Law Of Mass Action
A silicon crystal with intrinsic carrier concentration (1.5 \times 10^{16},\text{m}^{-3}) is doped so that its free electron density rises to (4.5 \times 10^{22},\text{m}^{-3}). What is the equilibrium hole concentration in this doped sample?
Select the correct option:
Solution
\(5 \times 10^{9}\,\text{m}^{-3}\)
The law of mass action states that, at thermal equilibrium and fixed temperature, the product of the electron and hole concentrations in a semiconductor stays constant and equals the square of the intrinsic concentration: (n_e , n_h = n_i^2). Doping changes the individual carrier densities but cannot change this product, because increased recombination quickly suppresses the minority carrier when the majority carrier rises. Physically, once donor doping floods the crystal with extra electrons, any hole is far more likely to meet an electron and recombine, so the hole population is actively driven down rather than left unchanged until the fixed product is restored. This is why the minority density depends jointly on the doping level and the intrinsic value. Substituting (n_i = 1.5 \times 10^{16},\text{m}^{-3}) gives (n_i^2 = 2.25 \times 10^{32},\text{m}^{-6}). Dividing by the electron density (n_e = 4.5 \times 10^{22},\text{m}^{-3}) yields (n_h = (2.25 \times 10^{32})/(4.5 \times 10^{22}) = 5 \times 10^{9},\text{m}^{-3}). The option (3 \times 10^{6}) results from a wrong exponent in the division. The option equal to (n_i) ignores doping entirely. The option equal to (n_e) wrongly assumes carriers stay balanced after doping. As a plausibility check, the hole density is far below (n_i), which is exactly right for an n-type sample where holes are strongly suppressed minority carriers.
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About This Question
- Subject
- physics
- Chapter
- semiconductor electronics
- Topic
- law of mass action
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
\(5 \times 10^{9}\,\text{m}^{-3}\)
The law of mass action states that, at thermal equilibrium and fixed temperature, the product of the electron and hole concentrations in a semiconductor stays constant and equals the square of the intrinsic concentration: (n_e , n_h = n_i^2). Doping changes the individual carrier densities but cannot change this product, because increased recombination quickly suppresses the minority carrier when the majority carrier rises. Physically, once donor doping floods the crystal with extra electrons, any hole is far more likely to meet an electron and recombine, so the hole population is actively driven down rather than left unchanged until the fixed product is restored. This is why the minority density depends jointly on the doping level and the intrinsic value. Substituting (n_i = 1.5 \times 10^{16},\text{m}^{-3}) gives (n_i^2 = 2.25 \times 10^{32},\text{m}^{-6}). Dividing by the electron density (n_e = 4.5 \times 10^{22},\text{m}^{-3}) yields (n_h = (2.25 \times 10^{32})/(4.5 \times 10^{22}) = 5 \times 10^{9},\text{m}^{-3}). The option (3 \times 10^{6}) results from a wrong exponent in the division. The option equal to (n_i) ignores doping entirely. The option equal to (n_e) wrongly assumes carriers stay balanced after doping. As a plausibility check, the hole density is far below (n_i), which is exactly right for an n-type sample where holes are strongly suppressed minority carriers.
This medium difficulty physics question is from the chapter semiconductor electronics, covering the topic of law of mass action. It appeared in the 2025 exam.
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