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Q.

In a p-n junction the depletion region is 400nm wide and electric field of 5 x 105 Vm-1 exists in it. The minimum energy of a conduction electron, which can diffuse from n-side to the p-side is

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a

4eV

b

5eV

c

0.4eV

d

0.2eV

answer is D.

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Detailed Solution

W=Vq=Edq=5×105×400×10−9×1.6×10−19J=5×105×400×10−9×1.6×10−19J1.6×10−19eV=2000×10−4eV=0.2eV

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