Question

In an intrinsic semiconductor band gap is $1.2eV$, then the ratio of number of charge carriers at $600K$ and $300K$ is of order:
A. ${10^4}$
B. ${10^7}$
C. ${10^{10}}$
D. ${10^3}$

Answer 1

Hint: The main key to these solutions is observation of the semiconductor nature, as we can deduce the impact of the temperature by the nature of semiconductor material. After that we can deduce the equations by comparisons.

Complete step by step answer:
The energy gap ${E_g} = 1.2eV$
Now the relation in charge carrier and temperature is
${n_i} = {n_o}\exp [ - {E_g}/2{K_B}T]$ where ${K_B}$is Boltzmann constant and its value is $8.62\times{10^{ - 5}}eV/K$
So at the given temperature of $300K$the relation is
${n_{i1}} = {n_o}\exp [ - {E_g}/2{K_B}\times300]$
Similarly at the given temperature of $600K$the relation is
${n_{i2}} = {n_o}\exp [ - {E_g}/2{K_B}\times600]$
Now the ratio of the given two is
$\dfrac{{{n_{i2}}}}{{{n_{i1}}}} = \dfrac{{{n_o}\exp [ - {E_g}/2{K_B}\times600]}}{{{n_o}\exp [ - {E_g}/2{K_B}\times300]}}$
$
   = \exp \dfrac{{{E_g}}}{{2{K_B}}}[\dfrac{1}{{300}} - \dfrac{1}{{600}}] \\
   = \exp [11.6] \\
   = 1.09\times{10^5} \\
   \approx {10^5} \\
 $

Note:- To deduce these relations we have to forget minor segments and values as they won’t affect anything at the large scale. But if anything isn’t at any decimal stage then they should be taken in account in finding the relations as they will have their consequences.