Referring to Fig. 8.11 and noting that the slope of the tangent at the point (x=xi) is

Question:

Referring to Fig. 8.11 and noting that the slope of the tangent at the point \(x=\xi\) is \(-1 / 4\), the approximate distribution is given by

\[
f(x)=\left\{\begin{array}{cc}
1 & 0 \leq x \leq(\xi-2) \\
(\xi+2-x) / 4 & (\xi-2) \leq x \leq(\xi+2) \\
0 & (\xi+0) \leq x
\end{array}ight.
\]

where \(x=\varepsilon / k T\) and \(\xi=\mu / k T\). Accordingly,

\[
N=g \cdot \frac{2 \pi V}{h^{3}}(2 m)^{3 / 2} \int_{0}^{\infty} n(\varepsilon) \varepsilon^{1 / 2} d \varepsilon=C \int_{0}^{\infty} f(x) x^{1 / 2} d x
\]

where \(C=g\left(2 \pi V / h^{3}ight)(2 m k T)^{3 / 2}\). After some algebra, one gets

\(N=\frac{1}{5} C\left\{(\xi+2)^{5 / 2}-(\xi-2)^{5 / 2}ight\}=\frac{2}{3} C \xi^{3 / 2}\left\{1+\frac{1}{2} \xi^{-2}+\ldotsight\} \quad(\xi \gg 1)\).

Comparing (1) with eqn. (8.1.24), which may be written as

\[
N=\frac{2}{3} C\left(\frac{\varepsilon_{F}}{k T}ight)^{3 / 2}
\]

we get

\[
\begin{equation*}
\xi=\frac{\varepsilon_{F}}{k T}\left\{1-\frac{1}{3}\left(\frac{k T}{\varepsilon_{F}}ight)^{2}+\ldotsight\} \tag{2}
\end{equation*}
\]

Similarly,

\[
\begin{align*}
U & =C k T \int_{0}^{\infty} f(x) x^{3 / 2} d x=\frac{1}{35} C k T\left\{(\xi+2)^{7 / 2}-(\xi-2)^{7 / 2}ight\} \\
& =\frac{2}{5} C k T \xi^{5 / 2}\left\{1+\frac{5}{2} \xi^{-2}+\ldotsight\} \tag{3}
\end{align*}
\]

Combining (1) and (3), and then making use of (2), we get

\[
U=\frac{3}{5} N k T \xi\left\{1+2 \xi^{-2}+\ldotsight\}=\frac{3}{5} N \varepsilon_{F}\left\{1+\frac{5}{3}\left(k T / \varepsilon_{F}ight)^{2}+\ldotsight\}
\]

It follows that, at temperatures much less than \(\varepsilon_{F} / k\),

\[
C_{\mathrm{V}}=2 N k\left(k T / \varepsilon_{F}ight),
\]

which is "correct" insofar as the dependence on \(T\) is concerned but is numerically less than the true value, given by eqn. (8.1.39), by a factor of \(4 / \pi^{2}\).

The reason for the numerical discrepancy lies in the fact that the present approximation takes into account only a fraction of the particles that are thermally excited; see Fig. 8.11. In fact, the ones that are not taken into account have a higher \(\Delta \varepsilon\) than the ones that are, which explains why the magnitude of the discrepancy is so large.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question
Question Posted: