Question

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

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

Answer 1

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

Given: $T=500\text{K}$,
$n_i=1.5\times {10}^{16}{\text{m}}^{-3}$,
$n_h=4.5\times {10}^{22}{\text{m}}^{-3}$

In an extrinsic semiconductor, $n_i^2=n_e{n}_h$
$(1.5\times {10}^{16}{)}^2=n_e(4.5\times {10}^{22})$
$n_e=0.5\times {10}^{10}$
$n_e=5\times {10}^9$
$n_h=4.5\times {10}^{22}$
$n_h>>n_2$
Semiconductor is p-type and $n_e=5\times {10}^9{\text{m}}^{-3}$

Hence, option $D$ is correct.