Question

Pure Si at 500 K has equal number of electron $\left(n_e\right)$ and hole $\left(n_h\right)$ concentrations of $1.5\times {10}^{16}{m}^{-3}$. Doping by indium increases $n_h$ to $4.5\times {10}^{22}{m}^{-3}$. The doped semiconductor is of :

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

Answer 1

The correct option is A p-type having electron concentrations $n_e=5\times {10}^9{m}^{-3}$

Pure Silicon,$T=500K$

$n_e$ concentration$=1.5\times {10}^{16}{m}^{-3}$

$n_h$ concentration$=1.5\times {10}^{16}{m}^{-3}$

Doped with indium,increase $n_h$

So, Now

$n_h=4.5\times {10}^{22}m-3$

To find $n_c=$elecron concentration after doping.

$n_i^2=n_e{n}_h$

$n_e=\frac{n_i^2}{n_h}$

$\Rightarrow n_e=\frac{{(1.5\times {10}^{16})}^2}{4.5\times {10}^{22}}$

$\Rightarrow n_e=5\times {10}^9{m}^{-3}$

so, $n_h>>n_e$ semiconductor is $p-$type.