Question

In a p-n junction the depletion region is $400nm$ wide and electric field of $5\times {10}^{5}V{m}^{-1}$ exists in it. The minimum energy of a conduction electron, which can diffuse from n-side to the p-side is

• A. $4eV$
• B. $5eV$
• C. $0.4eV$
• D. $0.2eV$

Answer 1

The correct option is D $0.2eV$

Given that:
$E=5\times {10}^5$
$d=400nm=4\times {10}^{-7}m$

We have,
$E=\frac{V}{d}$

where, $V$ is the electric potential.
$V=Ed=5\times {10}^5\times 4\times {10}^{-7}$
$V=20\times {10}^{-2}=0.2V$
Hence, the minimum kinetic energy $E_{min}$ of a conduction electron, which can diffuse from n-side to the p-side is equal to the potential of the field.
$E_{min}=0.2eV$