Chapter 38, Problem 31P

(a)

To determine

The free-electron density in gold.

Answer to Problem 31P

The free-electron density in goldis $5.90\times {10}^{28}\text{e}/{\text{m}}^{\text{3}}$ .

Explanation of Solution

Given:

The density of gold is $\rho =19.3\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ .

The atomic mass is $m=196.97\text{g}/\text{mol}$ .

Formula used:

The expression for free-electron density is given by

  $n_e=\frac{\rho N_A{N}_{\text{atom}}}{m}$

Here, $N_A$ is Avogadro’s constant and $N_{\text{atom}}$ is number of electron per atom.

Calculation:

The free electron density of goldis calculated as,

  $\begin{array}{c}{n}_e=\frac{\left(19.3\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(6.022\times {10}^{23}\text{atoms}/\text{mol}\right)\left(\frac{1\text{e}}{1\text{atom}}\right)}{\left(196.97\text{g}/\text{mol}\right)}\\ =\frac{\left(19.3\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(6.022\times {10}^{23}\text{atoms}/\text{mol}\right)\left(\frac{1\text{e}}{1\text{atom}}\right)}{\left(196.97\text{g}/\text{mol}\right)\left(\frac{{10}^{-3}\text{kg}/\text{mol}}{1\text{g}/\text{mol}}\right)}\\ =5.90\times {10}^{28}\text{e}/{\text{m}}^{\text{3}}\end{array}$

Conclusion:

Therefore, the free electron density in gold is $5.90\times {10}^{28}\text{e}/{\text{m}}^{\text{3}}$ .

(b)

To determine

The Fermi energy for gold.

Answer to Problem 31P

The Fermi energy for goldis $5.50\text{eV}$ .

Explanation of Solution

Given:

The Fermi speed for goldis $v_F=1.39\times {10}^6\text{m}/\text{s}$ .

Formula used:

The expression for Fermi energy is given by,

  $E_F=\frac{1}{2}{m}_e{v}_F^2$

Here, $m_e$ is the mass of the electron.

Calculation:

The Fermi energy for gold is calculated as,

  $\begin{array}{c}{E}_F=\frac{1}{2}\left(9.109\times {10}^{-31}\text{kg}\right){\left(1.39\times {10}^6\text{m}/\text{s}\right)}^2\\ =\left(8.7997\times {10}^{-19}\text{J}\right)\left(\frac{1\text{eV}}{1.602\times {10}^{-19}\text{J}}\right)\\ =5.4929\text{eV}\\ \approx \text{5}\text{.50}\text{eV}\end{array}$

Conclusion:

Therefore, the Fermi energy for gold is $5.50\text{eV}$ .

(c)

To determine

The factor between Fermi energy and $kT$ energy.

Answer to Problem 31P

The factor between Fermi energy and $kT$ energyis $212$ .

Explanation of Solution

Given:

The $kT$ energy at room temperature is $kT=0.026\text{eV}$ .

Formula used:

The expression for required factor is given by,

  $f=\frac{E_F}{kT}$

Calculation:

The required factor is calculated as,

  $\begin{array}{c}f=\frac{5.50\text{eV}}{0.026\text{eV}}\\ =211.5\\ \approx 212\end{array}$

Conclusion:

Therefore, the factor by between Fermi energy and $kT$ energy is $212$ .

(d)

To determine

The difference between Fermi energy and $kT$ energy.

Explanation of Solution

Introduction:

The difference between higher and lower energy level that is occupied by the charged particle of material at $0\text{K}$ is called Fermi energy.

$kT$ energy of the average conduction electron at any temperature $T$ and it is always less than or equal to Fermi energy.

At absolute zero, the energy available at conduction electron in a higher energy state is termed as Fermi energy. It is higher than or equal to $kT$ energy because no two electrons can occupy the same higher energy state simultaneously.

When the electron does not obey the exclusion principle, the energy of average conduction electrons at any temperature $T$ is called $kT$ energy.

Conclusion:

Therefore, the Fermi energy is always greater than or equal to $kT$ energy.