Chapter 14.12, Problem 69AAP

To determine

The electrical conductivity of pure silicon.

Answer to Problem 69AAP

The electrical conductivity of pure silicon is $17.42{\left(\text{Ω}\cdot \text{m}\right)}^{-1}$.

Explanation of Solution

Write the expression for electrical conductivity,

    ${\sigma }_T={\sigma }_o\exp \left[\left(\frac{-E_g}{2k}\right)\frac{1}{T_T}\right]$                                                  (I)

Here, temperature is $T$, Boltzmann constant is $k$, band energy gap is $-E_g$, conductivity at $0\text{°C}$ is ${\sigma }_o$ and the conductivity at $T°\text{C}$ is ${\sigma }_T$.

Write the expression for electrical conductivity,

    $\sigma =\frac{1}{\rho }$                                                                                (II)

Here, the electrical conductivity is $\sigma$ and electrical resistivity of material is $\rho$.

Write the expression for temperature in $\text{K}$.

    $T\left(\text{K}\right)=273+T\left(°\text{C}\right)$                                                         (III)

Here, temperature in Kelvin is $T\left(\text{K}\right)$ and temperature in centigrade is $T\left(°\text{C}\right)$.

Conclusion:

Substitute $325°\text{C}$ for $T\left(°\text{C}\right)$ in Equation (III).

    $\begin{array}{c}T=\left(273+325\text{°C}\right)\text{K}\\ =\text{598}\text{K}\end{array}$

Substitute ${\sigma }_{598}$ for ${\sigma }_T$ and $T_{598}$ for $T_T$ at Equation (I).

    ${\sigma }_{598}={\sigma }_0\exp \left[\left(\frac{-E_g}{2k}\right)\frac{1}{T_{598}}\right]$                                  (IV)

Substitute ${\sigma }_{300}$ for ${\sigma }_T$ and $T_{300}$ for $T_T$ at Equation (I).

    ${\sigma }_{300}={\sigma }_0\exp \left[\left(\frac{-E_g}{2k}\right)\frac{1}{T_{300}}\right]$                          (V)

Divide Equation (IV) by (V).

    $\frac{{\sigma }_{598}}{{\sigma }_{300}}=\frac{{\sigma }_0\exp \left[\left(\frac{-E_g}{2k}\right)\frac{1}{T_{598}}\right]}{{\sigma }_0\exp \left[\left(\frac{-E_g}{2k}\right)\frac{1}{T_{300}}\right]}$

    $\frac{{\sigma }_{598}}{{\sigma }_{300}}=\exp \left[\left(\frac{-E_g}{2k}\right)\left(\frac{1}{T_{598}}-\frac{1}{T_{300}}\right)\right]$      (VI)

Substitute $-1.1\text{eV}$ for $E_g$, $8.62\times {10}^{-5}\frac{\text{eV}}{\text{K}}$ for $k$ and $598\text{K}$ for $T_{598}$ and $300\text{K}$ for $T_{300}$ in Equation (VI).

    $\begin{array}{c}\frac{{\sigma }_{598}}{{\sigma }_{300}}=\exp \left[\left(\frac{-\left(-1.1\text{eV}\right)}{2\times \left(8.62\times {10}^{-5}\frac{\text{eV}}{\text{K}}\right)}\right)\left(\frac{1}{598\text{K}}-\frac{1}{300\text{K}}\right)\right]\\ =\exp \left[10.5986\left(\frac{\text{eV}\cdot \text{K}}{\text{eV}\cdot \text{K}}\right)\right]\\ =\exp \left[10.5986\right]\end{array}$

    ${\sigma }_{598}={\sigma }_{300}\times \exp \left[10.5986\right]$                        (VII)

Substitute $2.3\times {10}^3\text{Ω}\cdot \text{m}$ for $\rho$ in Equation (II) for $300\text{K}$ temperature.

    $\begin{array}{c}{\sigma }_{300}=\frac{1}{2.3\times {10}^3\text{Ω}\cdot \text{m}}\\ =0.4347\times {10}^{-3}{\left(\text{Ω}\cdot \text{m}\right)}^{-1}\end{array}$

Substitute $0.4347\times {10}^{-3}{\left(\text{Ω}\cdot \text{m}\right)}^{-1}$ for ${\sigma }_{300}$ in Equation (VI).

    $\begin{array}{c}{\sigma }_{598}=0.4347\times {10}^{-3}{\left(\text{Ω}\cdot \text{m}\right)}^{-1}\times \exp \left[10.5986\right]\\ =17.42{\left(\text{Ω}\cdot \text{m}\right)}^{-1}\end{array}$

The electrical conductivity of pure silicon is $17.42{\left(\text{Ω}\cdot \text{m}\right)}^{-1}$.