Solution:
According to mass action law
\[ n_{e}\times n_{h}= n_{i}^{2} \]
\[ n_{h}=10^{21} ; n_{i}= 10^{19}; n_{e}=n_{i}^{2}/n_{h} = n_{e}= 10^{38}/10^{21}= 10^{17} \]
A Ge specimen is doped with Al. The concentration of acceptor atoms is \[\sim 10^{21} atom/m^{3}\]. Given that the intrinsic concentration of electron hole pairs is \[\sim 10^{19}/m^{3}\], the concentration of electrons in the specimen is :
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