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In a pure intrinsic semiconductor at temperature $T > 0\,\text{K}$, the relation between the number density of free electrons $n_e$ and the number density of holes $n_h$ is
A$n_e \gg n_h$
B$n_e = n_h$
C$n_e \ll n_h$
D$n_e = n_h^2$
Answer & Solution
Correct answer: B. $n_e = n_h$
1. In an INTRINSIC (pure, undoped) semiconductor, free electrons are generated ONLY by thermal excitation of valence electrons across the band gap into the conduction band.
2. Each electron that crosses the gap leaves behind ONE hole in the valence band.
3. So electrons and holes are created in PAIRS, one for one: $n_e = n_h \equiv n_i$, the intrinsic carrier concentration.
4. At room temperature, $n_i \approx 1.5\times 10^{10}\,\text{cm}^{-3}$ for Si, much smaller than typical doping densities.
5. Options A and C describe EXTRINSIC semiconductors (n-type or p-type respectively) — not intrinsic. Option D is dimensionally wrong.
_Source: NCERT Class 12 Physics Part 2, Ch 14, §14.3 (Intrinsic Semiconductor — pair production), p. 5._
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