Question

A semiconductor has equal electron and hole concentration of $2 \times 10^{8} \,m ^{-3}$. On doping with a certain impurity, the electron concentration increases to $4 \times 10^{10} \,m ^{-3}$, then the new hole concentration of the semiconductor is

• A. $10^{6} \, m ^{-3}$
• B. $10^{8} \, m ^{-3}$
• C. $10^{10} \,m ^{-3}$
• D. $10^{12} \,m ^{-3}$

Answer 1

Answer: (A)

Electron concentration, $n_{e}=2 \times 10^{8}\, m ^{-3}$
Hole concentration, $n_{h}=2 \times 10^{8} \,m ^{-3}$
After doping with a impurity, the new electron concentration, $n_{e}^{'}=4 \times 10^{10}\, m ^{-3}$
New hole concentration, $n_{h}^{'}=?$
We know that, $n_{e} \times n_{h}=n_{i}^{2}$
and $n_{e}^{'} \times n_{h}^{'}=n_{i}^{2} $
$\therefore n_{e} \times n_{h}=n_{e}^{'} \times n_{h}^{'}$
$n_{h}^{'}=\frac{n_{e} \times n_{h}}{n_{e}^{\prime}}=\frac{2 \times 10^{8} \times 2 \times 10^{8}}{4 \times 10^{10}} / m ^{3}=10^{6} / m ^{3}$
$\Rightarrow n_{h}^{'}=10^{6} \,m ^{-3}$