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 $2.3 \,m^2 /Vs$ it and $0.01 \,m^2 /Vs$ 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}=2.3 \,m ^{2} / V - s , $
$\mu_{h}=0.01\, m ^{2} / V - s .$
As $n_{ e } > n_{h}$, so semiconductor is $n$-type.
Also conductivity $(\sigma)=\frac{1}{\text { Resistivity }(\rho)}$
$=e\left(n_{e} \mu_{e}+n_{h} \mu_{h}\right)$
$\Rightarrow \rho=1.6 \times 10^{-19}$
$\left[8 \times 10^{18} \times 2.3+5 \times 10^{18} \times 0.01\right]$
$\Rightarrow =0.34\, \Omega - m$