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In a p–n junction, the depletion region is 400 nm wide and an electric field of $5 \times 10^{5} V / m$ exists in it. What should be the minimum kinetic energy of a conduction electron which can diffuse from the n – side to the p – side?

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0.1 eV

0.3 eV

0.2 eV

0.4 eV

answer is C.

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Detailed Solution

d = 400×10^–9m; $E = 5 \times 10^{5} V / m$

$K E = e V = e \times E \times d J$ = E × deV

$= 5 \times 10^{5} \times 400 \times 10^{- 9}$

= 2000 × 10^–4 = 0.2eV

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