Q 12-14-202JEE MainJEE Main 2017 (8 Apr)Easy
The conductivity of a semiconductor sample having electron concentration of $5\times10^{18}\ \text{m}^{-3}$, hole concentration of $5\times10^{19}\ \text{m}^{-3}$, electron mobility of $2.0\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$ and hole mobility of $0.01\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$ is: (Take charge of an electron as $1.6\times10^{-19}$ C)
Answer: (B) $1.65\ (\Omega\,\text{m})^{-1}$
$$\sigma = e(n_e\mu_e + n_h\mu_h) = 1.6\times10^{-19}\left(5\times10^{18}\times2.0 + 5\times10^{19}\times0.01\right)$$
$$= 1.6\times10^{-19}\left(1.0\times10^{19} + 0.05\times10^{19}\right) = 1.6\times1.05 \approx 1.68\ (\Omega\,\text{m})^{-1}$$
The closest option is $1.65\ (\Omega\,\text{m})^{-1}$.
Solution by Sreeraj P, M.Sc Physics