Question

Pure Si at $500K$ has equal number of electrons$\left(n_e\right)$ and hole $\left(n_h\right)$ concentrations of $1.5\times {10}^{16}{m}^{-3}$. Doping by indium increases $n_{h}to4.5\times {10}^{22}{m}^{-3}$. The doped semiconductor is of :

• A. n-type with electron concentration $n_e=5\times {10}^{22}{m}^{-3}$
• B. p-type with electron concentration $n_e=2.5\times {10}^{23}{m}^{-3}$
• C. n-type with electron concentration $n_e=2.5\times {10}^{10}{m}^{-3}$
• D. p-type with electron concentration $n_e=5\times {10}^9{m}^{-3}$

Answer 1

The correct option is D p-type with electron concentration $n_e=5\times {10}^9{m}^{-3}$

We know that
$n_i^2=n_e{n}_h\Rightarrow n_e=\frac{{\left(n_i\right)}^2}{n_h}\Rightarrow n_e=\frac{{\left(1.5\times {10}^{16}\right)}^2}{\left(}\Rightarrow n_e=5\times {10}^9{m}^{-3}So{n}_h>n_e$
Therefore semiconductor is p-type.