Arrange the following materials in increasing order of their number density (per \(m^{-3}\)) of electrons.
A. n-type Silicon semiconductor
B. metal
C. pure Silicon semiconductor
D. p-type Silicon semiconductor
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → D, C, A, B
The electron density of a semiconductor depends on its doping, and by the mass-action law \( n_e n_h = n_i^{2} \); a metal has about one free electron per atom.
\( \text{p-type Silicon semiconductor (D)}: n_e = n_i^{2}/n_h \ll n_i \)
\( \text{pure Silicon semiconductor (C)}: n_e = n_i \approx 1.5\times10^{16}\,\mathrm{m^{-3}} \)
\( \text{n-type Silicon semiconductor (A)}: n_e \approx N_D \sim 10^{22}\,\mathrm{m^{-3}} \)
\( \text{metal (B)}: n_e \sim 10^{28}\,\mathrm{m^{-3}} \)
In increasing order: D, C, A, B.
Note: It seems that there is a mistake in the original NTA paper, which prints the unit of number density as “per \( m^{-3} \)” when it should read “\( \mathrm{m^{-3}} \)” (number per cubic metre). The unit does not affect the order, so the question can still be solved.
NCERT (2026-27):
p-type Semiconductor – “… For p-type semiconductors, the recombination process will reduce the number (ni) of intrinsically generated electrons to ne.”
Extrinsic Semiconductor – “… Example 14.2 … Solution Note that thermally generated electrons (ni ~1016 m–3) are negligibly small as compared to those produced by doping.”
Drift of Electrons and the Origin of Resistivity – “… The density of conduction electrons, n is equal to the number of atoms per cubic metre (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). A cubic metre of copper has a mass of 9.0 × 103 kg. Since 6.0 × 1023 copper atoms have a mass of 63.5 g,
\( \displaystyle n = \frac{6.0 \times 10^{23}}{63.5} \times 9.0 \times 10^{6} = 8.5 \times 10^{28}\ \mathrm{m^{-3}} \)”