ωRC 1.
(d) For the highest achievable capacity (when the container is full of fluid) we have
ωRC = 104 s−1 × 100Ω × 8.85 × 10−12
As
2m × 0.1m
× (80 − 1) ×
= 0.028 1
Vm
0.005m
(21)
Therefore,
vout (h) = vin ωRC
"
#
h
0 wd
1 + (r − 1)
= vin ωR
s
d
:= v0 + ξh
(22)
Therefore:
0 wd
s
As
100V × 104 s−1 × 100Ω × 8.85 × 10−12 Vm
× 0.1m × 2m
=
0.005m
= 35.4mV.
(23)
0 w(r − 1)
s
As
4 −1
100V × 10 s × 100Ω × 8.85 × 10−12 Vm
× 0.1m × (80 − 1)
=
0.005m
= 1.398V/m.
(24)
v0 = vin ωR
and
ξ = vin ωR
Common Exam - 2009
Department of Physics
University of Utah
August 22, 2009
Examination booklets have been provided for recording your work and your solutions.
Please note that there is a separate booklet for each numbered question (i.e., use
booklet #1 for problem #1, etc.).
To receive full credit, not only should the correct solutions be given, but a sufficient
number of steps should be given so that a faculty grader can follow your reasoning.
Define all algebraic symbols that you introduce. If you are short of time it may be helpful
to give a clear outline of the steps you intended to complete to reach a solution. In some
of the questions with multiple parts you will need the answer to an earlier part in order to
work a later part. If you fail to solve the earlier part you may represent its answer with an
algebraic symbol and proceed to give an algebraic answer to the later part. This is a
closed book exam: No notes, books, or other records should be consulted. YOU MAY
ONLY USE THE CALCULATORS PROVIDED. The total of 250 points is divided
equally among the ten questions of the examination.
All work done on scratch paper should be NEATLY transferred to answer booklets.
SESSION 2
COMMON EXAM DATA SHEET
e = - 1.60 × 10-19 C = - 4.80 × 10 -10 esu
c = 3.00 × 108 m/s = 3.00 × 10 10 cm/s
h = 6.64 × 10 -34 JAs = 6.64 × 10 -27 ergAs = 4.14 × 10 -21 MeVAs
S = 1.06 × 10 -34 JAs = 1.06 × 10 -27 ergAs = 6.59 × 10 -22 MeVAs
k = 1.38 × 10 -23 J/K = 1.38 × 10 -16 erg/K
g = 9.80 m/s2 = 980 cm/s 2
G = 6.67 × 10-11 NAm2/kg2 = 6.67 × 10 -8 dyneAcm 2/g 2
NA = 6.02 × 1023 particles/gmAmole = 6.02 × 10 26 particles/kgAmole
go(SI units) = 8.85 × 10 -12 F/m
:o(SI units) = 4B × 10 -7 H/m
m(electron) = 9.11 × 10 -31 kg = 9.11 × 10 -28 g= 5.4859 × 10 -4 AMU = 511 keV
M(proton) 1.673 × 10 -27 kg = 1.673 × 10 -24 g = 1.0072766 AMU = 938.2 MeV
M(neutron) 1.675 × 10 -27 kg = 1.675 × 10 -24 g = 1.0086652 AMU = 939.5 MeV
M(muon) = 1.88 × 10 -28 kg = 1.88 × 10 -25 g
1 mile = 1609 m
1 m = 3.28 ft
1 eV = 1.6 × 10 -19 J = 1.6 × 10 -12 ergs
hc = 12,400 eVAD
Table of Integrals and Other Formulas
Spherical Harmonics
Conic Section in Polar Coordinates
(origin at the focus; e = 1 for parabola)
Normal Distribution
Cylindrical Coordinates (orthonormal bases)
Spherical Coordinates (orthonormal bases)
Maxwell’s Equations (Rationalized MKS)
Maxwell’s Equations (Gaussian Units)
Problem 6 – Electromagnetism
An infinitely long conducting sheet of width d carries electric current I which is homogeneously distributed
across the width of the sheet. A very small loop of wire with area A and resistance R is placed some distance
h from the axis of the sheet so that the loop and the sheet are within the same plane xy, as shown in the left
picture. At some point the sheet is rotated by 90 degrees around its axis (x-axis) so that the sheet is now
within the xz plane (see the right picture).
~ 1 (h) (both the absolute value B1 (h) and the direction) at the loop prior
(a) [5 pts.] Find magnetic field B
to the rotation.
~ 2 (h) after the rotation?
(b) [6 pts.] What is the value of magnetic field B
(c) [4 pts.] Sketch the functions B1 (h) and B2 (h) (on the same plot) when h changes from d/2 to infinity.
(d) [6 pts.] What is the total amount of charge that passes around the loop following the rotation?
(e) [4 pts.] In which way does it pass the loop (clockwise or counterclockwise according to the picture)?
Explain your answer.
Problem 6 – Electromagnetism – sample solution
The simplest way to solve this problem is by noticing that the field can be calculated as the sum of the
fields created by infinitely long and thin wires with width dy each carrying current dI = (I/d)dy. Magnetic
field created by such a wire at some point distance r away is dB =
µ0 dI
.
2πr
When the sheet is within xy plane all
elementary wires create fields along the same direction (positive direction of z axis), with distance r = h − y
to the point of observation:
!
d/2
dy
µ0 I
h + d/2
µ0 I Z
=
log
.
B1 (h) =
2πd
h−y
2πd
h − d/2
−d/2
~ 2 is still pointed in the same direction. However, the
After the sheet is rotated the total magnetic field B
individual fields of elementary wires (each carrying current dI = (I/d)dz) are now tilted by the angle θ,
√
√
determined by cos θ = h/ h2 + z 2 . The distance to the point of observation, r = h2 + z 2 . The total filed
along the z-axis is now,
!
d/2
d/2
µ0 Ih Z
µ0 I
µ0 I Z
cos θ
dz
d
=
=
arctan
.
B2 (h) =
dz √ 2
2πd
2πd
h2 + z 2
πd
2h
h + z2
−d/2
−d/2
At large distances h d the two expressions approach the same limit, which is simply the filed of a thin wire
with current I:
B(h d) ≈
µ0 I
.
2πh
At small distances h → d/2 the value B2 (h) remains finite while B1 (h) diverges logarithmically. Therefore,
the sketch looks as follows
The flux through the wire loop, therefore, decreases by ∆Φ = (B1 − B2 )A upon the rotation. According to
the Lenz’s rule the induced EMF works in the direction of increasing the flux through the loop, i.e. to the
counterclockwise current. The total amount of charge
Q=
∆Φ 1
∆Φ
(B1 − B2 )A
∆t =
=
.
∆t R
R
R
Problem 7 – Quantum Mechanics
A ground state in a spherical potential well


 0,
r < R,
 ∞,
r > R,
U (~r) = 
is approximated by the trial wavefunction


 R − r,
r < R,
 0,
r > R.
ψ(~r) = const 
(a) [5 pts.] Normalize this wavefunction.
(b) [9 pts.] Find the approximate value of the ground state energy calculated from this trial wavefunction
for a particle of mass m.
(c) [6 pts.] Find the exact wavefunction and energy of the ground state.
(d) [5 pts.] Compare the approximate value with the exact ground state energy. What is the relative error?
Problem 7 – Quantum Mechanics – sample solution
Normalization gives
const2
Z
R
s
dr(4πr2 )(R − r)2 = 1,
→ const =
0
Since in spherical coordinates the radial part of the Laplacian, ∇2 =
the ground state is
1 ∂
r 2 ∂r
1 d
dψ(r)
h̄2 Z R
dr(4πr2 ) ψ(r) 2
r2
E=−
2m 0
r dr
dr
!
=
15
.
2πR5
∂
r2 ∂r
, the approximate energy of
5h̄2
.
mR2
The exact ground state energy E0 and wavefunction ψ0 (r) are found from the Schrödinger equation
1 d
dψ0
r2
2
r dr
dr
!
+
2mE0
ψ0 = 0.
h̄2
The solution finite at r = 0 and vanishing at r = R that has the lowest energy is
ψ0 (r) = A
sin (πr/R)
,
r
where the constant is found similar to the above A =
√1 .
2πR
E0 =
The ratio of the two energies,
10
E
= 2 ≈ 1.013.
E0
π
The relative error is about 1%.
π 2 h̄2
,
2mR2
Problem 8 – Lagrangian Mechanics
A point like bead with mass m is sliding frictionless on a wire that is described by a continuous and
differentiable function h(x) in a constant and uniform gravitational field with acceleration g.
(a) [3 pts.] What is the potential energy V (z) of the bead and what is the potential energy V (x) due to the
constraint imposed by the wire?
(b) [2 pts.] Is the system described above conservative? Explain your answer.
(c) [3 pts.] What would be the Lagrange-function L(x, ẋ, z, ż) of the bead if the constraint imposed by the
wire did not exist (the bead would be an unconstrained object in a constant and uniform gravitational
field)?
(d) [5 pts.] Under consideration of the constraint imposed by the wire, the Lagrangian of the bead as written
down in (c) changes to a Lagrangian that depends only on one coordinate x. Find this constrained
Lagrangian L(x, ẋ)?
(e) [7 pts.] Use the Lagrangian L(x, ẋ) obtained in (d) to derive an equation of motion of the bead for the
coordinate x.
(f) [5 pts.] The wire is a straight inclined line following the function h(x) = z0 − λx. Write down the
equation of motion using the general form of the equation of motion derived in (e) and solve it for the
initial conditions x(t = 0) = x0 and v(t = 0) = v0 .
z
gravitational force mg
bead with mass m
slides on wire
frictionless in constant
gravitational force
wire described by
arbitrary function
z = h(x)
x
Problem 8 – Lagrangian Mechanics – sample solution
(a) The gravitational force in z-direction is Fz = −mg.
−
Rz
0
The potential function is therefore V (z) =
(−)mgdz 0 = mgz. Since z = h(x), we know that V (x) = V [z(x)] = mgh(x).
(b) Since a potential function exists, the system is conservative.
(c) The kinetic energy of a free particle in to dimensions (x,z) is T (ẋ, ż) = 21 m (ẋ2 + ż 2 ). Given the potential
as found in (a) we obtain a free Lagrangian L (x, ẋ, z, ż) = T (ẋ, ż) − V (z) = 21 m (ẋ2 + ż 2 ) − mgz.
(d) We know that z = h(x) and therefore, ż =
d
dt
∂h
ẋ.
∂x
2 ∂h
∂x
[h(x)] =
Thus, the Lagrangian becomes L (x, ẋ) = 12 mẋ2 1 +
− mgh(x)
(e) The equation of motion is
d ∂L
∂L
−
= 0
∂x dt ∂ ẋ


"
#2  
2
1
∂h
∂h
∂
h
∂h
d
1
 2ẋ
= mẋ2 2
− mg
−  m 1 +
2
∂x ∂x2
∂x dt 2
∂x


#2 

∂h ∂ 2 h
∂h  
∂h 1  ∂h ∂ 2 h
= mẋ2
2ẍ
−
mg
− m 2
ẋ2ẋ + 1 +
2
2
∂x ∂x
∂x 2
∂x ∂x
∂x

∂h ∂ 2 h
∂h
∂h
= −mẋ2
− mg
− mẍ 1 +
2
∂x ∂x
∂x
∂x
"
!2 
.
Simplified, this equation assumes the form

∂h
∂h ∂ 2 h
ẋ2
+ ẍ 1 +
2
∂x ∂x
∂x
(f) With h(x) = z0 − λx we know that
∂h
∂x
= −λ and
!2 
∂h
+g
=0
∂2h
∂x2
∂x
(1)
= 0. Thus, the equation of motion becomes
λ
ẍ (1 + λ2 ) − gλ = 0 and hence ẍ = g 1+λ
2 . With the given boundary conditions, the system is obviously
a constantly accelerated object and the solution is therefore easily integrated:
x(t) =
Z
0
t
Z
0
t0
0
ẍdt dt =
Z
0
t
Z
0
t0
g
λ
gλt2
0
dt
dt
=
x
+
v
t
+
0
0
1 + λ2
2 (1 + λ2 )
(2)
Problem 9 – Statistical Mechanics
(Effusion is a process by which a gas escapes from a very small hole.) Consider a container of an ideal
monoatomic gas (molecule mass m) with a small hole of area A in the side. Gas effuses (leaks) though the hole,
which is so small that the equilibrium of the gas in the container is not disturbed. The gas in the container is
described by Maxwell-Boltzmann distribution,
m 3/2
m(vx2 + vy2 + vz2 )
mv 2
m 3/2
2
]dvx dvy dvz = 4π
]dv.
exp[−
v exp[−
f (v)dv =
2πkB T
2kB T
2πkB T
2kB T
The container is kept at a constant temperature T .
(a) [6 pts.] Work out an expression for the number of molecules with speeds in the interval from v to v + dv
moving in an angular interval θ to θ + dθ, and ϕ to ϕ + dϕ. Here θ and ϕ are the standard spherical
angles as defined with respect to the normal to the hole.
(b) [4 pts.] Among the considered in Part (a) molecules, how many escape through the hole during time
interval dt?
(c) [5 pts.] Using results from Parts (a) and (b), calculate the rate (total number of molecules per unit
time) at which gas molecules escape the container.
(d) [5 pts.] Calculate the rate of change of mass M of the gas inside the container. Using the ideal gas law
relate
dM
dt
to the temperature T and pressure P of the gas.
(e) [5 pts.] What is reactive (jet) force produced by effusing molecules? Express your result in terms of the
pressure P and the area A of the hole. Hint: This question can be answered by clear physical reasoning,
with no integrals involved. The answer can also be derived by extending the analysis in the previous
parts, and performing an additional integration.
A
T, P
FIG. 1: Container with a small hole of area A in the side.
Problem 9 – Statistical Mechanics – sample solution
Consider a container of an ideal monoatomic gas (molecule mass m) with a small hole of area A in the side.
Gas effuses (leaks) through the hole, which is so small that the equilibrium of the gas in the container is not
disturbed. The gas in the container is described by Maxwell-Boltzmann distribution,
m 3/2
m(vx2 + vy2 + vz2 )
m 3/2
mv 2
]dvx dvy dvz = 4π
]dv.
f (v)dv =
exp[−
v 2 exp[−
2πkB T
2kB T
2πkB T
2kB T
The container is kept at a constant temperature T .
(a) [6 pts.] Work out an expression for the number of molecules with speeds in the interval from v to v + dv
moving in an angular interval θ to θ + dθ, and ϕ to ϕ + dϕ, to the normal to the hole. Here θ and ϕ are
the standard spherical angles.
We have n = N/V = P/(kB T ) molecules of ideal gas per unit volume.
Number of them with speeds between v and v + dv: nf (v)dv, where
m 3/2 2
4
) v exp[−mv 2 /(2kB T )]
f (v) = √ (
π 2kB T
is the Maxwell distribution in spherical coordinates.
Number of molecules travelling an angular interval (θ, θ + dθ), (ϕ, ϕ + dϕ), to a given direction is
dΩ
nf (v)dv.
4π
Here dΩ = sin θdθdϕ is the spherical angle.
(b) [4 pts.] Among the considered in Part (a) molecules, how many escape through the hole during time
interval dt?
Molecules that hit the hole at angle θ during time interval dt are located within cylinder of volume
A cos θvdt. Note that A cos θ is the apparent area of the hole that molecules see.
Thus, per unit time, A cos θv dΩ
nf (v)dv of them escape.
4π
(c) [5 pts.] Using results from Parts (a) and (b), calculate the rate (total number of molecules per unit
time) at which gas molecules escape the container.
Putting all of the above together we have that the rate dN/dt of escape is given by
Z π/2
Z ∞
dN
nA Z 2π
nAhvi
=
dϕ
dθ cos θ sin θ
dvvf (v) =
.
dt
4π 0
4
0
0
Here the average speed is
2
m 3/2 Z ∞
4
2
dv v · v 2 e−mv /(2kB T ) = √
)
hvi = √ (
π 2kB T
π
0
s
2kB T
.
m
√
Using n = P/(kB T ) we get dN/dt = AP/ 2πmkB T .
(d) [5 pts.] Calculate the rate of change of mass M of the gas inside the container. Using the ideal gas law
relate
dM
dt
to the temperature T and pressure P of the gas.
dM
dN
=m
= AP
dt
dt
s
m
.
2πkB T
(e) [5 pts.] What is reactive (jet) force produced by effusing molecules? Express your result in terms of the
pressure P and the area A of the hole.
Physical argument: the jet force produced by effusing molecules is just 12 P A. This is so because pressure
is the force per unit area. An important factor of
1
2
appears because effusing molecule “carries its
momentum” away, while molecules hitting the walls of the container transmit twice the value of the
normal component of the momentum to the wall. Thus effusing molecules “take” 1/2 less than that with
them. The force is directed opposite to the direction of effusion.
More formal argument: let us calculate the rate of momentum reduction dp/dt due to escaping molecules.
Each of them carries momentum mv cos θ: we only need to account for the normal component of the
momentum, components parallel to the walls average out to zero.
Thus
Z ∞
Z π/2
dp
nA Z 2π
dv(mv cos θ) cos θvf (v).
=
dϕ
dθ sin θ
dt
4π 0
0
0
Angular
R∞
integrals
give
dvv 2 exp[−mv 2 /2kB T ]
√
Γ[5/2] = 3 π/4.
0
2π/3,
=
while
the
integral
R
5/2 ∞
(1/2)(2kB T /m)
0
over
du u3/2 e−u .
the
speed
is
proportional
to
The last integral (over u) is
The result is, accounting for the normalization factor of f (v),
√
dp
mnA 2π 3 π 2kB T 5/2 4 m 3/2 1
1
√
=
= nAkB T = P A.
dt
4π 3 8
m
π 2kB T
2
2
Problem 10 – Modern Physics
Electromagnetic radiation of frequency ν flows in the (−x) direction. Free electrons (rest mass m, energy E)
moving in the (+x) direction undergo a relativistic inverse Compton scattering e(E, p~) + γ(hν) → e(E 0 , p~ 0 ) +
γ(hν 0 ).
(a) [10 pts.] In terms of E, pc and hν, express the energy hν 0 of the outgoing photon in the case of head on
collisions with the hν 0 photon being scattered in the forward direction and hence having the maximal
possible energy.
m
m
E, P
h!
E’, P’
h !’
(b) [4 pts.] Rewrite the energy hν 0 of the scattered photons in terms of just E and hν when E >> mc2 >>
hν.
(c) [7 pts.] Considering a photon of energy hν 0 propagating in the (+x) direction, write the threshold
condition on hν 0 for these photons to interact with an hν photon to create a e+ /e− pair.
m
h !’
h!
m
(d) [4 pts.] Combining your answers to the two previous questions express the minimal energy of electrons
whose inverse Compton scattering may result in photons with sufficient energy to participate in e+ /e−
pair creations.
Problem 10 – Modern Physics – sample solution
Electromagnetic radiation of frequency ν flows in the (−x) direction. Free electrons (rest mass m, energy E)
moving in the (+x) direction undergo a relativistic inverse Compton scattering e(E, p~) + γ(hν) → e(E 0 , p~ 0 ) +
γ(hν 0 ).
(a) [10 pts.] In terms of E, pc and hν, express the energy hν 0 of the outgoing photon in the case of head on
collisions with the hν 0 photon being scattered in the forward direction and hence having the maximal
possible energy.
m
m
E, P
E’, P’
h!
h !’
Using the conservation of energy E + hν = E 0 + hν 0 , the conservation of momentum in the (x) direction
pc−hν = p0 c+hν 0 and the relation E 02 = p02 c2 +m2 c4 , we have (p0 c)2 = (pc−hν −hν 0 )2 = (E+hν −hν 0 )2 −
m2 c4 which after developments and simplification gives −pchν − pchν 0 + hνhν 0 = Ehν − Ehν 0 − hνhν 0
which can be solved for the energy hν 0 of the scattered photon: hν 0 =
E+pc
hν.
E−pc+2hν
(b) [4 pts.] Rewrite the energy hν 0 of the scattered photons in terms of just E and hν when E >> mc2 >>
hν.
We can rewrite our result as hν 0 =
mc2 >> hν, hν 0 ≈
2E(E+pc)
hν
m2 c4
≈
√ E+pc
hν
E− E 2 −m2 c4 +2hν
=
√
E(1−
E+pc
hν
1−m2 c4 /E 2 )+2hν
and, so, when E >>
4E 2
hν.
m2 c4
(c) [7 pts.] Considering a photon of energy hν 0 propagating in the (+x) direction, write the threshold
condition on hν 0 for these photons to interact with an hν photon to create a e+ /e− pair.
m
h !’
h!
m
The energy invariant must exceed twice the rest energy of the electron: (hν + hν 0 )2 − (hν 0 − hν)2 > 4m2 c4
and hν 0 >
m2 c4
.
hν
(d) [4 pts.] Combining your answers to the two previous questions express the minimal energy of electrons
whose inverse Compton scattering may result in photons with sufficient energy to participate in e+ /e−
pair creations.
Combining our two previous results
4E 2
hν
m2 c4
>
m2 c4
hν
or E >
m2 c4
.
2hν