Question

In a semiconductor, the concentrations of electrons and holes are $8\times {10}^{18}{\text{/m}}^3$ and $5\times {10}^{18}{\text{/m}}^3$ respectively. If the mobilities of electrons and hole are $2.3{\text{m}}^2/\text{volt-sec}$ and $0.01{\text{m}}^2/\text{volt-sec}$ respectively, then the semiconductor is

• A. $N$-type and its resistivity is $0.34\text{ohm-metre}$
• B. $P$-type and its resistivity is $0.034\text{ohm-metre}$
• C. $N$-type and its resistivity is $0.034\text{ohm-metre}$
• D. $P$-type and its resistivity is $3.40\text{ohm-metre}$

Answer 1

The correct option is A $N$-type and its resistivity is $0.34\text{ohm-metre}$

Given

$n_e=8\times {10}^{18}{\text{m}}^3$,

$n_h=5\times {10}^{18}{\text{m}}^3$

${\mu }_e=2.3{\text{m}}^2/\text{volt-sec}$, ${\mu }_h=0.01{\text{m}}^2/\text{volt-sec}$

As, $n_e>n_h$, it is a $N-$type semiconductor.

Now, $\sigma =q_e(n_e{\mu }_e+n_h{\mu }_h)$

And, $\rho =\frac{1}{\sigma }=\frac{1}{q_e(n_e{\mu }_e+n_h{\mu }_h)}$

$=\frac{1}{1.6\times {10}^{-19}(8\times {10}^{18}\cdot (2.3)+5\times {10}^{18}\cdot (0.01\left)\right)}$

$\therefore \rho \approx 0.34\Omega -\text{m}$

Hence, $\left(A\right)$ is the correct answer.