As the reverse bias voltage across a p-n junction diode is increased…

The correct answer is Option (2) → even a small voltage is sufficient to sweep the minority carriers across the junction

Option (1) the depletion layer stops widening beyond a small reverse voltage — Incorrect. Under reverse bias, the applied voltage acts in the same direction as the barrier potential. So the depletion region goes on widening as the reverse voltage is raised.

Option (2) even a small voltage is sufficient to sweep the minority carriers across the junction — Correct. The reverse current is the drift of minority carriers, electrons from the p-side and holes from the n-side. A small reverse voltage already sweeps across every minority carrier that reaches the junction, so a larger voltage adds almost nothing. The size of this current is set by the concentration of minority carriers, and the voltage does not change it.

Option (3) the potential barrier at the junction falls to zero under reverse bias — Incorrect. Reverse bias raises the barrier height to \( V_0+V \). Only forward bias lowers it.

Option (4) the majority carriers cross the junction at a constant drift speed — Incorrect. The higher barrier greatly reduces the diffusion of majority carriers (electrons from n to p and holes from p to n). So they carry almost no current under reverse bias.

NCERT (2026-27):
p-n Junction Diode under Reverse Bias – “The diode reverse current is not very much dependent on the applied voltage. Even a small voltage is sufficient to sweep the minority carriers from one side of the junction to the other side of the junction. The current is not limited by the magnitude of the applied voltage but is limited due to the concentration of the minority carrier on either side of the junction.”