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In a p–n junction, the depletion region is 400 nm wide and an electric field of 5 × 10⁵ 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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0.3 eV

0.1 eV

0.2 eV

0.4 eV

answer is C.

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Detailed Solution

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

$Work done W = qV$

$V = 5 \times 10^{5} \times 400 \times 10^{- 9} = 0 .2 V$

$W = 1.6 \times 10^{- 19} \times 0.2 J o r 0.2 eV$

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