Q.

A semiconductor has equal electron and hole concentration of 2 x 108 m-3. On doping with a certain impurity, the electron concentration increases to 4 x 1010 m-3, then the new hole concentration of the semiconductor is 

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a

106 m-3

b

1010 m-3

c

108 m-3

d

1012 m-3

answer is A.

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Detailed Solution

n=nHnenenH=ne1nH1nH1=2×108×2×1084×1010=106/m3

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