Statement I : When a Si sample is doped with Boron, it becomes P type and when doped by Arsenic it becomes N-type semi conductor such that P-type has excess holes and N-type has excess electrons.
Statement II : When such P-type and N-type semi-conductors, are fused to make a junction, a current will automatically flow which can be detected with an externally connected ameter.
In the light of above statements, choose the most appropriate answer from the options given below

Solution

Physics sir · 4 steps

Doping is fine, but a junction left alone settles into balance, so the ammeter reads zero.

1. Check Statement I. Boron is trivalent, so each B atom leaves one bond short of an electron, which creates a hole. That gives a p-type semiconductor. Arsenic is pentavalent, so each As atom donates one extra electron, which gives n-type. Statement I is correct.
2. Junction formation. When p and n are joined, holes diffuse into the n side and electrons into the p side for a very short time. This leaves fixed ions behind, forming a depletion layer with a barrier potential V₀.
3. Equilibrium. The field of this layer drives a drift current opposite to the diffusion current. They quickly become equal, so the net current is I_(diff) - I_(drift) = 0. The barrier potential also cannot push current through an external wire, because the contact potentials at the metal leads cancel it.
4. So. No steady current flows and an external ammeter shows nothing, so Statement II is incorrect. Statement I is correct and Statement II is incorrect, which is option (c).

The idea to keep. An unbiased p-n junction is in equilibrium: diffusion and drift currents cancel. You need an external battery, light or heat to get a net current out of it.

Answer from the NTA final answer key. Solution by PrepWiser.