Pure Si at 500K has equal number of electron $(n_e)$ and hole $(n_h)$ concentrations of $1.5 × 10^{16} m^{–3}$. Doping by indium increases nh to $4.5 × 10^{22} m^{–3}$. The doped semiconductor is of
Answer & explanation
Correct answer: option 4
$n_i^2= n_en_h$
$(1.5×10^{16})^2 = n_3(4.5×10^{22})$
$⇒ n_e = 0.5×10^{10}$
or $n_e = 5×10^9$
Given $n_h = 4.5×10^{22}$
$⇒ n_h = 4.5×10^{22}$