Current Electricity 1 Question 10

10. Mobility of electrons 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 $10^{19} m^{-3}$ and their mobility is $1.6 m^{2}(V-s)$, then the resistivity of the semiconductor (since, it is an $n$-type semiconductor contribution of holes is ignored) is close to

(2019 Main, 9 Jan I)

(a) $2 \Omega-m$

(b) $0.2 \Omega-m$

(c) $0.4 \Omega-m$

(d) $4 \Omega-m$

Show Answer

Solution:

  1. Since, it is an $n$-type semiconductor and concentration of the holes has been ignored. So, its conductivity is given as

$$ \sigma=n _e e \mu _e $$

where, $n _e$ is the number density of electron, $e$ is the charge on electron and $\mu _e$ is its mobility.

Substituting the given values, we get

$$ \sigma=10^{19} \times 1.6 \times 10^{-19} \times 1.6=2.56 $$

As, resistivity, $\rho=\frac{1}{\sigma}=\frac{1}{2.56}$

or

$$ \rho=0.39 \simeq 0.4 \Omega-m $$



NCERT Chapter Video Solution

Dual Pane