Question

The electron concentration in an $n-$type semiconductor is the same as the hole concentration in a $p-$type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them

• A. No current will flow in $p-$type, but current will only flow in $n-$type.
• B. Current in $n-$type $=$ current in $p$-type
• C. Current in $p-$type $>$ current in $n-$type
• D. Current in $n-$type $>$ current in $p-$type.

Answer 1

The correct option is D Current in $n-$type $>$ current in $p-$type.

The current through a semiconductor is,

$I=neA{v}_d=neA\mu E[\because v_d=\mu E]$

Where, $\mu$ is the mobility of charge carriers.

$\Rightarrow \frac{I_n}{I_p}=\frac{n_{e}eA{\mu }_{e}E}{n_{h}eA{\mu }_{h}E}$

$\Rightarrow \frac{I_n}{I_p}=\frac{{\mu }_p}{{\mu }_h}$

$\because {\mu }_e>{\mu }_h$

$\Rightarrow I_n>I_p$

Hence, $\left(D\right)$ is the correct answer.