A pure silicon crystal with \( 5\times10^{28} \) atoms \( \mathrm{m^{-3}} \) has \( n_i = 1.5\times10^{16}\,\mathrm{m^{-3}} \). It is doped with a concentration of 1 in \( 10^{5} \) pentavalent atoms, the number density of holes (per \( \mathrm{m^3} \)) in the doped semiconductor will be :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 4.5\times10^{8} \)
Each pentavalent atom donates one free electron, so the electron density is about equal to the donor density. The holes then follow from the mass action law \( n_e n_h = n_i^{2} \).
\( \displaystyle N_D = \frac{5\times10^{28}}{10^{5}} = 5\times10^{23}\,\mathrm{m^{-3}} \)
Since \( N_D \gg n_i \), the electron density is \( n_e \approx N_D = 5\times10^{23}\,\mathrm{m^{-3}} \).
\( \displaystyle n_h = \frac{n_i^{2}}{n_e} = \frac{(1.5\times10^{16})^{2}}{5\times10^{23}} = \frac{2.25\times10^{32}}{5\times10^{23}} \)
\( \displaystyle n_h = 4.5\times10^{8}\,\mathrm{m^{-3}} \)
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 – …
