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Question #43660

Difficulty: 8/10 | 🟢 Free

The Question

A free, initially stationary atom of mass $M$ (in its ground state) undergoes an electronic transition from an excited state to the ground state. The energy difference between the excited state and the ground state is $\Delta E$ . Assuming $\Delta E \ll Mc^2$ , where $c$ is the speed of light in vacuum, which of the following expressions best represents the exact energy $E_\gamma$ of the emitted photon, accounting for the recoil of the atom?

Options

A.
$\Delta E$
B.
$\Delta E - \frac{(\Delta E)^2}{2Mc^2}$
C.
$\Delta E - \frac{(\Delta E)^2}{Mc^2}$
D.
$\Delta E + \frac{(\Delta E)^2}{2Mc^2}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Recoil of Atom|Reduced Mass Effect|Approximation (Binomial)

Why This Question Is Tricky

Common Mistakes

This question tests the application of momentum and energy conservation to atomic transitions, specifically targeting the 'recoil energy shift' trap.

Students commonly assume that the emitted photon's energy exactly equals the electronic transition energy ( $E_\gamma = \Delta E$ ), completely ignoring the atom's recoil.

However, because the photon carries momentum, the atom must recoil in the opposite direction to conserve momentum.

This recoil imparts kinetic energy to the atom, which is siphoned directly from the available transition energy.

Failing to account for this macroscopic kinetic energy shift leads to the incorrect idealized answer, missing the critical first-order relativistic correction.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Atoms - Class 12
Concept	Recoil of Atom
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free