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CUET · PHYSICS · PYQ PAPER 2025

Match List-I with List-II
List-IList-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
Choose the correct answer from the options given below:

  1. A (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
  2. B (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  3. C (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
  4. D (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Verified Solution

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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