Chapter 38, Problem 66P

(a)

To determine

Show that the total number of energy states is $\frac{2}{3}A{E}_F^{\frac{3}{2}}$ .

Answer to Problem 66P

It is showed that the total number of energy states is $\frac{2}{3}A{E}_F^{\frac{3}{2}}$ .

Explanation of Solution

Given:

The density of the electron states in a metal is $g\left(E\right)=A{E}^{\frac{1}{2}}$ .

Formula used:

The number of energy states is given by,

  $N={\displaystyle \underset{0}{\overset{E_F}{\int }}g\left(E\right)}dE$

Calculation:

The number of energy states in a metal is calculated as:

  $\begin{array}{c}N={\displaystyle \underset{0}{\overset{E_F}{\int }}g\left(E\right)}dE\\ ={\displaystyle \underset{0}{\overset{E_F}{\int }}\left(A{E}^{\frac{1}{2}}\right)}dE\\ =\frac{2A}{3}{\left[E^{\frac{3}{2}}\right]}_0^{E_F}\\ =\frac{2A{E}_F^{\frac{3}{2}}}{3}\end{array}$

Conclusion:

Therefore, it is showed that the total number of energy states is $\frac{2}{3}A{E}_F^{\frac{3}{2}}$ .

(b)

To determine

The fraction of the conduction electrons that are within $kT$ of Fermi energy

Answer to Problem 66P

The fraction of the conduction electrons that are within $kT$ of Fermi energy is $\frac{3}{2}\frac{KT}{E_F}$ .

Explanation of Solution

Formula used:

The number of energy states is given by,

  $N={\displaystyle \underset{0}{\overset{E_F}{\int }}g\left(E\right)}dE$

Calculation:

The fraction of number of states that is within $kT$ of the Fermi energy is calculated as:

  $\begin{array}{c}\frac{kTg\left(E_F\right)}{N}=\frac{kT{\left(A{E}_F\right)}^{\frac{1}{2}}}{\frac{2}{3}A{E}_F^{\frac{3}{2}}}\\ =\frac{3}{2}\frac{KT}{E_F}\end{array}$

Conclusion:

Therefore, the fraction of the conduction electrons that are within $kT$ of the Fermi energy is $\frac{3}{2}\frac{KT}{E_F}$ .

(c)

To determine

The fraction of the conduction electrons of copper that are within $kT$ of Fermi energy at $T=30\text{0}\text{K}$ .

Answer to Problem 66P

The fraction of the conduction electrons of copper that are within $kT$ of Fermi energy at $T=30\text{0}\text{K}$ is $5.51\times {10}^{-3}$ .

Explanation of Solution

Formula used:

The expression forthe fraction of the conduction electrons that are within $kT$ of the Fermi energyis given by,

  $\frac{kTg\left(E_F\right)}{N}=\frac{3}{2}\frac{KT}{E_F}$

Calculation:

The value of Fermi energy of copper is $E_F=7.04\text{eV}$ . The fraction of the conduction electrons of copper that are within $kT$ of Fermi energy at $T=30\text{0}\text{K}$ is calculated as:

  $\begin{array}{c}\frac{kTg\left(E_F\right)}{N}=\frac{3}{2}\frac{KT}{E_F}\\ =\frac{3}{2}\frac{\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)\times \left(300\text{K}\right)}{\left(7.04\text{eV}\right)}\\ =5.51\times {10}^{-3}\end{array}$

Conclusion:

Therefore, the fraction of the conduction electrons of copper that are within $kT$ of Fermi energy at $T=30\text{0}\text{K}$ is $5.51\times {10}^{-3}$ .