A hole diffuses from the $P -$ side to the $N -$ side in a $P - N$ junction. This means that

A bond is broken on the $N -$ side and the electron free from the bond jumps to the conduction band

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A conduction electron on the $P -$ side jumps to a broken bond to complete it

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A bond is broken on the $N -$ side and the electron free from the bond jumps to a broken bond on the $P -$ side to complete it.

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A bond is broken on the $P -$ side and the electron free from the bond jumps to a broken bond on the $N -$ side complete it

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Solution

The correct option is C
A bond is broken on the $N -$ side and the electron free from the bond jumps to a broken bond on the $P -$ side to complete it.

Whenever a hole diffuses from the $P -$ side to the $N -$ side in a $P - N$ junction. A bond is broken on the $N -$ side and the electron free from the bond jumps to a broken bond on the $P -$ side to complete it. Thus maintaining the balance.

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Similar questions

A hole diffuses from the P-side to the N-side in a P-N junction. This means that-

p - n

0.2 V

p -

n -

0.1 V

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