Pure Si at 300 K has equal electron (ni) concentrations of 1.5 x1016m-3 . Doping by indium increases nh 4.5 x 1022 m-3. ne in the doped Si is
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a
5 x 109
b
7 x 109
c
9 x 109
d
8 x 109
answer is A.
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Detailed Solution
We know that of r a doped semiconductor in thermal equilibrium, we have nenh=ni2As per given data, nl=1.5×1016m−3nn=4.5×1022m−3Thus ne=ni2nh=1.5×10162m−64.5×1022m−3=5.0×109m−3