Question:
A bar of silicon is doped with boron concentration of \( 10^{16} \, \text{cm}^{-3} \) and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of \( 10^{20} \, \text{cm}^{-3} \, \text{s}^{-1} \). If the recombination lifetime is 100 µs, intrinsic carrier concentration of silicon is \( 10^{10} \, \text{cm}^{-3} \) and assuming 100\% ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is
A bar of silicon is doped with boron concentration of \( 10^{16} \, \text{cm}^{-3} \) and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of \( 10^{20} \, \text{cm}^{-3} \, \text{s}^{-1} \). If the recombination lifetime is 100 µs, intrinsic carrier concentration of silicon is \( 10^{10} \, \text{cm}^{-3} \) and assuming 100\% ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is
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In semiconductor physics, the steady-state product of electron and hole concentrations is equal to the square of the intrinsic carrier concentration, and can be derived from the generation and recombination rates.
Updated On: Dec 26, 2025
- \( 10^{20} \, \text{cm}^{-6} \)
- \( 2 \times 10^{20} \, \text{cm}^{-6} \)
- \( 10^{32} \, \text{cm}^{-6} \)
- \( 2 \times 10^{32} \, \text{cm}^{-6} \)
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The Correct Option is D
Solution and Explanation
To solve this, we use the equation for steady-state electron-hole concentration product in a semiconductor:
\[
n_i^2 = n_e \cdot n_h,
\]
where \(n_i\) is the intrinsic carrier concentration, and \(n_e\) and \(n_h\) are the electron and hole concentrations.
The light exposure rate is given as \( \frac{d \, n_e}{d \, t} = \text{generation rate} = 10^{20} \, \text{cm}^{-3} \, \text{s}^{-1} \).
The steady-state concentrations \( n_e \) and \( n_h \) will be such that the generation rate equals the recombination rate, which is:
\[
\frac{n_e}{\tau} = \text{generation rate},
\]
where \( \tau \) is the recombination lifetime.
Given \( \tau = 100 \, \mu \text{s} = 10^{-4} \, \text{s} \), we solve for \( n_e \) and \( n_h \) (since \( n_e = n_h \) in steady state):
\[
n_e = n_h = \sqrt{2 \times 10^{32}} = 2 \times 10^{32} \, \text{cm}^{-6}.
\]
Thus, the product of steady-state electron and hole concentrations is \( 2 \times 10^{32} \, \text{cm}^{-6} \).
Final Answer: \( 2 \times 10^{32} \, \text{cm}^{-6} \)
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