A semiconductor has an electron concentration of \( 8\times10^{13}\,\mathrm{m^{-3}} \) and hole concentration of \( 2\times10^{3}\,\mathrm{m^{-3}} \). The semiconductor behaves as:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → n-type semiconductor
The type of a semiconductor is decided by its majority charge carriers. Here the electron concentration (\( 8\times10^{13}\,\mathrm{m^{-3}} \)) is far greater than the hole concentration (\( 2\times10^{3}\,\mathrm{m^{-3}} \)), so electrons are the majority carriers.
Option (1) n-type semiconductor — Correct. With \( n_e \gg n_h \), electrons are the majority carriers, which is the case of a semiconductor doped with a pentavalent (donor) impurity.
Option (2) p-type semiconductor — Incorrect. A p-type semiconductor has holes as the majority carriers, \( n_h \gg n_e \).
Option (3) intrinsic semiconductor — Incorrect. An intrinsic semiconductor has equal numbers of electrons and holes, \( n_e = n_h \).
Option (4) p-n junction — Incorrect. A p-n junction is a device formed by joining p-type and n-type regions, not a single material with one pair of carrier concentrations.
NCERT (2026-27):
Extrinsic Semiconductor – “… Thus, with proper level of doping the number of conduction electrons can be made much larger than the number of holes. Hence in an extrinsic semiconductor doped with pentavalent impurity, electrons become the majority carriers and holes the minority carriers. These semiconductors are, therefore, known as n-type semiconductors. For n-type semiconductors, we have,
\( \displaystyle n_e \gg n_h \qquad (14.3) \) ”