In a semiconductor, the number density of intrinsic charge carriers at 27 ° C is 1 . 5 × 10 ^ 16 m ^ - 3 . If the semiconductor is doped with an impurity atom, the hole density increases to 4 . 5 × 10 ^ 22 m ^ - 3 . The

Correct answer: 0 n _ e n _ h = n _ i ^ 2 n _ e = n _ i ^ 2 n _ h = (1 . 5 × 10 ^ 16 ) ^ 2 4 . 5 × 10 ^ 22 = 1 . 5 × 1 . 5 × 10 ^ 32 4 . 5 × 10 ^ 22 5 × 10 ^ 9 m ^ - 3

JEE Main 202125 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 25 July
Shift: Evening Shift
Subject: Physics
Chapter: Semiconductors

JEE Main 2021 · 25 July · Evening Shift · Physics · Semiconductors

In a semiconductor, the number density of intrinsic charge carriers...

In a semiconductor, the number density of intrinsic charge carriers at

27 ° C

1 . 5 \times {10}^{16}{m}^{- 3}

. If the semiconductor is doped with an impurity atom, the hole density increases to

4 . 5 \times {10}^{22}{m}^{- 3}

. The electron density in the doped semiconductor is _______

\times {10}^{9}{m}^{- 3}

Correct answer

Step-by-step solution

${n}_{e}{n}_{h} = {n}_{i}^{2}$

${n}_{e} = {n}_{i}^{2}{n}_{h} = \dfrac{{\left(1 . 5 \times {10}^{16}\right)}^{2}}{4 . 5 \times {10}^{22}}$

$= \dfrac{1 . 5 \times 1 . 5 \times {10}^{32}}{4 . 5 \times {10}^{22}}$

$5 \times {10}^{9}{m}^{- 3}$

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