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In a p-n junction the depletion region is $200 nm$ wide and a electric field of $10^6 vm^{-1}$ exists in it. What is the minimum kinetic energy of conducting electron to diffuse through the barriers.

$(a)\;0.1 \;ev \\ (b)\;1\;ev \\ (c)\;0.2 \;ev \\ (d)\;2 ev $

1 Answer

We know that,
Potential barrier $=E \times d$
Where $E$ - electric field
$d$ - width of depetion region
$\therefore v - 200 \times 10^{-9}$
$E= 10^6 vm^{-1}$
$\therefore v= 0.2 v$
Kinetic energy = Potential barrier $\times$ charge of electron
$\qquad= 0.2 v \times e$
$\qquad= 0.2 \;ev$
Hence c is the correct answer.
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