Current Electricity Ques 4
- 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 $ $\Omega-m$
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Answer:
Correct Answer: 4.(c)
Solution:
Formula:
Electric Current in a Conductor:
- 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 $
BETA
