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-cype 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

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-cype 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 (A) Statenent I is correct but statement II is incorrect (B) Both Statement I and Statement II are incorrect (C) Statenent I is incorrect but statement II is correct (D) Both Statenent I and statement II are correct

Tardigrade Admin · Accepted Answer

Statement - I is correct
When P-N junction is formed an electric field is generated form N-side to P-side due to which barrier potential arises & majority charge carrier can not flow through the junction due to barrier potential so current is zero unless we apply forward bias voltage.

Statenent I is correct but statement II is incorrect

Both Statement I and Statement II are incorrect

Statenent I is incorrect but statement II is correct

Both Statenent I and statement II are correct

Statement - I is correct When P-N junction is formed an electric field is generated form N-side to P-side due to which barrier potential arises & majority charge carrier can not flow through the junction due to barrier potential so current is zero unless we apply forward bias voltage.

Statement - I is correct
When P-N junction is formed an electric field is generated form N-side to P-side due to which barrier potential arises & majority charge carrier can not flow through the junction due to barrier potential so current is zero unless we apply forward bias voltage.