91Ó°ÊÓ

This problem concerns a collection of N identical harmonic oscillators (perhaps an Einstein solid or the internal vibrations of gas molecules) at temperature T. As in Section 2.2, the allowed energies of each oscillator are 0, hf, 2hf, and so on.

(a) Divide 1 by 1 - x using long division, yielding:

(b) Assume that the oscillator's ground state energy is zero; the energies are provided by:

$E_n=nhfn=0,1,2,…$

The partition function is:

$Z=\underset{n}{∑}{e}^{-\mathrm{β}{E}_n}$

Substitute with n:

$$\begin{gathered} Z=\underset{n}{∑}{e}^{-n\mathrm{β}hf} \\ Z=\underset{n}{∑}{\left(e^{-\mathrm{β}hf}\right)}^n \end{gathered}$$

From part (a),

$\frac{1}{1-x}=1+x+x^2+x^3+…=\underset{n}{∑}{x}^n$

Using the above expression, the partition function is:

$$\begin{gathered} Z=\underset{n}{∑}{\left(e^{-\mathrm{β}hf}\right)}^n=\frac{1}{1-e^{-\mathrm{β}hf}} \\ Z=\frac{1}{1-e^{-\mathrm{β}hf}}\left(1\right) \end{gathered}$$

(c)The average energy is given by:

$\stackrel{¯}{E}=-\frac{1}{Z}\frac{∂Z}{∂\mathrm{β}}$

Substitute from part (b)

$$\begin{gathered} \stackrel{¯}{E}=-\left(1-e^{-\mathrm{β}hf}\right)\frac{∂}{∂\mathrm{β}}\left(\frac{1}{1-e^{-\mathrm{β}hf}}\right) \\ \stackrel{¯}{E}=\left(1-e^{-\mathrm{β}hf}\right)\frac{hf{e}^{-\mathrm{β}hf}}{{\left(1-e^{-\mathrm{β}hf}\right)}^2} \\ \stackrel{¯}{E}=\frac{hf{e}^{-\mathrm{β}hf}}{1-e^{-\mathrm{β}hf}} \\ \stackrel{¯}{E}=\frac{hf}{e^{\mathrm{β}hf}-1} \end{gathered}$$

(d) The average energy multiplied by the number of oscillators equals the total energy of the system, so:

$U=N\stackrel{¯}{E}$

Substitute from part (c):

$U=\frac{Nhf}{e^{\mathrm{β}hf}-1}$

(e) The heat capacity is equal to the partial derivative of total energy in terms of temperature, i.e.

$$\begin{gathered} C=\frac{∂U}{∂T}=\frac{∂\mathrm{β}}{∂T}\frac{∂U}{∂\mathrm{β}} \\ C=\frac{∂\mathrm{β}}{∂T}\frac{∂}{∂\mathrm{β}}\left(\frac{Nhf}{e^{\mathrm{β}hf}-1}\right) \\ C={\left(\frac{∂T}{∂\mathrm{β}}\right)}^{-1}\left(-\frac{N(hf{)}^2{e}^{\mathrm{β}hf}}{{\left(e^{\mathrm{β}hf}-1\right)}^2}\right) \end{gathered}$$

But, $\mathrm{β}=1/kT→T=1/k\mathrm{β}$,So:

${\left(\frac{∂T}{∂\mathrm{β}}\right)}^{-1}=-k{\mathrm{β}}^2$

Therefore,

$C=\frac{Nk(\mathrm{β}hf{)}^2{e}^{\mathrm{β}hf}}{{\left(e^{\mathrm{β}hf}-1\right)}^2}$

We can express the exponential as: at a high temperature $\mathrm{β}→0$, we can write it as:

$e^{\mathrm{β}hf}=1+\mathrm{β}hf$

Therefore,

$$\begin{gathered} C=\frac{Nk\left(\mathrm{β}hf{)}^2\right(1+\mathrm{β}hf)}{(\mathrm{β}hf{)}^2} \\ C=Nk(1+\mathrm{β}hf) \end{gathered}$$

Since $\mathrm{β}$is very small, we can approximate:

$C(\mathrm{β}→0)≈Nk$

As the temperature goes to zero, the Boltzmann factor therefore$\mathrm{β}→∞,\text{so}{e}^{\mathrm{β}hf}≫1$hence we can approximate:

$$\begin{gathered} C≈\frac{Nk(\mathrm{β}hf{)}^2{e}^{\mathrm{β}hf}}{e^{2\mathrm{β}hf}} \\ C(\mathrm{β}→∞)≈Nk(\mathrm{β}hf{)}^2{e}^{-\mathrm{β}hf} \end{gathered}$$