Suppose a pure Si semiconductor has 2 x 1028 atoms per m3. It is doped with 5 ppm concentration of pentavalent atoms. If the intrinsic concentration of the electrons in the semiconductor is 2 x 1016 m−3, what will be the concentration of holes after doping?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4 x 109 m−3
Each pentavalent atom donates one free electron, so after doping the electron concentration is practically the donor concentration. The hole concentration then follows from the mass-action law \( n_e n_h = n_i^{2} \).
\( \displaystyle N_D = 5\,\mathrm{ppm} \times 2\times10^{28}\,\mathrm{m^{-3}} = 5\times10^{-6} \times 2\times10^{28}\,\mathrm{m^{-3}} \)
\( \displaystyle N_D = 1\times10^{23}\,\mathrm{m^{-3}} \)
The donor electrons (\( 10^{23} \)) far outnumber the intrinsic electrons (\( 2\times10^{16} \)), so \( n_e \approx N_D \).
\( \displaystyle n_h = \frac{n_i^{2}}{n_e} = \frac{(2\times10^{16}\,\mathrm{m^{-3}})^{2}}{1\times10^{23}\,\mathrm{m^{-3}}} \)
\( \displaystyle n_h = \frac{4\times10^{32}}{1\times10^{23}}\,\mathrm{m^{-3}} = 4\times10^{9}\,\mathrm{m^{-3}} \)
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 – …
