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 $0.2eV$

Potential Difference = $V=Ed$; where $E$ is electric filed and $d$ is depletion region width

Work done $W=qV$

$\begin{array}{c}V=5\times {10}^5\times 400\times {10}^{-19}=0.2\\ W=1.6\times {10}^{-19}\times 0.2=0.2eV\end{array}$