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A semi conductor has an electron concentration of 0.60×1011m3 and a hole concentration of 6×1019m3 . Calculate its conductivity. Given electron mobility is 0.125m2v1s1  , hole  mobility is  0.032m2v1s1

a
0.3052sm−1
b
0.3072sm−1
c
0.3062sm−1
d
0.3082sm−1

detailed solution

Correct option is B

Conductivity of semiconductor is the sum of the conductivities due to electrons and holes.σ=σe+σhσ=neeμe+nheμhGiven : ne=0.60×1011m−3nh=6×1019m−3μe=0.125m2v−1s−1μh=0.032m2v−1s−1σ=neeμe+nheμhne<<<

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