A hole diffuses from the p-side to the n-side in a p-n junction. … - Vidyadip

MCQ

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

A
A bond ia broken on the $n-$side and the electron freed from the bond jumps to the conduction band.
B
A conduction electron on the $p-$side jumps to a broken bond to complete it.
✓
A bond is broken on the $n-$side and the electron freed from the bond jumps to a broken bond on the $p-$side to complete it.
D
A bond is broken on the $p-$side and the electron freed from the bond jumps to a broken bond on the $n-$side to complete it.

✓

Answer

Correct option: C.

A bond is broken on the $n-$side and the electron freed from the bond jumps to a broken bond on the $p-$side to complete it.

A hole diffuses from the $p$ side to the n side in a $p−n$ junction; that is, an electron moves from the n side to the $p$ side. This implies that a bond is broken on the $n$ side. As the electron travels towards the $p$ side, which is rich in holes, it combines with a hole. A hole is created because of the deficiency of one electron. So, when an electron combines with a hole, it completes that bond.

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