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

The correct option is A ${10}^6{m}^{-3}$

Given - Intrinsic carrier concentration $n_i=2\times {10}^8{m}^{-3}$

Electron concentration $n_e=4\times {10}^{10}{m}^{-3}$

Hole concentration $n_h=?$

We have , $n_e{n}_h=n_i^2$

or $n_h=\frac{n_i^2}{n_e}=\frac{(2\times {10}^8{)}^2}{4\times {10}^{10}}={10}^6{m}^{-3}$