Question

Mobility of electron in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 10¹⁹ m⁻³
and their mobility is 1.6 m²/(V.s) then the resistivity of the semiconductor(since it is an n-type semiconductor contribution of holes is ignored) is close to:
(A) 2 Ωm (B) 4 Ωm
(C) 0.4 Ωm (D) 0.2 Ωm
[JEE Main 2019]

Answer 1

The resistivity of the n-type semiconductor is close to 0.4 Ωm

Option (c) is correct.

Explanation:

Given :

The carrier density $n = 10^{19} m^{-3}$.

Mobility  $\mu = 1.6 \frac{m^{2} }{Vs}$.

From equation of conductivity $\sigma = nq\mu$.

But, resistivity is inversely proportional to the conductivity.

    $\rho = \frac{1}{nq\mu}$

   $\rho = \frac{1}{1.6\times 10^{-19} \times 1.6 \times 10^{19} }$

   $\rho = \frac{1}{2.56} = 0.39$ Ωm

So, the resistivity of the n-type semiconductor is close to 0.4 Ωm