The conductivity of a semiconductor sample having electron concentration of 5 × 10 18 e m - 3 , hole concentration of 5 × 10 19 holes m - 3 , electron mobility of 2 .0 m 2 V - 1 s - 1 and hole mobilit

Question

The conductivity of a semiconductor sample having electron concentration of $5 \times 10^{18} e m^{- 3}$ , hole concentration of $5 \times 10^{19} holes m^{- 3}$ , electron mobility of $2 .0 m^{2} V^{- 1} s^{- 1}$ and hole mobility of $0.01 m^{2} V^{- 1} s^{- 1}$ is

(Take charge of an electron as $1 .6 \times 10^{- 19} C$ )

TestHub · Accepted Answer

The conductivity of a semiconductor is given by

$σ = e (n_{e} μ_{e} + n_{h} μ_{h})$

$= 1 .6 \times 10^{- 19} (5 \times 10^{18} \times 2 + 5 \times 10^{19} \times 0 .01)$

$= 1.6 \times 10^{- 19} (10^{19} + 0.05 \times 10^{19})$

$= 1.6 + 1.05$

$= 1 .65 {(Ω m)}^{- 1}$

Physics - Semiconductors Question with Solution | TestHub

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