Q51E

Page 406

There is a simple argument, practically by inspection, that distributions(9-31),(9-32), and(9-33)should agree whenever occupation number is much less than 1. Provide the argument.

Q52E

Page 406

Equation (9-27) gives the density of states for a system of oscillators but ignores spin. The result, simply one state per energy change ofbetween levels, is incorrect if particles are allowed different spin states at each level, but modification to include spin is easy. From Chapter 8, we know that a particle of spinis allowedspin orientations, so the number of states at each level is simply multiplied by this factor. Thus,

D(E)=(2s+1)/hω0.

(a) Using this density of states, the definitionNhω0/(2s+1)=ε1, and

N=0N(E)D(E)dE

calculate the parameterin the Boltzmann distribution (9-31) and show that the distribution can thus be rewritten as

N(E)Boltz=εkBT1eE/kBT

(b) Argue that ifkBT>>ε,the occupation number is much less than 1 for all E.

Q53E

Page 406

Using density of states D(E)=(2s+1)0, which generalizes equation (9-27) to account for multiple allowed spin states (see Exercise 52), the definition Nhω0/(2s+1)=εand N=0N(E)D(E)dE. Solve for Bin distributions (9-32) and (9-33) careful use of ±will cut your work by about half. Then plug back in and show that for a system of simple harmonic oscillators, the distributions become γ(E)BE=1eE/kBT1eδ/kBT1andN(E)FD=1eE/kBTe+δ/kBT1+1.

You will need the following integral:0(Be2±1)1dz=±ln(1±1B).

Q54E

Page 407

Exercise 52 gives the Boltzmann distribution for the special case of simple harmonic oscillators, expressed in terms of the constant ε=Nhω0/(2s+1). Exercise 53 gives the Bose-Einstein and Fermi-Dirac distributions in that case. Consider a temperature low enough that we might expect multiple particles to crowd into lower energy states:kBT=15ε. How many oscillators would be expected in a state of the lowest energy,E=0? Consider all three-classically distinguishable. boson, and fermion oscillators - and comment on the differences.

Q55E

Page 407

Exercise 52 gives the Boltzmann distribution for the special case of simple harmonic oscillators, expressed in terms of the constantε, Nω0/(2s+1)and Exercise 53 gives the two quantum distributions in that case. Show that both quantum distributions converge to the Boltzmann in the limitkBT.

Q57E

Page 407

Exercise 54 calculates the three oscillator distributions'E=0values in the special case wherekBTis15ξ. Using a very common approximation technique. show that in the more general low-temperature limit,kBTε,theoccupation numbers becomeδ/kBT,e5/kBT, and 1, for the distinguishable. boson. and fermion cases, respectively. Comment on these results. (Note: Although we assume thatkBTNω0/(2s+1). we also still assume that levels are closely spaced-that is kBhω0.),

Q58E

Page 407

Show that. using equation(936), density of states(938)follows fromlocalid="1658380849671" (937)

Q59E

Page 407

Density of states (9-39) does not depend on N, the total number of particles in the system; neither does the density of states in equation (9-27). Why not?

Q5CQ

Page 402

Defend or refuel the following claim: An energy distribution, such as the Boltzmann distribution. specifies the microstate of a thermodynamic system.

Q60E

Page 407

For a particle in a one-dimensional (ID) box, Enis proportional to a single quantum number n. Let us simplify things by ignoring the proportionality factor: En=n2 . For a 3D box, Enx,ny,nz=nx2+ny2+nz2, and the 2D box is fairly obvious.

(a) The table shows a start on accounting for allowed states. Complete the table, stopping after the 10th state (state, not energy) for all three cases.

(b) Find the number of states per energy difference for the first five states and the last five states for all three cases. For instance, for the first five in the ID case, it is 5 states per energy difference of 24, or5/24 .

(c) Overlooking the obviously crude aspects of this accounting, does the "density of states" seem to increase with energy, decrease with energy, or stay about the same?

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