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Question #43647

Difficulty: 8/10 | 🟢 Free

The Question

A hypothetical atom has a ground state rest mass $M$ . It is initially at rest in a highly excited state. The difference in rest mass energy between this excited state and the ground state is $\Delta E = \frac{3}{2}Mc^2$ . The atom emits a single photon and transitions directly to the ground state. Taking into account exact relativistic energy and momentum conservation, what is the energy $E_\gamma$ of the emitted photon? ( $c$ is the speed of light in vacuum)

Options

A.
$\frac{21}{20} Mc^2$
B.
$Mc^2$
C.
$\frac{3}{2} Mc^2$
D.
$\frac{3}{8} Mc^2$

Check Your Answer

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

Energy Level Scaling|Recoil of Atom|Impulse-Momentum Theorem

Why This Question Is Tricky

Common Mistakes

This question tests the limits of the standard atomic recoil approximation.

Students typically either ignore recoil entirely ( $E_\gamma = \Delta E$ , Option C) or apply the non-relativistic kinetic energy formula for the recoiling atom ( $E_\gamma + E_\gamma^2 / 2Mc^2 = \Delta E$ ).

Because the transition energy $\Delta E$ is comparable to the rest mass energy $Mc^2$ , the recoil is highly relativistic.

Applying the non-relativistic recoil formula incorrectly yields $E_\gamma = Mc^2$ (Option B).

The correct approach requires using the exact relativistic energy-momentum invariant and acknowledging that the excited atom has a greater rest mass $M^* = M + \Delta E/c^2$ than the ground state atom.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Atoms - Class 12
Concept	Energy Level Scaling
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free