When a semiconductor is doped with a donor impurity, the number of __…

The correct answer is Option (3) → the holes decrease and the electrons increase

A donor impurity is pentavalent (such as P or As). Four of its electrons form covalent bonds with the neighbouring atoms, and the fifth is set free, so the number of free electrons increases and the semiconductor becomes n-type.

With many more free electrons around, the rate at which electrons meet holes and recombine goes up, so the number of holes decreases below its intrinsic value.

Option (1) the electrons and holes decrease — Incorrect. The donor adds electrons, so their number cannot decrease.

Option (2) the electrons and holes increase — Incorrect. The holes do not increase; the extra electrons fill some of them.

Option (3) the holes decrease and the electrons increase — Correct. Electrons increase from the donors, and holes decrease through recombination.

Option (4) the holes increase and the electrons decrease — Incorrect. This is the opposite case, an acceptor (trivalent) impurity that makes a p-type semiconductor.

NCERT (2026-27):
Extrinsic Semiconductor – “… Thus, the pentavalent dopant is donating one extra electron for conduction and hence is known as donor impurity.”

Extrinsic Semiconductor – “… But the rate of recombination of holes would increase due to the increase in the number of electrons. As a result, the number of holes would get reduced further. Thus, with proper level of doping the number of conduction electrons can be made much larger than the number of holes.”