When a p-n junction is reverse biased:

The correct answer is Option (4) → The diffusion of electrons and holes stops

In reverse bias the applied voltage adds to the barrier potential. The depletion region widens and the barrier rises to \( V_0 + V \). The higher barrier stops the diffusion of majority carriers across the junction, while minority carriers still drift across and give the small reverse current.

Option (1) Width of depletion region decreases — Incorrect. The depletion region widens in reverse bias; it narrows in forward bias.

Option (2) The drift of electrons and holes stops. — Incorrect. The drift of minority carriers continues; it is the small reverse (saturation) current, of the order of microamperes.

Option (3) Potential barrier decreases — Incorrect. The barrier increases, from \( V_0 \) to \( V_0 + V \).

Option (4) The diffusion of electrons and holes stops — Correct. The raised barrier practically stops majority carriers from diffusing across the junction.

NCERT (2026-27):
p-n Junction Diode Under Reverse Bias – “… The direction of applied voltage is same as the direction of barrier potential. As a result, the barrier height increases and the depletion region widens due to the change in the electric field. The effective barrier height under reverse bias is (V₀ + V), [Fig. 14.15(b)]. This suppresses the flow of electrons from n → p and holes from p → n. Thus, diffusion current, decreases enormously compared to the diode under forward bias.”

p-n Junction Diode Under Reverse Bias – “… The drift current is of the order of a few μA.”