In an extrinsic semiconductor, the number density of holes is $4\times10^{20}\ \mathrm{m^{-3}}$. If the number density of intrinsic carriers is $1.2\times10^{15}\ \mathrm{m^{-3}}$, the number density of electrons in it is

(C) $3.6\times10^{9}\ \mathrm{m^{-3}}$

(A) $1.8\times10^{9}\ \mathrm{m^{-3}}$

(B) $2.4\times10^{10}\ \mathrm{m^{-3}}$

(D) $3.2\times10^{9}\ \mathrm{m^{-3}}$

CBSE Class 12 · 2023 · SemiconductorsMultiple Choice

In an extrinsic semiconductor, the number density of holes is

4 \times 1 0^{20} m^{- 3}

. If the number density of intrinsic carriers is

1.2 \times 1 0^{15} m^{- 3}

, the number density of electrons in it is

Answer: (C)

The mass-action law gives

n_{e} n_{h} = n_{i}^{2}

n_{e} = \frac{n _{i}^{2}}{n _{h}} = \frac{( 1.2 \times 1 0 ^{15} ) ^{2}}{4 \times 1 0 ^{20}} = \frac{1.44 \times 1 0 ^{30}}{4 \times 1 0 ^{20}} .

n_{e} = 3.6 \times 1 0^{9} m^{- 3} .

n_{e} = 3.6 \times 1 0^{9} m^{- 3}

CBSE Class 12 Board · 2023 · Single correct option