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Q.
At room temperature, number of electrons in a pure Ge crystal is $10^{15} m ^{-3}$. If $10^{21}$ In atoms are mixed per $m ^{3}$ in the crystal then the number of electrons in the extrinsic semiconductor is :-
Solution:
$n _{ i }^{2}= n _{ e } \times n _{ n }$
$\Rightarrow \left(10^{15}\right)^{2}= n _{ e } \times 10^{21}$
$\Rightarrow n _{ e }=10^{9} m ^{-3}$