Q. In semiconductor the concentration of electrons and holes are $8 \times 10^{18}/m^{3}$ and $5 \times 10^{18}/m$ respectively. If the mobilities of electrons and holes are $2.3\ m^{2}/volt$-$s$ and $0.01\ m^{2}/vol$-$s$ respectively, then semiconductor is
A) $N$-type and its respectively is 0.34 $ohm$-$metre$
B) $P$-type and its respectively is 0.034 $ohm$-$metre$
C) $N$-type and its respectively is 0.034 $ohm$-$metre$
D) $P$-type and its respectively is 3.40 $ohm$-$metre$
Explanation
(a) $n_{e} = 8 \times 10^{18}/m^{3},$ $n_{h} = 5 \times 10^{18}/m^{3}$
$\mu_{e} = 2.3\frac{m^{3}}{volt - \sec},\mu_{h} = 0.01\frac{m^{2}}{volt - \sec}$
$\because n_{e} > n_{h}$ so semiconductor is $N$ type
Also conductivity $\sigma = \frac{1}{Resistivity(\rho)} = e(n_{e}\mu_{e} + n_{h}\mu_{h})$
$\Rightarrow \frac{1}{\rho} = 1.6 \times 10^{- 19}\lbrack 8 \times 10^{18} \times 2.3 + 5 \times 10^{18} \times 0.01\rbrack$
$\Rightarrow \rho = 0.34\ \Omega$-$m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.