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Question #42570

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The Question

A solid sample initially contains $N_0$ identical, independent radioactive nuclei at fixed lattice sites. The decay constant of the nuclei is $\lambda$ . The decayed daughter nuclei remain at their original sites. Assuming $N_0 \gg 1$ and that the only change in the system's entropy arises from the number of possible microscopic configurational arrangements of decayed and undecayed nuclei, which of the following expressions represents the entropy $S(t)$ of the sample at time $t$ ? (Here $k_B$ is the Boltzmann constant).

Options

A.
$N_0 k_B [\lambda t e^{-\lambda t} - (1 - e^{-\lambda t}) \ln(1 - e^{-\lambda t})]$
B.
$k_B (\ln N_0 - \lambda t)$
C.
$N_0 k_B \lambda t e^{-\lambda t}$
D.
$N_0 k_B [\lambda t e^{-\lambda t} + (1 - e^{-\lambda t}) \ln(1 - e^{-\lambda t})]$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Entropy and Reversibility|Statistical Nature of Decay

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental statistical definition of entropy ( $S = k_B \ln \Omega$ ) in a non-thermal, time-varying process (radioactive decay).

Students often incorrectly associate entropy strictly with heat transfer ( $dS = dQ/T$ ) and fail to recognize that the irreversible decay process generates entropy through the increasing number of configurational microstates.

The trap lies in assuming entropy is proportional to the remaining nuclei or failing to account for the indistinguishability of the decayed/undecayed subsets, leading to incorrect combinatorial terms.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Entropy and Reversibility
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free