Consider the following statements about the formation of a p-n juncti…

The correct answer is Option (4) → (B) and (D) only

A junction forms as carriers diffuse across it. The fixed ions left behind set up an electric field, and this field drives a drift current in the opposite direction.

(A) Holes diffuse from the p-side to the n-side because of the concentration gradient. — Correct statement. The p-side has far more holes than the n-side. So holes spread from the p-side into the n-side.

(B) An ionised donor left behind on the n-side is free to move through the crystal. — Incorrect statement. The ionised donor is bonded to the surrounding atoms, so it stays fixed. These fixed positive ions form the space-charge layer on the n-side.

(C) The thickness of the depletion region is of the order of one-tenth of a micrometre. — Correct statement. The depletion region is very thin, about \( 10^{-7}\ \mathrm{m} \).

(D) In a p-n junction at equilibrium, the diffusion current is larger than the drift current. — Incorrect statement. Diffusion goes on only until the diffusion current becomes equal to the drift current. At equilibrium the two are equal, so there is no net current.

NCERT (2026-27):
p-n junction formation – “… due to the concentration gradient across p-, and n- sides, holes diffuse from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p).”

p-n junction formation – “This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms.”

p-n junction formation – “The thickness of depletion region is of the order of one-tenth of a micrometre.”

p-n junction formation – “This process continues until the diffusion current equals the drift current. Thus a p-n junction is formed. In a p-n junction under equilibrium there is no net current.”