# Atom emit photon

## Homework Statement

An atom initially at rest emits a photon. Explain why the mass of the atom decreases, taking into account energy and momentum conservation.

## The Attempt at a Solution

I get the energy conservation part. E=delta(m)c^2=hf. But how would momentum conservation cause an atom at rest to decrease in mass? Unless it's talking about the zeroth component of a 4-vector - in which case that's just energy, surely?

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Delta m c^2 = h f is not exactly valid, because then you are ignoring the change in momentum of the atom. If you pretend that the atom is in rest before and after the emission of the photon, then the change in energy of the atom is delta m c^2 and conservation of energy implies that this equals up to a minus sign h f.

You can say that the intial energy of the system is m c^2 and the final energy is gamma m' c^2 + hf. Now, we can write

gamma m' c^2 = m' c^2 + E_k

We know that E_k > 0 due to recoil. Energy conservation then implies that:

m c^2 = m' c^2 + E_k + h f

Then it follows that m' < m

Good exercise: Compute m' in terms of m and f

Ok. My explanation of the mass deficit due to energy conservation was oversimplified. But the problem lies in the momentum conservation. Any idea how I would explain that?

Ok. My explanation of the mass deficit due to energy conservation was oversimplified. But the problem lies in the momentum conservation. Any idea how I would explain that?
You simply have to say that the photon has a momentum of hf/c, so the atom must have the same mometum, but in the opposite direction, This then means that the kinetic energy E_k is nonzero. Then since:

m c^2 = m' c^2 + E_k + h f

You have:

m' = m - [E_k + hf]/c^2

The contents of the square brackets is positive so this proves that
m' < m

But I think it is far more interesting to actually compute m'. That is a good exercise to practice the algebra of four vectors.

Duh! I couldn't see the wood for the trees! :P Thanks!