A pure silicon crystal has \( 5 \times 10^{28} \) atoms \( \mathrm{m^{-3}} \). It is doped by 2 ppm concentration of pentavalent arsenic. The number of holes are: Consider \( n_i = 1.5 \times 10^{16}\,\mathrm{m^{-3}} \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2.25 \times 10^{9}\,\mathrm{m^{-3}} \)
Pentavalent arsenic is a donor: each arsenic atom gives one free electron, so the doped crystal is n-type and the electron density is almost equal to the donor density. The hole density then follows from the mass-action relation of a semiconductor in equilibrium:
\( \displaystyle n_e\,n_h = n_i^{2} \)
2 ppm means 2 arsenic atoms per \( 10^{6} \) silicon atoms:
\( \displaystyle n_e \approx N_D = 2\times10^{-6} \times 5\times10^{28} = 1\times10^{23}\,\mathrm{m^{-3}} \)
\( \displaystyle n_h = \frac{n_i^{2}}{n_e} = \frac{(1.5\times10^{16})^{2}}{1\times10^{23}} = \frac{2.25\times10^{32}}{10^{23}} \)
\( \displaystyle n_h = 2.25\times10^{9}\,\mathrm{m^{-3}} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Extrinsic Semiconductor – …
