Question

The hole and the free electron concentrations in pure silicon at room temperature are given by $1.4 \times 10^{16}\ \mathrm{m}^{-3}$. When doped with indium, and the hole concentration becomes $4 \times 10^{22}\ \mathrm{m}^{-3}$, the electron concentration is

• A. $0.49 \times 10^{10}\ \mathrm{m}^{-3}$
• B. $0.14 \times 10^{10}\ \mathrm{m}^{-3}$
• C. $0.36 \times 10^{10}\ \mathrm{m}^{-3}$
• D. $0.72 \times 10^{10}\ \mathrm{m}^{-3}$

Hint:

Use mass action law: $n_i^2 = n_e n_h$ to compute carrier concentrations in semiconductors.

Answer 1

The Correct Option is A

We use the mass action law: $n_i^2 = n_e \cdot n_h$
Given: $n_i = 1.4 \times 10^{16},\ n_h = 4 \times 10^{22}$
So, $n_e = \frac{n_i^2}{n_h} = \frac{(1.4 \times 10^{16})^2}{4 \times 10^{22}} = 0.49 \times 10^{10}$