Match List-I with List-II for doped silicon at room temperature.
| List-I | List-II |
|---|---|
| (A) Ionised donor atom in n-type Si | (I) An immobile core with one negative charge |
| (B) Ionised acceptor atom in p-type Si | (II) Mostly supplied by the impurity atoms |
| (C) Holes in n-type Si | (III) An immobile core with one positive charge |
| (D) Electrons in the conduction band of n-type Si | (IV) Reduced in number by recombination with majority carriers |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(A) Ionised donor atom in n-type Si — This matches (III) an immobile core with one positive charge. The donor gives away its fifth electron, but the atom stays bonded in the lattice.
(B) Ionised acceptor atom in p-type Si — This matches (I) an immobile core with one negative charge. The acceptor takes an electron from a neighbour to complete its fourth bond.
(C) Holes in n-type Si — This matches (IV). Holes are the minority carriers here. The large number of electrons raises the rate of recombination, so the number of holes falls further.
(D) Electrons in the conduction band of n-type Si — This matches (II). Most donor atoms are ionised at room temperature, while very few Si atoms are. So most conduction electrons come from the donors.
NCERT (2026-27):
p-n junction formation – “… This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. …”
p-type semiconductor – “… Therefore, the dopant atom of p-type material can be treated as core of one negative charge along with its associated hole …”
n-type semiconductor – “… But the rate of recombination of holes would increase due to the increase in the number of electrons. As a result, the number of holes would get reduced further.”
Extrinsic Semiconductor – “… So the conduction band will have most electrons coming from the donor impurities …”