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A semi conductor has an electron concentration of 0.60×1011m−3 and a hole concentration of 6×1019m−3 . Calculate its conductivity. Given electron mobility is 0.125m2v−1s−1 , hole mobility is 0.032m2v−1s−1

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a

0.3052sm−1

b

0.3072sm−1

c

0.3062sm−1

d

0.3082sm−1

answer is B.

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Detailed Solution

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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