Pure Si at 500 K has equal number of electron $(n_e)$ and hole $(n_h)$ concentrations of $1.5×10^{16}m^{-3}$. Doping by indium increases $n_h$ to $4.5×10^{22}m^{-3}$. The doped semiconductor is of:
Answer & explanation
Correct answer: option 4
As $n_i^2=n_e×n_h$
$(1.5×10^{16}m^{-3})^2=n_e×(4.5×10^{22}m^{-3})$
$∴n_e=\frac{(1.5×10^{16}m^{-3})^2}{(4.5×10^{22}m^{-3})}=5×10^{9}m^{-3}$
As $n_h >> n_e$, so semiconductor is p – type and $n_e=5×10^{9}m^{-3}$