First slide

Electrons and holes in semiconductors

Question

Calculate the conductivity of silicon crystal at 300 K. It is given that $μ_{e} = 1350 c m^{2} / V s$ , $μ_{n} = 480 c m^{2} / V s$ and at 300 K, the electron hole pair concentration is $1.072 \times 10^{10} p e r c m^{3}$

Easy

A

$3.14 \times 10^{- 4} s i e m e n s / m$
B

$3.14 \times 10^{- 5} s i e m e n s / m$
C

$3.14 \times 10^{- 3} s i e m e n s / m$
D

$3.14 s i e m e n s / m$

Solution

$C o n d u c t i v i t y : σ = e (n_{e} μ_{e} + n_{n} μ_{n}) [F o r i n t r i n s i c s e m i c o n d u c t o r : n_{e} = n_{n}]$
$= 1.6 \times 10^{- 19} \times 1.072 \times 10^{10} (1350 + 480)$
$= 3.14 \times 10^{- 6} s i e m e n s / c m = 3.14 \times 10^{- 4} s i e m e n s / m$

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