Question

There are two silicon samples (one is $p-$ type and other is $n-$type) which are doped with ${10}^{14}/c{m}^3$ acceptor impurities and ${10}^{16}/c{m}^3$ donor impurities. It is found that conductivity in $n-$ type silicon sample will be two times that of $p-$ type silicon sample. Then the obtained ratio of hole mobility to electron mobility will be________.
(Assume complete ionization of impurities).

Answer 1

$\begin{array}{cc}\text{Given},& \\ \text{Acceptor impurities,}{N}_A& ={10}^{14}/c{m}^3\\ \text{Donor impurities,}{N}_D& ={10}^{16}/c{m}^3\\ \text{We know that conductivity,}\sigma & =nqu\\ \text{for p-type,}{\sigma }_p& =N_{A}q{\mu }_p\\ \text{for n-type,}{\sigma }_n& =N_{D}q{\mu }_p\\ \text{given,}{\sigma }_n=2{\sigma }_p& \\ \therefore 2\left(N_{A}q{\mu }_p\right)& =N_{D}q{\mu }_n\\ \Rightarrow \frac{{\mu }_p}{{\mu }_n}& =\left(\frac{N_D}{2N_A}\right)=\frac{{10}^{16}}{2\times {10}^{14}}=50\end{array}$