An intrinsic semiconductor has \( 5\times10^{28}\,\mathrm{atoms/m^3} \). It is doped by 0.01 ppm. concentration of arsenic. If \( n_i = 1.5\times10^{16}\,\mathrm{m^3} \), then number of holes in the n-type semiconductor will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 4.5\times10^{11}/\mathrm{m^3} \)
Arsenic is pentavalent, so each arsenic atom donates one free electron. In the doped semiconductor the electron and hole densities obey the mass action law \( n_e n_h = n_i^2 \).
A doping of 0.01 ppm means 0.01 arsenic atom per \( 10^6 \) atoms of the semiconductor:
\( \displaystyle N_D = 5\times10^{28}\times\frac{0.01}{10^{6}} = 5\times10^{20}\,\mathrm{m^{-3}} \)
The donor electrons far outnumber the intrinsic carriers (\( 5\times10^{20} \gg 1.5\times10^{16} \)), so \( n_e \approx N_D \):
\( \displaystyle n_h = \frac{n_i^2}{n_e} = \frac{(1.5\times10^{16})^2}{5\times10^{20}} = \frac{2.25\times10^{32}}{5\times10^{20}} \)
\( \displaystyle n_h = 4.5\times10^{11}\,\mathrm{m^{-3}} \)
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 – …
