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Physics Question on Semiconductor electronics: materials, devices and simple circuits

What is the conductivity of a semiconductor sample having electron concentration of 5×1018 m−35 \times 10^{18} \, m^{-3}, hole concentration of 5×1019 m−35 \times 10^{19} \, m^{-3} , electron mobility of 2.0 m2 V−1 s−12.0 \, m^2 \, V^{-1} \, s^{-1} and hole mobility of 0.01 m2 V−1 s−10.01 \, m^2 \, V^{-1} \, s^{-1} ? (Take charge of electron as 1.6×10−19C1.6 \times 10^{-19} C)

A

1.68(Ω−m)−11.68 (\Omega - m)^{-1}

B

1.83(Ω−m)−11.83 (\Omega - m)^{-1}

C

0.59(Ω−m)−10.59 (\Omega - m)^{-1}

D

1.20(Ω−m)−11.20 (\Omega - m)^{-1}

Answer

1.68(Ω−m)−11.68 (\Omega - m)^{-1}

Explanation

Solution

s=e(neμe+nnμn)s = e \left(n_{e}\mu_{e} + n_{n}\mu_{n}\right)
=1.6?10−19(5?1018?2+5?1019?0.01)= 1.6 ? 10^{-19} \left(5 ? 10^{18} ? 2 + 5 ? 10^{19} ? 0.01\right)
=1.6?10−19(1019+0.05?1019)= 1.6 ? 10^{-19} \left(10^{19} + 0.05 ? 10^{19}\right)
=1.6?1.05= 1.6 ? 1.05
=1.68= 1.68