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

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When an atom emits a photon, it loses mass due to the energy-mass equivalence principle, E=mc^2. The emitted photon carries energy quantified by E=hf, leading to a decrease in the atom's mass, but this loss is greater than hf/c^2 because the atom also recoils. The recoil momentum of the atom must be considered, as it conserves momentum alongside energy. Thus, the mass loss calculation must account for both the energy of the emitted photon and the recoil effect. Understanding these principles clarifies why the mass loss exceeds the initial photon energy divided by c^2.
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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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