If ne and nh are the number of electrons and holes concentrations in a semiconductor heavily doped with phosphorus, then
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → ne >> nh
Phosphorus is pentavalent. Each phosphorus atom uses four electrons in covalent bonds with the neighbouring silicon or germanium atoms and donates the fifth as a free electron, so the doped crystal is an n-type semiconductor.
With heavy doping the donated electrons far outnumber the thermally generated electron-hole pairs. Moreover, by the law of mass action the extra electrons reduce the hole concentration below its intrinsic value:
\( n_e n_h = n_i^{2} \)
So \( n_e \) rises far above \( n_i \) while \( n_h \) falls below it, and electrons are the majority carriers.
Why the other options are incorrect
Option (2) ne << nh — Incorrect. Holes greatly outnumber electrons in a p-type semiconductor, doped with a trivalent impurity such as boron.
Option (3) ne ≤ nh — Incorrect. Holes can never equal or exceed the electrons in a crystal doped with a donor impurity.
Option (4) ne = nh — Incorrect. Equal concentrations of electrons and holes occur only in an intrinsic (pure) semiconductor.
NCERT (2026-27):
Extrinsic Semiconductor – “… Thus, with proper level of doping the number of conduction electrons can be made much larger than the number of holes. Hence in an extrinsic semiconductor doped with pentavalent impurity, electrons become the majority carriers and holes the minority carriers. These semiconductors are, therefore, known as n-type semiconductors. For n-type semiconductors, we have,
\( \displaystyle n_e \gg n_h \qquad (14.3) \)”