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Question #43816

Difficulty: 8/10 | 🟢 Free

The Question

An intrinsic semiconductor sample with an intrinsic carrier concentration $n_i = 1.0 \times 10^{10} \text{ cm}^{-3}$ is illuminated by a short laser pulse. This generates an initial excess electron-hole pair concentration of $\Delta n_0 = \Delta p_0 = 1.0 \times 10^{16} \text{ cm}^{-3}$ . The recombination rate of electrons and holes is given by $R = rnp$ , where $n$ and $p$ are the total electron and hole concentrations respectively, and the recombination rate constant is $r = 2.0 \times 10^{-10} \text{ cm}^3 \text{ s}^{-1}$ . The thermal generation rate is constant at $G_{th} = r n_i^2$ . If the laser is turned off at $t=0$ , what is the time $t_{1/2}$ required for the excess carrier concentration to drop to half of its initial value? (Assume uniform generation and no drift or diffusion currents).

Options

A.
$0.50 \, \mu\text{s}$
B.
$0.35 \, \mu\text{s}$
C.
$1.00 \, \mu\text{s}$
D.
$0.17 \, \text{s}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Statistical Nature of Decay|Law of Mass Action|Differential Equations

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between first-order probabilistic decay (like radioactivity) and second-order macroscopic decay governed by the Law of Mass Action.

A common trap is to mindlessly apply the standard radioactive half-life formula $t_{1/2} = \ln(2)/\lambda$ , assuming a constant decay probability.

However, in the high-injection regime of a semiconductor, the recombination rate depends on the product of electron and hole concentrations ( $R \propto n p \approx (\Delta n)^2$ ), leading to a non-linear, second-order differential equation.

Students who fail to set up the proper rate equation $\frac{d(\Delta n)}{dt} = -r(\Delta n)^2$ will incorrectly calculate $0.35 \, \mu\text{s}$ (using $\ln(2)$ ) or use the low-injection lifetime to get $0.17 \, \text{s}$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Nuclei - Class 12
Concept	Statistical Nature of Decay
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free