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Question #43702

Difficulty: 8/10 | 🟢 Free

The Question

In a theoretical model of nuclear decay, a stationary heavy nucleus emits a nucleon. The total energy released due to the mass defect is $Q = 15.00 \text{ MeV}$ . However, before emission, the nucleon must be localized within a surface layer of thickness $\Delta x = 1.25 \text{ fm}$ due to the short-range saturation of the strong nuclear force. The kinetic energy required for this localization is provided entirely by the decay energy $Q$ . Assuming the daughter nucleus is infinitely massive and using the minimum uncertainty $\Delta x \Delta p = \hbar/2$ , calculate the final kinetic energy of the emitted nucleon far from the nucleus. (Given: $\hbar c = 200 \text{ MeV fm}$ , rest mass energy of the nucleon $mc^2 = 1000 \text{ MeV}$ ).

Options

A.
$11.80 \text{ MeV}$
B.
$15.00 \text{ MeV}$
C.
$2.20 \text{ MeV}$
D.
$18.20 \text{ MeV}$

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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

Nuclear Force Range and Saturation|Heisenberg's Limitation|Statistical Nature of Decay

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to integrate quantum mechanical uncertainty with nuclear decay energetics.

The primary trap (Option B) exploits the classical 'billiard ball' misconception, where students assume the entire mass defect energy Q is converted into the kinetic energy of the emitted particle.

However, the finite range of the strong nuclear force requires the nucleon to be localized before emission, imposing a mandatory zero-point confinement energy cost due to Heisenberg's Uncertainty Principle ( $\Delta p \ge \hbar/2\Delta x$ ).

By failing to account for this quantum localization bottleneck, students ignore the energy 'cost' of forming the particle state, leading to an overestimation of the emission kinetic energy.

Option C traps students who incorrectly use $\Delta x \Delta p = \hbar$ , while Option D traps those who erroneously add the confinement energy instead of subtracting it.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Nuclei - Class 12
Concept	Nuclear Force Range and Saturation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free