Let $△E$ denotes the energy gap between the valence band and the conduction band. The population of conduction electrons (and of the holes) is roughly proportional to  $e^{-△E/2kT}$ . Find the ratio of the concentration of conduction electrons in diamond to that in silicon at room temperature 300 K . $△E$  for silicon is 1.1 eV and for diamond is 6.0 eV.

Given  $\eta =e^{-△E/2kT}$ 
   $△E$  (diamond) = 6 eV
             $△E$  (Si)  = 1.1 eV
 Now, ${\eta }_1 \left(Diamond\right)=e^{\frac{-△E_1}{2kT}}$     

$$\begin{gathered} =e^{\frac{-6 eV}{2\times 1.38\times {10}^{-23}\times 300}} \\ =e^{\frac{-6\times 1.6\times {10}^{-19}}{2\times 1.38\times {10}^{-23}\times 300},} \end{gathered}$$

${\eta }_1=e^{-\frac{6}{2\times 300\times 8.62\times {10}^{-5}}} = e^{-116}=4.14722\times {10}^{-51}$                                     

$$\begin{gathered} Similarly , {\eta }_2 \left(Silicon\right)=e^{\frac{-△E_2}{2kT}}=e^{\frac{-1.1}{2\times 1.38\times {10}^{-23}\times 300}} \\ =e^{\frac{-1.1\times 1.6\times {10}^{-19}}{2\times 1.38\times {10}^{-23}\times 300}} \\ =e^{\frac{-1.1}{2\times 8.62\times {10}^{-5}\times 300}} = e^{-21.2683}= 5.7979\times {10}^{-10} \end{gathered}$$

$\therefore ratio = \frac{{\eta }_1}{{\eta }_2}=\frac{4.14722\times {10}^{-51}}{5.7979\times {10}^{-10}}=7.15\times {10}^{-42}$