# 物理代写|固体物理代写Solid-state physics代考|KYA322

## 物理代写|固体物理代写Solid-state physics代考|Electrical Conductivity

In the absence of an electric field, the free electron moves randomly in the metal. During the motion, they collide with fixed positive ions as well as with other electrons. Since the motion is completely random, therefore the average velocity of electrons in any direction is zero. If a constant electric field $\boldsymbol{E}$ is applied inside a metal, the electrons experience a force $\boldsymbol{F}=-e \boldsymbol{E}$. As a result, they move in a direction opposite to the direction of the electric field. This electron undergoes frequent collisions, and it is assumed that immediately after collision the electron velocities are completely random; that is, electron from a collision does not remember if it had been previously accelerated or not. Thus, the momentum gained under the influence of electric field is lost. As a result of the application of electric field and random motion, the electrons are subjected to a very slow directional motion. This motion is called drift, and the average velocity of this motion is called drift velocity $v_d$

When an electric field $\boldsymbol{E}$ is applied to the metal, the electrons are accelerated in the direction of the field and acquire an average drift velocity and momentum $\boldsymbol{p}$ parallel to $\boldsymbol{E}$. In time $\mathrm{d} t$, an electron of charge $-e$ acquires an additional momentum $-e \boldsymbol{E} \mathrm{d} t$ through the acceleration by the field $\mathbf{E}$. Let in time $\mathrm{d} t$, a fraction $\mathrm{d} n$ of the total number of electrons $n$ per unit volume makes collision where
$$\frac{\mathrm{d} n}{n}=\frac{\mathrm{d} t}{\tau}$$
where $\tau$ is mean time between collisions. Immediately after the collision, the electron velocities are completely random, and hence, the momentum gained under the influence of the electric field is lost. The momentum gained in time $\mathrm{d} t$ is
$$-n e \boldsymbol{E} \mathrm{d} t$$
The momentum destroyed in collision from Eq. (8.1) is
$$p \mathrm{~d} n=n p \frac{\mathrm{d} t}{\tau}$$
For equilibrium, there must be balance. From Eqs. (8.2) and (8.3)

## 物理代写|固体物理代写Solid-state physics代考|Wiedemann and Franz Law

Since metals are much better conductors of heat than electrical insulators, therefore it is assumed that the thermal conduction in a metal is also mainly due to free electrons. From Eq. (7.71), the thermal conductivity is
$$K=\frac{1}{3} C_e v l$$
The electronic specific heat $C_e$ for electron gas
$$C_e=\frac{1}{3} N v l \frac{d E}{d T}$$
The kinetic energy $E$ is
$$E=\frac{1}{2} m v^2=k_B T$$

From Eqs. (8.6) and (8.9)
$$\frac{K}{\sigma}=\frac{(1 / 3) N v l(\mathrm{~d} E / \mathrm{d} T)}{N\left(e^2 / m\right) \tau}=\frac{m v^2}{3 e^2} \frac{\mathrm{d} E}{\mathrm{~d} T}$$
If the electron obeys classical statistics
$$\begin{gathered} E=\frac{1}{2} m v^2=\frac{3}{2} k_B T \ m v^2=3 k_B T \end{gathered}$$
From Eqs. (8.12)-(8.14)
$$\frac{K}{\sigma}=\frac{3 k_B T}{3 e^2} \frac{\mathrm{d}}{\mathrm{d} T}\left(\frac{3}{2} k_B T\right)=\frac{3}{2}\left(\frac{k_B}{e^2}\right)^2 T$$
This equation shows that the ratio of the thermal and electrical conductivity should be oroportional to absolute temperature for a given metal and it should be the same for all metals at a given temperature. This is Wiedemann and Franz law. The numerical value of $(K / \sigma T)$ given by Eq. (8.15) is in good agreement with the experimental values for copper, silver and gold over the limited temperature range of the experiments. The experimental values of $\mathrm{K}$ and $\sigma$ themselves and their variation with temperature do not, however, fit he theory.

# 固体物理代写

## 物理代写|固体物理代写Solid-state physics代考|Electrical Conductivity

$$\frac{\mathrm{d} n}{n}=\frac{\mathrm{d} t}{\tau}$$

$$-n e \boldsymbol{E} \mathrm{d} t$$

$$p \mathrm{~d} n=n p \frac{\mathrm{d} t}{\tau}$$

## 物理代写|固体物理代写Solid-state physics代考|Wiedemann and Franz Law

$$K=\frac{1}{3} C_e v l$$

$$C_e=\frac{1}{3} N v l \frac{d E}{d T}$$

$$E=\frac{1}{2} m v^2=k_B T$$

$$\frac{K}{\sigma}=\frac{(1 / 3) N v l(\mathrm{~d} E / \mathrm{d} T)}{N\left(e^2 / m\right) \tau}=\frac{m v^2}{3 e^2} \frac{\mathrm{d} E}{\mathrm{~d} T}$$

$$E=\frac{1}{2} m v^2=\frac{3}{2} k_B T m v^2=3 k_B T$$

$$\frac{K}{\sigma}=\frac{3 k_B T}{3 e^2} \frac{\mathrm{d}}{\mathrm{d} T}\left(\frac{3}{2} k_B T\right)=\frac{3}{2}\left(\frac{k_B}{e^2}\right)^2 T$$

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