Semiconductors

JEE Main level, single correct, medium

22 of 209

Question 22 of 209

The conductivity of a semiconductor sample having electron concentration of 5×1018electronsm−35\times {10}^{18}electrons{m}^{-3}, hole concentration of 5×1019holesm−35\times {10}^{19}holes{m}^{-3}, electron mobility of 2.0m2V−1s−12.0{m}^{2}{V}^{-1}{s}^{-1} and hole mobility of 0.01m2V−1s−10.01{m}^{2}{V}^{-1}{s}^{-1} is

(Take charge of an electron as 1.6×10−19C1.6\times {10}^{-19}C )

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Answer: (B) 1.65(Ωm)−11.65{\left(\Omega m\right)}^{-1}

The conductivity of a semiconductor is given by

σ=e(neμe+nhμh)\sigma =e\left({n}_{e}{\mu }_{e}+{n}_{h}{\mu }_{h}\right)

=1.6×10−19(5×1018×2+5×1019×0.01)=1.6\times {10}^{-19}\left(5\times {10}^{18}\times 2+5\times {10}^{19}\times 0.01\right)

=1.6×10−19(1019+0.05×1019)=1.6\times {10}^{-19}\left({10}^{19}+0.05\times {10}^{19}\right)

=1.6+1.05=1.6+1.05

=1.65(Ωm)−1=1.65{(\Omega m)}^{-1}

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