When a small forward voltage is applied to a p-n junction diode, the current through it is small. This is because
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → only a few carriers, those in the uppermost energy levels, have enough energy to cross the junction
Option (1) only a few carriers, those in the uppermost energy levels, have enough energy to cross the junction — Correct. A small forward voltage lowers the barrier only slightly below its equilibrium value. Only the carriers in the uppermost energy levels can get over it, so the current stays small. A larger voltage lowers the barrier more, and the current rises.
Option (2) the depletion layer becomes wider under forward bias — Incorrect. A forward voltage opposes the built-in potential \( V_0 \). So the depletion layer becomes narrower.
Option (3) most of the applied voltage drops across the p-side and the n-side — Incorrect. The depletion region has a much higher resistance than the p-side and n-side. So the applied voltage drops mostly across the depletion region.
Option (4) the diode offers a higher resistance in forward bias than in reverse bias — Incorrect. The forward bias resistance of a diode is low compared with its reverse bias resistance.
NCERT (2026-27):
p-n junction diode under forward bias – “If the applied voltage is small, the barrier potential will be reduced only slightly below the equilibrium value, and only a small number of carriers in the material—those that happen to be in the uppermost energy levels—will possess enough energy to cross the junction. So the current will be small.”