Chapter 38, Problem 47P

(a)

To determine

The mean free path of the electrons and the copper

Answer to Problem 47P

The mean free path of the electrons and the copper is. $37.1\text{nm}$ and $38.8\text{nm}$ respectively.

Explanation of Solution

Given:

The concentration of doped n-type silicon sample is $n=1.0\times {10}^{16}{\text{cm}}^{\text{-3}}$

The resistivity at $\text{300}\text{K}$ is $\rho =5\times {10}^{23}\text{Ω}\text{.m}$ .

The effective mass of the electrons is $m_{eff}=0.2m_e$ .

The density and resistivity of the copper is ${\rho }_c$ and $\rho =1.7\times {10}^{-8}\text{Ω}\text{.m}$

Formula used:

The expression for the mean free path in terms of average velocity is given by,

  $\lambda =\frac{m_{eff}{v}_m}{n{e}^2\rho }$

The expression of the mean velocity of the electrons is given by:

  $v_m=\sqrt{\frac{3kT}{m_{eff}}}$

Here, K is Boltzmann’s constant $\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)$ and $T$ is absolute temperature.

The expression for the Fermi velocity is given by:

  $u_F=\sqrt{\frac{2E_F}{m_e}}$

Calculation:

The mean free path of the conduction electrons is calculated as,

  $\begin{array}{c}\lambda =\frac{m_{eff}{v}_m}{n{e}^2\rho }\\ =\frac{m_{eff}}{n{e}^2\rho }\sqrt{\frac{3kT}{m_{eff}}}\\ =\frac{\sqrt{3kT{m}_{eff}}}{n{e}^2\rho }\end{array}$

Substitute values in above expression,

  $\begin{array}{c}\lambda =\frac{\sqrt{\text{3}\left(\text{1}{\text{.38×10}}^{\text{-23}}\text{J}/\text{K}\right)\left(\text{300}\text{K}\right)\left(\text{0}\text{.2×9}{\text{.11×10}}^{\text{-31}}\text{kg}\right)}}{\left(\text{1}{\text{.0×10}}^{\text{16}}{\text{cm}}^{\text{-3}}\text{×}\frac{{\left(\text{100}\text{cm}\right)}^{\text{3}}}{\text{1}{\text{m}}^{\text{3}}}\right){\left(\text{1}{\text{.6×10}}^{\text{-19}}\text{C}\right)}^{\text{2}}\left({\text{5×10}}^{\text{23}}\text{Ω}\text{.m}\right)}\\ =3.71\times {10}^{-8}\text{m}\times \left(\frac{\text{1}\text{nm}}{{10}^{-9}\text{m}}\right)\\ =37.1\text{nm}\end{array}$

The Fermi velocity of electrons in the copper is calculated as:

  $\begin{array}{c}{u}_F=\sqrt{\frac{2E_F}{m_e}}\\ =\sqrt{\frac{2\times \left(7.03\text{eV}\right)\times \left(\frac{1.6\times {10}^{-19}\text{J}}{\text{1}\text{}\text{eV}}\right)}{\left(9.11\times {10}^{-31}\text{kg}\right)}}\\ =0.157\times {10}^7\text{m}/\text{s}\end{array}$

The number density of the copper is calculated as:

  $\begin{array}{c}n=\frac{{\rho }_{c}N}{M}\\ =\frac{\left(8.9\text{g}/{\text{cm}}^{\text{3}}\right)\times \left(6.023\times {10}^{23}{\text{mole}}^{\text{-1}}\right)}{\left(63.5\text{g}/\text{mole}\right)}\\ =0.847\times {10}^{23}{\text{cm}}^{\text{-3}}\times \left(\frac{\text{1}{\text{cm}}^{\text{3}}}{{\text{10}}^{\text{-6}}{\text{m}}^{\text{3}}}\right)\\ =8.47\times {10}^{28}{\text{m}}^{\text{-3}}\end{array}$

The free mean path is calculated as:

  $\begin{array}{c}\lambda =\frac{v_m{m}_{eff}}{\rho n{e}^2}\\ =\frac{\left(0.157\times {10}^7\text{m}/\text{s}\right)\times \left(\text{9}{\text{.11×10}}^{\text{-31}}\text{kg}\right)}{\left(1.7\times {10}^{-8}\text{Ω}\text{.m}\right)\times \left(8.47\times {10}^{-19}\text{electron}/{\text{m}}^{\text{3}}\right)\times \left(1.6\times {10}^{-31}\text{kg}\right)}\\ =3.88\times {10}^{-8}\text{m}\times \left(\frac{\text{1}\text{nm}}{{10}^{-9}\text{m}}\right)\\ =38.8\text{nm}\end{array}$

Both these mean free paths are within the $4\%$ of the mean free path of the copper at $300\text{K}$ .

Conclusion:

Therefore, the mean free path of the electrons and the copper is $37.1\text{nm}$ and $38.8\text{nm}$ respectively.