Question

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

• A. N-type and its resistivity is $0.34\text{ohm-m}$
• B. P-type and its resistivity is $0.34\text{ohm-m}$
• C. N-type and its resistivity is $3.4\text{ohm-m}$
• D. P-type and its resistivity is $3.4\text{ohm-m}$

Answer 1

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

$n_e=8\times {10}^{18}/m^3,n_h=5\times {10}^{18}/m^3$
${\mu }_e=2.3\frac{m^2}{\text{volt -sec}},{\mu }_n=0.01\frac{m^2}{\text{volt -sec}}$
$\because n_e>n_h$ so semiconductor is N - type
Also conductivity $\sigma =\frac{1}{\text{Resistivity}\left(\rho \right)}$
$=e(n_e{\mu }_e+n_h{\mu }_h)$
$\Rightarrow \frac{1}{\rho }=1.6\times {10}^{-19}[8\times {10}^{18}\times 2.3+5\times {10}^{18}\times 0.01]$
$\Rightarrow \rho =0.34\Omega -\text{m}$