What is the conductivity of a semiconductor sample having electron concentration of 5 × 10¹⁸ m⁻³, hole concentration of 5 × 10¹⁹ m⁻³, electron mobility of 2.0 m2 V ⁻¹ s-1 and hole mobility of 0.01 m2 V ⁻¹ s ⁻¹? (Take charge of electron as 1.6 × 10 - 19 C) (a) 1.68 ( Ω -m) ⁻¹ (b) 1.83 ( Ω -m) ⁻¹ (c) 0.59 ( Ω -m) ⁻¹ (d) 1.20 ( Ω -m) ⁻¹

Answer: (a). σ = e(ne μe + nh μh) ne μe = 5 × 10¹⁸ × 2.0 = 1.0 × 10¹⁹, nh μh = 5 × 10¹⁹ × 0.01 = 0.05 × 10¹⁹ σ = 1.6 × 10⁻¹⁹ × 1.05 × 10¹⁹ = 1.68 (Ω-m)⁻¹ So the answer is (a) 1.68 (Ω-m)⁻¹.

What is the conductivity of a semiconductor sample having electron… | JEE Main 2017 (8 Apr, Shift 1)

Q47JEE Main 2017 · 8 APR · SHIFT 1Medium

Intrinsic & Extrinsic Semiconductors

What is the conductivity of a semiconductor sample having electron concentration of $5 \times 10^{18}\text{ m}^{-3},$ hole concentration of $5 \times 10^{19}\text{ m}^{-3},$ electron mobility of 2.0 m2 V ${}^{-1}$ s-1 and hole mobility of 0.01 m2 V ${}^{-1}$ s ${}^{-1}$ ?

(Take charge of electron as 1.6 $\times$ 10 $-$ 19 C)

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Answer: (a)

$\sigma = e(n_e \mu_e + n_h \mu_h)$
$n_e \mu_e = 5 \times 10^{18} \times 2.0 = 1.0 \times 10^{19}$ , $n_h \mu_h = 5 \times 10^{19} \times 0.01 = 0.05 \times 10^{19}$
$\sigma = 1.6 \times 10^{-19} \times 1.05 \times 10^{19} = 1.68\ (\Omega\text{-m})^{-1}$
So the answer is (a) $1.68\ (\Omega\text{-m})^{-1}$ .

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