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APhO 2007, Теория — Задача 3A. Average contribution of each electron to specific heat of free electron gas at constant volume

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Всички материали от APhO 2007

Theoretical Problem 3 · 5 т.

Условие

1. According to the classical physics the conduction electrons in metals constitute free electron gas like an ideal gas. In thermal equilibrium their average energy relates to temperature, therefore they contribute to the specific heat. The average contribution of each electron to the specific heat of free electron gas at constant volume is defined as

c_V=\frac{d\overline{E}}{dT}\ , \qquad (1)

where $\overline{E}$ is the average energy of each electron. However the value of the specific heat at constant volume is a constant, independent of temperature. Please calculate $\overline{E}$ and the average contribution of each electron to the specific heat at constant volume $c_V$ . [1 т.]

2. (1) Experimentally it has been shown that the specific heat of the conduction electrons at constant volume in metals depends on temperature, and the experimental value at room temperature is about two orders of magnitude lower than its classical counterpart. This is because the electrons obey the quantum statistics rather than classical statistics. According to the quantum theory, for a metallic material the density of states of conduction electrons (the number of electronic states per unit volume and per unit energy) is proportional to the square root of electron energy $E$ , then the number of states within energy range $\mathrm{d}E$ for a metal of volume $V$ can be written as

\mathrm{d}S=CVE^{1/2}\mathrm{d}E \qquad (2)

where $C$ is the normalization constant, determined by the total number of electrons of the system.

The probability that the state of energy $E$ is occupied by electron is

f(E)=\frac{1}{1+\exp(\frac{E-E_F}{k_BT})}\ , \qquad (3)

where $k_B=1.381\times10^{-23}\,\mathrm{J\cdot K^{-1}}$ is the Boltzmann constant and $T$ is the absolute temperature, while $E_F$ is called Fermi level. Usually at room temperature $E_F$ is about several eVs for metallic materials (1eV= $1.602\times10^{-19}$ J). $f(E)$ is called Fermi distribution function shown schematically in the figure below.

Schematic graph of the Fermi distribution function f(E) against E: the curve stays at a constant level, then falls smoothly to zero; three dotted vertical lines are marked on the E axis as E_F-k_BT, E_F and E_F+k_BT; axes labelled f(E) and E.

Please calculate $c_V$ at room temperature according to $f(E)$ [3,5 т.]

2. (2) Please give a reasonable explanation for the deviation of the classical result from that of quantum theory. [0,5 т.]

Note: In your calculation the variation of the Fermi level $E_F$ with temperature could be neglected, i.e. assume $E_F=E_F^0$ , $E_F^0$ is the Fermi level at 0K. Meanwhile the Fermi distribution function could be simplified as a linearly descending function within an energy range of $2k_BT$ around $E_F$ , otherwise either 0 or 1, i.e.

f(E)=\begin{cases}1 & E<E_F-k_BT\\ \text{linearly descending function} & E_F-k_BT<E<E_F+k_BT\\ 0 & E>E_F+k_BT\end{cases}

At room temperature $k_BT\ll E_F$ , therefore calculation can be simplified accordingly. Meanwhile, the total number of electrons can be calculated at 0K.

Отговори

Решение

3A. Average specific heat of each free electron at constant volume

(1). Each free electron has 3 degrees of freedom. According to the equipartition of energy theorem, at temperature $T$ its average energy $\overline{E}$ equals to $\frac{3}{2}k_BT$ , therefore the average specific heat $c_v$ equals to

c_V=\frac{\mathrm{d}\overline{E}}{\mathrm{d}T}=\frac{3}{2}k_B

(2). Let $U$ be the total energy of the electron gas, then

U=\int_0^S Ef(E)\mathrm{d}S

where $S$ is the total number of the electronic states, $E$ the electron energy.
Substitution of (1) for $\mathrm{d}S$ in the above expression gives

U=CV\int_0^{\infty}E^{3/2}f(E)dE=CVI\ ,

where $I$ represents the integral

I=\int_0^{\infty}E^{3/2}f(E)dE\ .

Usually at room temperature $k_BT\ll E_F$ . Therefore, with the simplified $f(E)$

f(E)=\begin{cases}1 & E<E_F-k_BT\\ -\dfrac{E-(E_F+k_BT)}{2k_BT} & E_F-k_BT<E<E_F+k_BT\\ 0 & E>E_F+k_BT\end{cases}

$I$ can be simplified as

I=\frac{2}{5}E_F^{\ 3/2}(1-k_BT/E_F)^{5/2}+\frac{E_F+k_BT}{5k_BT}E_F^{\ 5/2}\left[(1+k_BT/E_F)^{5/2}-(1-k_BT/E_F)^{5/2}\right]

-\frac{1}{7k_BT}E_F^{\ 7/2}\left[(1+k_BT/E_F)^{7/2}-(1-k_BT/E_F)^{7/2}\right]

\approx E_f^{\ 5/2}\left[\frac{2}{5}+\frac{3}{4}\left(k_BT/E_F\right)^2\right]

Therefore

U=CVE_F^{\ 5/2}\left[\frac{2}{5}+\frac{3}{4}\left(k_BT/E_F\right)^2\right]

However the total electron number

N=CV\int_0^{E_F^0}E^{1/2}\mathrm{d}E=\frac{2}{3}CVE_F^{0\,3/2}

CV=\frac{3}{2}N\left(E_F^0\right)^{-3/2}

where $E_F^0$ is the Fermi level at 0K, leading to

U=\frac{3}{2}N\left(E_F^0\right)^{-3/2}E_F^{\ 5/2}\left[\frac{2}{5}+\frac{3}{4}\left(k_BT/E_F\right)^2\right]

Taking $E_F\approx E_F^0$ , and $U=N\overline{E}$ ,

\overline{E}\approx\frac{3}{2}E_F\left[\frac{2}{5}+\frac{3}{4}\left(k_BT/E_F\right)^2\right]

c_v=\frac{\partial\overline{E}}{\partial T}=\frac{9}{4}k_B\frac{k_BT}{E_F}\ll\frac{3}{2}k_B

(3). Because at room temperature $k_BT$ =0.026eV while Fermi level of metals at room temperature is generally of several eVs, it can be seen from the above expression that according to the quantum theory the calculated average specific heat of each free electron at constant volume is two orders of magnitude lower than that of the classical theory. The reason is that with the temperature increase the energy of those electrons whose energy is far below Fermi level (several times of $k_BT$ less than $E_F$ ) does not change obviously, only those minor electrons of energy near $E_F$ contribute to the specific heat, resulting in a much less value of the average specific heat.

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf