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.
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.
- (A)Both Statement I and statement II are incorrect
- (B)Statement I is incorrect but statement II is correct
- (C)Both Statement I and statement II are correct
- (D)Statement I is correct but statement II is incorrect✓ Correct
Answer: (D)
Solution
Boron is trivalent, so doping silicon with it supplies acceptor levels and makes the crystal p-type with holes in excess; arsenic is pentavalent, so it supplies donor levels and makes the crystal n-type with electrons in excess, which makes the first statement sound. When the two are joined, carriers diffuse only until the depletion region and its built-in potential halt any further net flow, so without an external supply no steady current circulates and an ammeter placed in the circuit reads zero. The first statement is correct while the second is not.
JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Single correct option