CUET · PHYSICS · PYQ PAPER 2025
Match List-I with List-II
| List-I | List-II |
| (A) Intrinsic semiconductor | (I) Number of conduction electrons >> number of holes |
| (B) n-type semiconductor | (II) Number of conduction electrons << number of holes |
| (C) p-type semiconductor | (III) Net current is zero |
| (D) p-n junction under equilibrium | (IV) Number of conduction electrons = number of holes |
- A (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
- B (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
- C (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
- D (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Answer & Solution
Correct Answer
(C) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Step-by-step Solution
Detailed explanation
(A) Intrinsic semiconductor: \(n_e = n_h\). Thus (A)-(IV). (B) n-type semiconductor: \(n_e \gg n_h\). Thus (B)-(I). (C) p-type semiconductor: \(n_e \ll n_h\). Thus (C)-(II). (D) p-n junction under equilibrium: Net current is zero. Thus (D)-(III). The correct match is (A)-(IV),…
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