4.1: Counting Oscillator States
a) omega = [(q+N-1)!]/[(N-1)!q!]  where q=U/epsilon
b) omega = [(q+N-1)!]/[(N-1)!q!]  where q=U/epsilon
c) [(U+1)-energy]/omega
d) [(U+1)-energy]/omega


4.2: Harmonic Oscillators in Thermal Equilibrium
a) P1=Ce-E1/kT   ::   E1 = epsilon = hf
   P0=Ce-E0/kT   ::   E0=0
   P1/P0 = e-hf/kT/1  :: solve for T
b) P1/P0 = e-hf/kT  :: use T from (a) and multiply by the percentage
c) P2/P1 = (e-2hf/kT)/(e-hf/kT)
d) <K>/kT = P1E1 = P1hf/kT   ::   get P1 from part (a) because you know P0=1


4.3: Spins in Thermal Equilibrium
a) Nup/Ndown=[% of spins up]/[100-% of spins up]
b) B=[kT*ln(N)]/[2u]
c) T=[2uB]/[k*ln(N)]


4.4: Maxwell-Boltzmann Distribution of Atoms
a) P(E)=C * E^(1/2) * e^(-E/kT) ; C=[2/sqrt(pi)]*(kT)^(-3/2)
   P(1.0kT)=[2/sqrt(pi)] * [kT]^(-1) * (1/e)

   P=delta.E * P(1.0kT) ; delta.E=0.1kt
   P=(0.1*2)/(e*sqrt(pi))

   N=No*P   where No is the number of atoms
b) do this by solving dP(E)/dE=0 for E
   complicated work: i got Epeak=(1/2)kT=0.013356


NOTE ABOUT HOMEWORK B:  THESE ANSWERS MAY NOT BE CORRECT.  USE THEM AT YOUR OWN RISK.
HomeworkB 04
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Consider configuration A which has three isolated systems of oscillators a, b and c as shown.

Each of the systems a, b and c has 2 oscillators with energy spacing ε.
The energy of system a is Ea = 3ε, while systems b and c have energy Eb = Ec = 2ε.

1. Compare the entropies of a and b, σa and σb. 
 X σb  - σa = -0.287
2. System a and system b are brought into contact and allowed to reach equilibrium. Call this configuration B:

Compare the total entropy of the configuration B, σB, including systems a, b and c, to the total entropy of the configuration A, σA. 
 X σB  - σA = 1.54
3. How would this difference, σB-σA, change if the energy of system c were Ec= 5ε rather than 2ε as it is in Question 2 above? 
 X The difference σB-σA with Ec= 5ε is the same for Ec = 2ε
4. What is the probability Pa(1ε) that in the combined system a+b (Ea+b = 5ε) the two oscillators on the left (i.e. system a) have total energy 1ε. The picture below shows an example of a state where the two oscillators on the left have total energy 1ε: 
 X Pa(ε) = 0.036
5. The magnetic moment of the electron is μ=9.285×10-24 J/Tesla. In many materials, the dipole moments are free to point along or against an applied magnetic field. Consider such a material at room temperature (T=300K) in the Earth's magnetic field (B=2.3×10-5T).

The Earth's Magnetic Field


What is the difference between the fraction of electrons pointing along the magnetic field (the low energy state) and the fraction pointing against the field (the high energy state)?
For example, if these fractions are 0.53 (along) and 0.47 (against), the difference is +0.06.

The difference between the fraction of electrons pointing along the magnetic field (the low energy state) and the fraction pointing against the field (the high energy state) is: 

 X -10.31×10-8