This question examines how the depletion region of an abrupt p-n junction distributes itself when the two sides are doped unequally. At thermal equilibrium, diffusion of carriers across the metallurgical junction leaves behind fixed, ionized dopant atoms: negative acceptor ions on the p-side and positive donor ions on the n-side. This exposed charge forms the depletion (space-charge) region and sets up the built-in potential.

The controlling principle is overall charge neutrality of the depletion region — the total exposed negative charge on the p-side must equal the total exposed positive charge on the n-side:

N_A × x_p = N_D × x_n

where x_p and x_n are the depletion widths penetrating the p-side and n-side respectively. Rearranging:

x_n / x_p = N_A / N_D

When N_A ≫ N_D (heavily doped p-side, lightly doped n-side), the ratio N_A/N_D is large, so x_n ≫ x_p. In words: the side with fewer dopant atoms per unit volume must extend much farther to expose an equal amount of total charge. Therefore the depletion region extends primarily into the lightly doped n-type region — the correct answer. (This is exactly the principle behind a one-sided or p⁺-n junction.)

Why the other options are wrong:

• Extends primarily into the heavily doped p-type region: this is the reverse of the physics — the heavily doped side exposes the required charge over a very short distance, so its depletion width is smallest, not largest.
• Extends equally into both regions: this only occurs for a symmetric junction where N_A = N_D, which is explicitly not the case here.
• Confined strictly to the junction with zero penetration: a real depletion region always has finite width on both sides to store the space charge and support the built-in field; zero width is unphysical.