The conductivity of a semiconductor is given by
σ=e(neμe+nhμh)
=1.6×10−19(5×1018×2+5×1019×0.01)
=1.6×10−19(1019+0.05×1019)
=1.6+1.05
=1.65(Ωm)−1
JEE Main 2017 — Physics Modern Physics
The conductivity of a semiconductor sample having electron concentration of 5×1018electronsm−3, hole concentration of 5×1019holesm−3, electron mobility of 2.0m2V−1s−1 and hole mobility of 0.01m2V−1s−1 is
(Take charge of an electron as 1.6×10−19C )
Held on 8 Apr 2017 · Verified 6 Jul 2026.
1.83(Ωm)−1
1.65(Ωm)−1
1.20(Ωm)−1
0.59(Ωm)−1
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