Question

If ${\mu _e}$ and ${\mu _h}$ electron and hole mobility, $E$ be the electric field, the current density $J$ for intrinsic semiconductor is equal to
A. ${n_i}e\left( {{\mu _e} + {\mu _h}} \right)E$
B. ${n_i}e\left( {{\mu _e} - {\mu _h}} \right)E$
C. $\dfrac{{{n_i}e\left( {{\mu _e} + {\mu _h}} \right)}}{E}$
D. $\dfrac{E}{{{n_i}e\left( {{\mu _e} + {\mu _h}} \right)}}$

Answer 1

Hint Using the given information, first we have to calculate the total current flowing through the semiconductor. Then we have to substitute the value of the drift velocity in the expression for the current density.
Formulas used
$I = {n_i}\alpha ve$ where $\alpha$ is the cross sectional area of the semiconductor, $e$ is the charge of the carriers, ${n_i}$is the intrinsic carrier concentration and $v$is the drift velocity.
$J = \dfrac{I}{\alpha }$ where $J$ is the current density
$\mu = \dfrac{v}{E}$ where $\mu$ is the mobility and $E$ is the applied electric field.

Complete step by step answer
When an electric field $E$ is applied across a semiconductor, the intrinsic carriers experience a force and start moving with a drift velocity $v$. If ${n_i}$be the intrinsic carrier concentration, then the electric current flowing through the semiconductor is,
$I = {n_i}\alpha ve$ where $\alpha$ is the cross sectional area of the semiconductor and $e$ is the charge of the carriers.
Now the current through unit cross sectional area, i.e. current density is,
$J = \dfrac{I}{\alpha }$
$\Rightarrow J = {n_i}ve$
The mobility of a carrier is defined to be the average drift velocity per unit electric field. If $\mu$ be the mobility then we have,
$\mu = \dfrac{v}{E}$
$\Rightarrow v = \mu E$
Let ${\mu _e}$ and ${\mu _h}$ be the electron and hole mobility respectively.
As the number concentration of electrons and holes are equal to the intrinsic carrier concentration for an intrinsic semiconductor, so the total current density must be equal to
$J = {J_e} + {J_h} \\\ \Rightarrow J = {n_i}e{\mu _e}E + {n_i}e{\mu _h}E \\\ \Rightarrow J = {n_i}e\left( {{\mu _e} + {\mu _h}} \right)E \\\$

Therefore, the correct option is A.

Additional information In the absence of an externally applied electric field, the current carriers( i.e. electrons and holes) in a semiconductor move in a random fashion due to their thermal energy.

Note In case of an extrinsic semiconductor such as n-type semiconductor,$n > > p$, so the current density is equal to $ne{\mu _n}E$ where $n$is the number concentration of electrons. Similarly for a p-type semiconductor, $n < < p$, the current density is equal to $pe{\mu _h}E$ where $p$ is the number concentration of holes.