JEE Physics 2023 Question with Solution

Given: Statement I says that doping Si with Boron gives P-type semiconductor and doping with Arsenic gives N-type semiconductor, with excess holes and excess electrons respectively.

Find: Which option correctly evaluates Statement I and Statement II.

From the solution text, Statement I is correct.

When a P-N junction is formed, an electric field is generated from N-side to P-side. Because of this, a barrier potential arises. Due to the barrier potential, majority charge carriers cannot flow through the junction on their own, so the current is zero unless a forward bias voltage is applied.

Therefore, Statement II is incorrect.

The correct option is D.

Given: A P-type semiconductor has excess holes and an N-type semiconductor has excess electrons.

Find: Whether current flows automatically when they are fused to form a junction.

In P-type semiconductor, acceptor impurity such as Boron creates holes as majority carriers. In N-type semiconductor, donor impurity such as Arsenic creates electrons as majority carriers. So Statement I is correct.

After joining P-type and N-type semiconductors, diffusion initially occurs and a depletion region is formed. This creates an internal electric field and a barrier potential. Once equilibrium is established, there is no net current through the junction without external bias.

So an ammeter connected externally will not detect an automatic sustained current unless a forward bias voltage is applied.

Hence, Statement I is correct but Statement II is incorrect, so the correct option is D.