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, $n_e^{{}^{\prime }}=4\times 10^{10}{m}^{-3}$
New hole concentration, $n_h^{{}^{\prime }}=?$
We know that, $n_e\times n_h=n_i^2$
and $n_e^{{}^{\prime }}\times n_h^{{}^{\prime }}=n_i^2$
$\therefore n_e\times n_h=n_e^{{}^{\prime }}\times n_h^{{}^{\prime }}$
$n_h^{{}^{\prime }}=\frac{n_e\times n_h}{n_e^{\prime }}=\frac{2\times 10^8\times 2\times 10^8}{4\times 10^{10}}/m^3=10^6/m^3$
$\Rightarrow n_h^{{}^{\prime }}=10^6{m}^{-3}$