The cause of the potential barrier in a p-n junction diode is

The correct answer is Option (4) → Immobile positive and negative ions on either side of the junction

When a p-n junction is formed, electrons diffuse from n to p and holes from p to n. Each electron that leaves the n-side leaves behind an ionised donor (positive), and each hole that leaves the p-side leaves behind an ionised acceptor (negative). These ions are bound in the crystal and cannot move.

Option (1) Depletion of positive charges near the junction — Incorrect. The loss of holes from the p-side alone would leave only a negative layer on that side. A potential difference also needs the positive layer on the n-side.

Option (2) Depletion of negative charges near the junction — Incorrect. The loss of free electrons from the n-side alone would leave only a positive layer on that side. The barrier comes from the fixed ions that stay behind on both sides.

Option (3) Concentration of positive charges near the junction — Incorrect. A single layer of positive charge cannot set up a field across the junction. The barrier needs a positive layer on the n-side facing a negative layer on the p-side.

Option (4) Immobile positive and negative ions on either side of the junction — Correct. The positive space charge of donor ions on the n-side and the negative space charge of acceptor ions on the p-side form the depletion layer. The field between these two layers makes the n-side positive relative to the p-side. This barrier potential opposes further diffusion of carriers.

NCERT (2026-27):
Summary – “12. p-n junction is the ‘key’ to all semiconductor devices. When such a junction is made, a ‘depletion layer’ is formed consisting of immobile ion-cores devoid of their electrons or holes. This is responsible for a junction potential barrier.”