Chapter 38, Problem 36P

To determine

The mean free paths for the conduction electrons for Na, Au and Sn.

Answer to Problem 36P

The mean free paths for the conduction electrons for Na, Au and Sn are $34.1\text{nm,}40.9\text{nm}\text{and}\text{4}\text{.27}\text{nm}$ respectively.

Explanation of Solution

Given:

The temperature is $T=273\text{K}$ .

The resistivity and Fermi speed of Na is ${\rho }_1=4.2\text{μΩ}\cdot \text{cm}\text{and}{v}_{F_1}=1.07\times {10}^6\text{m}/\text{s}$ .

The resistivity and Fermi speed of Au is ${\rho }_2=2.04\text{μΩ}\cdot \text{cm}\text{and}{v}_{F_2}=1.39\times {10}^6\text{m}/\text{s}$

The resistivity and Fermi speed of Sn is ${\rho }_3=10.6\text{μΩ}\cdot \text{cm}\text{and}{v}_{F_3}=1.89\times {10}^6\text{m}/\text{s}$

Formula used:

The expression of mean free path is given by,

  $\lambda =\frac{m_e{v}_{\text{F}}}{n{e}^2\rho }$

Here, $m_e$ is mass of electron, $n$ is number of electrons per unit volume and $e$ is the charge of electron.

Calculation:

Refer to the Table 38-1, the value of $n$ for given elements is,

  $n_{\text{Na}}=26.5\text{electrons}/{\text{nm}}^{\text{3}}$ , $n_{\text{Au}}=59.0\text{electrons}/{\text{nm}}^{\text{3}}$ and $n_{\text{Sn}}=148\text{electrons}/{\text{nm}}^{\text{3}}$ .

The mean free paths for the conduction electrons for Na is calculated as,

  $\begin{array}{c}{\lambda }_{\text{Na}}=\frac{\left(9.109\times {10}^{-31}\text{kg}\right)\left(1.07\times {10}^6\text{m}/\text{s}\right)}{\left(26.5\text{electrons}/{\text{nm}}^{\text{3}}\right){\left(1.602\times {10}^{-19}\text{C}\right)}^2\left(4.2\text{μΩ}\cdot \text{cm}\right)}\\ =\frac{\left(9.109\times {10}^{-31}\text{kg}\right)\left(1.07\times {10}^6\text{m}/\text{s}\right)}{\left(26.5\times {10}^{22}\text{electrons}/{\text{cm}}^{\text{3}}\right){\left(1.602\times {10}^{-19}\text{C}\right)}^2\left(4.2\text{μΩ}\cdot \text{cm}\right)}\\ =34.1\text{nm}\end{array}$

The mean free paths for the conduction electrons for Au is calculated as,

  $\begin{array}{c}{\lambda }_{\text{Au}}=\frac{\left(9.109\times {10}^{-31}\text{kg}\right)\left(1.39\times {10}^6\text{m}/\text{s}\right)}{\left(59.0\text{electrons}/{\text{nm}}^{\text{3}}\right){\left(1.602\times {10}^{-19}\text{C}\right)}^2\left(2.04\text{μΩ}\cdot \text{cm}\right)}\\ =\frac{\left(9.109\times {10}^{-31}\text{kg}\right)\left(1.39\times {10}^6\text{m}/\text{s}\right)}{\left(5.9\times {10}^{22}\text{electrons}/{\text{cm}}^{\text{3}}\right){\left(1.602\times {10}^{-19}\text{C}\right)}^2\left(2.04\text{μΩ}\cdot \text{cm}\right)}\\ =40.9\text{nm}\end{array}$

The mean free paths for the conduction electrons for Sn is calculated as,

  $\begin{array}{c}{\lambda }_{\text{Sn}}=\frac{\left(9.109\times {10}^{-31}\text{kg}\right)\left(1.89\times {10}^6\text{m}/\text{s}\right)}{\left(148\text{electrons}/{\text{nm}}^{\text{3}}\right){\left(1.602\times {10}^{-19}\text{C}\right)}^2\left(10.6\text{μΩ}\cdot \text{cm}\right)}\\ =\frac{\left(9.109\times {10}^{-31}\text{kg}\right)\left(1.89\times {10}^6\text{m}/\text{s}\right)}{\left(14.8\times {10}^{22}\text{electrons}/{\text{cm}}^{\text{3}}\right){\left(1.602\times {10}^{-19}\text{C}\right)}^2\left(10.6\text{μΩ}\cdot \text{cm}\right)}\\ =4.27\text{nm}\end{array}$

Conclusion:

Therefore, the mean free paths for the conduction electrons for Na, Au and Sn are $34.1\text{nm,}40.9\text{nm}\text{and}\text{4}\text{.27}\text{nm}$ respectively.