Suppose a pure Si crystal has \( 9\times10^{28} \) atoms per \( \mathrm{m^{3}} \). It is doped by \( 1 \) ppm concentration of pentavalent Arsenic (As). Then, the number of holes (per \( \mathrm{m^{3}} \)) in the crystal will be
(Given the intrinsic concentration of charge carriers: \( n_{i} = 1.5\times10^{16}\ \mathrm{m^{-3}} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2.5\times10^{9} \)
In an n-type semiconductor each pentavalent atom donates one electron, so \( n_{e} \approx N_{D} \), which is much larger than \( n_{i} \). The mass-action law \( n_{e}n_{h} = n_{i}^{2} \) then gives the minority carrier (hole) density.
\( N_{D} = 1\,\mathrm{ppm}\times9\times10^{28}\,\mathrm{m^{-3}} = 10^{-6}\times9\times10^{28}\,\mathrm{m^{-3}} = 9\times10^{22}\,\mathrm{m^{-3}} \)
\( \displaystyle n_{h} = \frac{n_{i}^{2}}{n_{e}} = \frac{(1.5\times10^{16}\,\mathrm{m^{-3}})^{2}}{9\times10^{22}\,\mathrm{m^{-3}}} \)
\( \displaystyle n_{h} = \frac{2.25\times10^{32}}{9\times10^{22}}\,\mathrm{m^{-3}} = 2.5\times10^{9}\,\mathrm{m^{-3}} \)
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 – …
