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 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Which carriers are in the majority depends on how the semiconductor is doped.
(A) Intrinsic semiconductor — this matches (IV) Number of conduction electrons = number of holes. In a pure semiconductor every broken bond gives one free electron and one hole, so \( n_e = n_h = n_i \).
(B) n-type semiconductor — this matches (I) Number of conduction electrons >> number of holes. The pentavalent donor atoms add free electrons, which become the majority carriers.
(C) p-type semiconductor — this matches (II) Number of conduction electrons << number of holes. The trivalent acceptor atoms create holes, which become the majority carriers.
(D) p-n junction under equilibrium — this matches (III) Net current is zero. In equilibrium the diffusion current is exactly balanced by the drift current across the junction.
NCERT (2026-27):
Intrinsic Semiconductor – “… In intrinsic semiconductors, the number of free electrons, ne is equal to the number of holes, nh. That is
\( \displaystyle n_e = n_h = n_i \qquad (14.1) \)”
Summary – “… 7. In n-type semiconductors, \( n_e \gg n_h \) while in p-type semiconductors \( n_h \gg n_e \).”
p-n Junction Formation – “… In a p-n junction under equilibrium there is no net current.”