JEE Main 2023 — Semiconductors Question with Solution
JEE Main 2023 (25 Jan Shift 2)
Question
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 ammeter.
In the light of above statements, choose the most appropriate answer from the options given below.
Choose an option
Show full solutionCorrect option: D
Step-by-step explanation
Doping silicon with boron leads to holes without any additional electron. This results in P-type semiconductors. Arsenic is pentavalent therefore when added with silicon it leaves one electron as a free electron. This results in N-type semiconductors.
When such P-type and N-type semiconductors are fused to make a junction, an electric field is produced from the N-side to the P-side, causing a barrier potential to develop. Because of the barrier potential, the majority of charge carriers are unable to flow through the junction, hence current is zero, unless we apply forward bias voltage. We cannot detect the current through ammeter as battery is not connected.
Hence, statement I is correct but statement II is incorrect.
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Semiconductors chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2023, covering the Semiconductors chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.