Question

In a $p-n$ junction, the depletion region is $500\text{nm}$ wide an electric field $8\times {10}^5{\text{Vm}}^{-1}$ exist in it. The minimum kinetic energy of a conduction electron which can diffuse from the $n-$side to the $p-$side is ____$\times {10}^{-2}\text{eV}.$
$\left(\text{e = Charge of electron}\right)$

Answer 1

Given that,
Electric field, $E=8\times {10}^5{\text{Vm}}^{-1}$

Width of the depletion region, $d=500\text{nm}=5\times {10}^{-7}\text{m}$

The potential barrier is given by, $V=E\times d$.

$\Rightarrow V=5\times {10}^{-7}\text{m}\times 8\times {10}^5{\text{Vm}}^{-1}$

$\Rightarrow V=40\times {10}^{-2}\text{V}$

$\Rightarrow V=0.4\text{V}$

Now,
The energy of any electron accelerated through a potential of $V$ is given by, $E=eV$

Also, the minimum energy of the conduction electron to diffuse from the $n-$side to the $p-$side is given by,

$E=0.4\text{eV}=40\times {10}^{-2}\text{eV}$

Hence, $40$ is the correct answer.