Question

Calculate the resistivity of an n-type semiconductor from the following data: density of conduction electrons and holes are $8 \times {10^{13}}c{m^{ - 3}}$ and $5 \times {10^{12}}c{m^{ - 3}}$ respectively, mobility of conduction electrons and holes are $2.3 \times {10^4}c{m^2}{V^{ - 1}}{s^{ - 1}}$ and $100c{m^2}{V^{ - 1}}{s^{ - 1}}$ respectively.

Answer 1

Hint: We know that resistivity of a semiconductor is a multiplicative inverse of conductivity of the semiconductor. Conductivity of a semiconductor is the sum of products of density and mobility of charge particles present in a semiconductor. First find conductivity then find resistivity by dividing 1 by conductivity.

Complete step by step answer:
Given, the density of conduction electrons is ${n_e} = 8 \times {10^{13}}c{m^{ - 3}}$.
Density of holes is ${n_h} = 5 \times {10^{12}}c{m^{ - 3}}$.
Mobility of conduction electrons is ${\mu _e} = 2.3 \times {10^4}c{m^2}{V^{ - 1}}{s^{ - 1}}$.
Mobility of holes is ${\mu _h} = 100c{m^2}{V^{ - 1}}{s^{ - 1}}$.
We know that, conductivity of a semiconductor is given by
$\sigma = e\left( {{n_e}{\mu _e} + {n_h}{\mu _h}} \right)$, where $e$ is charge on electron $e = 1.6 \times {10^{ - 19}}C$.
$\sigma = 1.6 \times {10^{ - 19}}(8 \times {10^{13}} \times 2.3 \times {10^4} + 5 \times {10^{12}} \times 100)$
\[\sigma = .294\]
We know that resistivity is a multiplicative inverse of conductivity.
Then resistivity $\rho = \dfrac{1}{\sigma } = \dfrac{1}{{.294}} = 3.401\Omega /cm = 0.03401\Omega /m$.

Note: Conductivity of a semiconductor is caused due to holes and conduction electrons present in it. In n-type semiconductor conduction electrons are in majority and holes are in minority but in p-type semiconductor holes are in majority and conduction electrons are in minority. These holes and conduction electrons are produced by adding impurities in pure semiconductor.