How much mass does an atom lose when it emits a photon?

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SUMMARY

An atom at rest emits a photon with frequency f, resulting in a decrease in its mass. This mass loss is greater than hf/c² due to the conservation of momentum, as the atom recoils in the opposite direction of the emitted photon. The relationship between energy and mass is governed by E=mc² and E=hf, leading to the conclusion that the atom's mass loss must account for both energy and momentum conservation. Thus, the correct mass loss calculation involves considering the recoil effect of the atom post-emission.

PREREQUISITES
  • Understanding of Einstein's mass-energy equivalence (E=mc²)
  • Familiarity with Planck's equation (E=hf)
  • Knowledge of momentum conservation principles
  • Basic concepts of atomic physics and photon emission
NEXT STEPS
  • Study the implications of momentum conservation in photon emission scenarios
  • Explore advanced topics in quantum mechanics related to energy and mass loss
  • Learn about the relationship between frequency, energy, and mass in particle physics
  • Investigate real-world applications of mass-energy equivalence in nuclear reactions
USEFUL FOR

Students of physics, particularly those focusing on quantum mechanics, atomic physics, and energy conservation principles. This discussion is also beneficial for educators seeking to clarify concepts related to photon emission and mass loss.

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Homework Statement



An atom initially at rest emits a photon with frequency f. Explain why the mass of the atom decreases, and by an amount greater than hf/c^2.

Homework Equations



E= mc^2
E = hf

The Attempt at a Solution



So I know that the atom must lose mass when the photon is emitted because in order for it to emit energy it must lose mass (by E = mc^2), but I don't understand why it should be more than hf/c^2, since from these 2 equations:

E = mc^2 = hf

therefore m = hf/c^2

Why is this wrong?

Thanks.
 
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The atom is initially at rest. Is it at rest after it emits the photon?
 
Ok, I see. So the atom recoils with the opposite momentum of the photon (-hf/c), since momentum is conserved. Thanks, I was just thinking about energy conservation.
 

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