Q.

A semiconductor has an electron concentration of  0.45×1012 m−3 and a hole concentration of  5×1020m−3. Calculate its conductivity. Given electron mobility is 0.135m2v−1s−1 . Hole mobility is 0.048m2v−1s−1

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a

3.84sm−1

b

3.86sm−1

c

3.82sm−1

d

3.80sm−1

answer is A.

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Detailed Solution

Conductivity of semiconductor is the sum of the conductivities due to electrons and holesGiven , ne=0.45×1012m−3nh=5×1020m−3μe=0.135m2v−1s−1μh=0.048m2v−1s−1Conductivity , σ=σe+σh ⇒σ=neeμe+nheμhne<<<
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