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

Statement I is correct but statement II is incorrect

Statement I is incorrect but statement II is correct

Both Statement I and statement II are correct

Both Statement I and Statement II are incorrect

Electronic Devices PYQ — JEE Main 2023 · 25 Jan · Shift 2 · Q60 with Solution

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Correct answer: C

Analyze Statement I
- Silicon $(\text{Si})$ is a group 14 semiconductor with 4 valence electrons.
- Boron is a group 13 element with 3 valence electrons. When Si is doped with boron, one bond remains deficient in an electron, creating a hole.
- Hence, boron doping produces a p-type semiconductor.
- Arsenic is a group 15 element with 5 valence electrons. When Si is doped with arsenic, it contributes an extra electron.
- Hence, arsenic doping produces an n-type semiconductor.
- In p-type semiconductors, holes are the majority carriers; in n-type semiconductors, electrons are the majority carriers.
Therefore, Statement I is correct.
Analyze Statement II
- When p-type and n-type semiconductors are joined, a p-n junction is formed.
- Due to concentration difference, electrons and holes initially diffuse across the junction.
- This creates a depletion region and an internal electric field.
- Very quickly, equilibrium is established and the diffusion current is balanced by drift current.
- So, in equilibrium, no net current flows through the junction.
- Therefore, if an ammeter is externally connected without any external bias, it will not detect a continuous current.
Therefore, Statement II is incorrect.
Choose the correct option
- Statement I: Correct
- Statement II: Incorrect
Hence, the correct option is: $\boxed{\text{C}}$
Comparison with stored answer
- Derived answer: $\text{C}$
- Stored correct answer: $\text{C}$
- They match.

View written solutionFree

Correct answer: C

Analyze Statement I
- Silicon $(\text{Si})$ is a group 14 semiconductor with 4 valence electrons.
- Boron is a group 13 element with 3 valence electrons. When Si is doped with boron, one bond remains deficient in an electron, creating a hole.
- Hence, boron doping produces a p-type semiconductor.
- Arsenic is a group 15 element with 5 valence electrons. When Si is doped with arsenic, it contributes an extra electron.
- Hence, arsenic doping produces an n-type semiconductor.
- In p-type semiconductors, holes are the majority carriers; in n-type semiconductors, electrons are the majority carriers.
Therefore, Statement I is correct.
Analyze Statement II
- When p-type and n-type semiconductors are joined, a p-n junction is formed.
- Due to concentration difference, electrons and holes initially diffuse across the junction.
- This creates a depletion region and an internal electric field.
- Very quickly, equilibrium is established and the diffusion current is balanced by drift current.
- So, in equilibrium, no net current flows through the junction.
- Therefore, if an ammeter is externally connected without any external bias, it will not detect a continuous current.
Therefore, Statement II is incorrect.
Choose the correct option
- Statement I: Correct
- Statement II: Incorrect
Hence, the correct option is: $\boxed{\text{C}}$
Comparison with stored answer
- Derived answer: $\text{C}$
- Stored correct answer: $\text{C}$
- They match.