Question

The mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for a 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:
a). $2\; \Omega m$
b). $0.4\; \Omega m$
c). $4\; \Omega m$
d). $0.2\; \Omega m$

Answer 1

Hint: To find out the resistivity of the semiconductor we need to find its conductivity first and then we can take its reciprocal to find out the required resistivity. While using the conductivity formula, we need to ignore the hole concentration as given in the question.

Formula used:
Conductivity of the semiconductor, $\sigma=(p{\mu}_{h}+n{\mu}_{e})e$, where p and n are the electron and hole density respectively, ${\mu}_{h}$ and ${\mu}_{e}$ are the mobility of holes and electrons and e is the charge on the electron.

Complete step by step answer:
We have been given the density of electrons, $n=10^{19}\; m^{-3}$ and mobility, ${\mu}_{e}=1.6\; m^2 /(V.s)$.
So, to find the conductivity we need to use the formula $\sigma=(p{\mu}_{h}+n{\mu}_{e})e$, where p and n are the electron and hole density respectively, ${\mu}_{h}$ and ${\mu}_{e}$ are the mobility of holes and electrons and $e=1.6 \times 10^{-19}$ C is the charge on the electron.
According to the question, the semiconductor is of n-type, so contribution of holes will be ignored.
So, conductivity $\sigma = n{\mu}_{e}e=10^{19}\times 1.6 \times 1.6 \times 10^{-19}\; (\Omega m)^{-1}=2.56\; (\Omega m)^{-1}$
Now, we know that resistivity, $\rho =\dfrac{1}{\sigma}$
Therefore. $\rho =\dfrac{1}{2.56}\; \Omega m=0.4\; \Omega m$
Hence, option b is the correct answer.

Note: There will be certain resistivity that will be provided by the holes too and we should take care until and unless it has been mentioned to ignore its effect, it effect should be considered by calculating the resistivity of the semiconductor.