Question

What is the conductivity of a semiconductor (in $\Omega^{-1} m ^{-1}$ ) if electron density $=5 \times 10^{12} / cm ^{3}$ and hole density $=8 \times 10^{13} / cm ^{3} ?$
$\left(\mu_{e}=2.3 m ^{2} V ^{-1} s ^{-1}, \mu_{h}=0.01 m ^{2} V ^{-1} s ^{-1}\right)$

• A. 5.634
• B. 1.968
• C. 3.421
• D. 8.964

Answer 1

Answer: (B)

Here, $\mu_{e}=2.3 m ^{2} V ^{-1} s ^{-1}$
$\mu _{h}=0.01 m ^{2} V ^{-1} s ^{-1}, n_{e}=5 \times 10^{12} / cm ^{3}=5 \times 10^{18} / m ^{3}$
$n_{h}=8 \times 10^{13} / cm ^{3}=8 \times 10^{19} / m ^{3}$
$\sigma=e\left(n_{e} \mu_{e}+n_{h} \mu_{h}\right)$
$=1.6 \times 10^{-19}\left[\left(5 \times 10^{18}\right) \times 2.3+\left(8 \times 10^{19}\right) \times 0.01\right]$
$=1.6 \times 10^{-1}[11.5+0.8]=1.968 \Omega^{-1} m ^{-1}$