Ratio of free electron density to hole density of a p-type semiconductor is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Less than one
A p-type semiconductor is made by doping with a trivalent impurity, which supplies holes. The holes are the majority carriers and the free electrons are the minority carriers.
\( \displaystyle n_{h} \gg n_{e} \;\Rightarrow\; \frac{n_{e}}{n_{h}} \lt 1 \)
Option (1) Equal to one — Incorrect. Equal densities hold only in an intrinsic (pure) semiconductor.
Option (2) Less than one — Correct. Holes are the majority carriers, so the electron density is smaller than the hole density.
Option (3) Greater than one — Incorrect. A ratio greater than one describes an n-type semiconductor.
Option (4) Can be less than or greater than one depending upon semiconductor material. — Incorrect. The ratio is set by the type of doping, not by the semiconductor material; for p-type it is always less than one.
NCERT (2026-27):
Extrinsic Semiconductor – “… These holes are in addition to the intrinsically generated holes while the source of conduction electrons is only intrinsic generation. Thus, for such a material, the holes are the majority carriers and electrons are minority carriers. Therefore, extrinsic semiconductors doped with trivalent impurity are called p-type semiconductors. For p-type semiconductors, the recombination process will reduce the number (ni)of intrinsically generated electrons to ne. We have, for p-type semiconductors
\( \displaystyle n_h \gg n_e \qquad (14.4) \) ”