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PHYSICS 115A
Solutions to the Final Exam
June 12, 2008
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R = 8.314 J/K-mol kB = 1. 38054 × 10 −^23 J/K 1 atm = 1.013 × 105 N/m^2 273 K = 0o^ C ¯h = 1.05 × 10 −^34 J s c = 3.00 × 108 m/s
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A MAX POINTS SCORE INITIALS
Problem 1 50
Problem 2 10
Problem 3 10
Problem 4 10
Problem 5 10
Problem 6 10
Total 100
- Short Answer(50 points)
(a) (5 points) What is phase space? For a system with N particles in three di- mensions, what is the dimension of phase space? Answer: Phase space has 6N dimensions. The coordinate axes of this space correspond to the positions and momenta of the N particles in the system. A point in phase space has the coordinates (q 1 x, q 1 y, q 1 z , q 2 x, ..., qN z , p 1 x, p 1 y, ..., pN x, pN y, pN z ). Each point in phase space corresponds to a possible configuration of the sys- tem. (b) (5 points) State the second law of thermodynamics. Second Law: An equilibrium macrostate of a system can be characterized by a quantity S (called “entropy”) which has the properties that i. In any process in which a thermally isolated system goes from one macrostate to another, the entropy tends to increase, i.e.,
∆S ≥ 0 (1)
ii. If the system is not isolated and undergoes a quasi–static infinitesimal process in which it absorbs heat dQ, then
dS =
dQ T
where T is a quantity characteristic of the macrostate of the system. T is called the “absolute temperature” of the system.
(d) (5 points) In the pressure versus temperature phase diagram for 3 He, the melting curve has a negative slope (dp/dT < 0). Like most substances, solid (^3) He is denser than liquid 3 He. Suppose we have an equilibrium mixture of solid and liquid 3 He. If a small amount of heat Q is added to this mixture, will some solid converted into liquid or will some liquid be converted into solid? Or neither? Assume the mixture is still a mixture of solid and liquid after the heat is added (though with possibly different proportions of liquid and solid). Explain your reasoning. Answer: Liquid 3 He will be converted into solid. To understand why, note that from the Clausius-Clapeyron equation:
dp dT
∆s ∆v
where s is the entropy per mole and v is the volume of a mole. Since the solid is denser than the liquid, the volume of 1 mole of the solid is less than the volume of 1 mole of the liquid. So
∆v = vsolid − vliquid < 0 (6)
Thus the Clausius-Clapeyron equation implies that
∆s = ssolid − sliquid > 0 ssolid > sliquid
So when heat Q is added, the entropy of the mixture increases by an amount
∆S =
Q
T
This added entropy goes into converting liquid into solid which is the higher entropy phase. Note that when solid and liquid coexist, the temperature T stays constant as long as P is constant.
(e) (5 points)
i. Is a neutral 4 He atom a boson or a fermion? Explain your reasoning. Answer: 4 He is a boson. A neutral atom has 2 protons, 2 neutrons, and 2 electrons. Adding the spin of 6 fermions gives an integer spin to 4 He. So 4 He is a boson because bosons have integer spin.
ii. Is a neutral 3 He atom a boson or a fermion? Explain your reasoning. Answer: 3 He is a fermion. A neutral 3 He atom has 2 protons, 1 neutron, and 2 electrons. Adding the spin of 5 fermions, each with spin 1/2, yields a half integer spin. So 3 He is a fermion because fermions have half integer spin.
(h) (5 points) What is the Fermi energy? If an external magnetic field H~ is applied along the z− axis to a noninteracting electron gas, is the Fermi energy E F+ of the spin up electrons equal to the Fermi energy E F− of the spin down electrons? Explain your reasoning. Each electron has a magnetic moment ~μB. Answer: In zero magnetic field, the Fermi energy EF = μ(T = 0) where μ is the chemical potential. The Fermi energy is the energy of the highest energy electrons at T = 0, i.e., it is the energy of the highest energy levels that are occupied by electrons at T = 0. In a magnetic field, E F+ = E− F. Suppose E F+ > E F− , then some up spin elec- trons could lower their energy by flipping their spin and becoming down spin electrons. So the lowest energy state has E F+ = E− F. It is true that a mag- netic field shifts the electron energy by ±μB H. As a result the effect of the magnetic field is seen in the shifting of energy levels below the Fermi energy. For example, an energy level with E = 0 when H = 0 becomes shifted to E = −μB H if it is occupied by an up spin electron pointing along the field. If it is occupied by a down spin electron pointing antiparallel to the field, the energy level is shifted to E = +μB H. So for H 6 = 0, there will be more up spin electrons than down spin electrons, and the electron gas will have a net magnetization.
Up Spins
Down Spins
Fermi Energy
E
States
Density of States
Density of
(i) (5 points) Let the density of states of the electrons in some sample be assumed to be a constant D for ε > 0 (D = 0 for ε < 0) and the total number of electrons to be equal to N. Calculate the chemical potential μ at T = 0 K. Answer: Method I: The chemical potential is set by the condition that if you integrate over the occupations of all the energy levels, you have to get the total number N of particles. Using the fact that the Fermi occupation factor is a step function at T = 0, we can write
N =
∫ (^) ∞
0
eβ(ε−μ)^ + 1
Ddε
= D
∫ (^) μ
0
dε
= Dμ
μ =
N
D
We assume that the density of states D includes spin. Method II: At T = 0, the energy levels are occupied up to energy ε = μ. So
Dμ = N
μ =
N
D
μ(T = 0) = N D
(j) (5 points) What is Bose-Einstein condensation?
Answer: In Bose-Einstein condensation, there is a macroscopic population of the ground state by bosons.
∆S = N kB ln
( (^) V f Vi
)
- (10 points) Consider a photon in a square box with periodic boundary conditions. Each side has length L and the area of the box is A = L^2. The potential inside the box is zero. The solutions of Schroedinger’s equation are wavefunctions that are plane waves: ψ(~r) = exp
( i~k · ~r
) = exp [i (kxx + kyy)] (9)
Derive the expression for the density of single particle states ρ(E)dE between energy E and E + dE if the dispersion relation is E = pc where p is the magnitude of the momentum and c is the speed of light. Answer: Periodic boundary conditions imply that
ψ(x, y) = ψ(x + L, y) ψ(x, y) = ψ(x, y + L) eikxx+i^2 πnx^ = eikx(x+L)^ eiky^ y+i^2 πny^ = eiky^ (y+L) 2 πnx = kxL 2 πny = kyL kx = 2 πn L x ky = 2 πn Ly nx = k 2 xπL ny = k 2 yπ^ L
The number of states in a volume element dkxdky is
ρkd^2 k = ∆nx∆ny =
( L 2 π
) 2 dkxdky =
L^2
(2π)^2
dkxdky =
A
(2π)^2
dkxdky (11)
For an isotropic system
ρkdk =
A
(2π)^2
2 πkdk =
A
2 π
kdk
ε = pc = ¯hck
k =
E
hc ¯ dE = ¯hcdk dk =
dE ¯hc ρkdk =
A
2 π
kdk
A
2 π
EdE (¯hc)^2
= ρ(E)dE
ρ(E)dE =
A
2 π
EdE (¯hc)^2
for 1 photon polarization
A
π
EdE (¯hc)^2
for 2 photon polarizations
- (10 points) At what temperature T does the mean pressure p of cavity radiation equal 1 atm? Cavity radiation is a photon gas in equilibrium in a 3D cavity. Answer: The mean pressure p exerted on the walls of the enclosure by the radiation is simply related to the total energy density:
p =
∑ s
ns
( −
∂εs ∂V
)
T
uo(T ) (12)
To see where this comes from, recall that
εs = ¯hωs ω = ck (13)
Reflective boundary conditions (ψ(0) = ψ(L) = 0) implies
ψ(x) = A sin(kxx)sin(kyy)sin(kz z) (14)
kx =
π Lx
nx ky =
π Ly
ny kz =
π Lz
nz (15)
to obtain
εs = ¯hck = ¯hc
√ k x^2 + k y^2 + k^2 z
= ¯hc
( (^) π
L
) √ n^2 x + n^2 y + n^2 z
= π¯hcV −^1 /^3
√ n^2 x + n^2 y + n^2 z
where V is the volume. So the pressure associated with state s is
ps = −
∂εs ∂V
π¯hcV −^4 /^3
√ n^2 x + n^2 y + n^2 z =
εs 3 V
So the average pressure for the system is
p =
∑ s
psns =
3 V
∑ s
nsεs =
E
3 V
uo (17)
From the formula sheet,
uo(T ) =
π^2 15
(kB T )^4 (¯hc)^3
p =
π^2 15
(kB T )^4 (¯hc)^3
π^2 45
(kB T )^4 (¯hc)^3
T =
[ 45 π^2
(¯hc)^3 p k^4 B
] 1 / 4
[ 45 (1. 05 × 10 −^34 Js · 3 × 108 m/s)^3 (1. 013 × 105 N/m^2 ) π^2 (1. 38054 × 10 −^23 J/K)^4
] 1 / 4
= 1. 41 × 105 K
T = 1. 41 × 105 K
S = N kB
{ ln
[ 2 cosh
( E kB T
)] −
( E kB T
) tanh
( E kB T
)}
- (10 points) What is the entropy S of 1 mole of a 3 dimensional noninteracting electron gas at temperature T? Let the Fermi temperature be TF. Answer: The specific heat CV is related to the entropy S by
CV =
dQ dT
∣∣ ∣∣ ∣V^ =^
T dS dT
∣∣ ∣∣ ∣V^ (18)
We assume from the third law of thermodynamics, that the entropy S = 0 at T = 0. So S =
∫ (^) T
0
CV (T ′)
T ′^
dT ′^ (19)
From the formula sheet, the molar specific heat for a free electron gas is
CV =
π^2 2
R ·
T
TF
So
S =
∫ (^) T
0
CV (T ′)
T ′^
dT ′
π^2 2
R
∫ (^) T
0
T ′
T ′TF
dT ′
π^2 2
R
∫ (^) T
0
dT ′ TF
=
π^2 2
R
T
TF
S = π
2 2 R^
T TF
