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 $2.3m^2/Vs$ it and $0.01m^2/Vs$ respectively, then semiconductor is

• A. n-type and its resistivity is 0.34 $\Omega$ -m
• B. p-type and its resistivity is 0.034 $\Omega$ -m
• C. n-type and its resistivity is 0.034 $\Omega$ -m
• D. p-type and its resistivity is 3.4 $\Omega$ -m

Answer 1

Answer: (A)

In $n$-type semiconductors electrons are majority carriers and holes are minority carriers.
$n_e=8\times 10^{18}/m^3,$
$n_h=5\times 10^{18}/m^3,$
${\mu }_e=2.3m^2/V-s,$
${\mu }_h=0.01m^2/V-s.$
As $n_e>n_h$, so semiconductor is $n$-type.
Also conductivity $(\sigma )=\frac{1}{\text{ Resistivity }(\rho )}$
$=e\left(n_e{\mu }_e+n_h{\mu }_h\right)$
$\Rightarrow \rho =1.6\times 10^{-19}$
$\left[8\times 10^{18}\times 2.3+5\times 10^{18}\times 0.01\right]$
$\Rightarrow =0.34\Omega -m$