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JEE Main 20178 AprilOnline
Exam
JEE Main
Year
2017
Date
8 April
Mode
Online
Subject
Physics
Chapter
Semiconductors

JEE Main 2017 · 8 April · Online · Physics · Semiconductors

The conductivity of a semiconductor sample having electron...

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 )

Options

  1. A1.83(Ωm)−11 .83 {\left(Ωm\right)}^{- 1}
  2. B1.65(Ωm)−11 .65 {\left(Ωm\right)}^{- 1}
  3. C1.20(Ωm)−11 .20 {\left(Ωm\right)}^{- 1}
  4. D0.59(Ωm)−10 .59 {\left(Ωm\right)}^{- 1}

Correct answer

B. 1.65(Ωm)−11 .65 {\left(Ωm\right)}^{- 1}

Step-by-step solution

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 {( Ωm )}^{- 1}

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