What is the Energy of a Photon Released in Atom Decay with a Change in Mass?

AI Thread Summary
The discussion centers on calculating the energy of a photon released during an atom's decay from an excited state to the ground state, considering a change in mass. The initial momentum of the atom is given, and after decay, the atom's mass decreases while it also moves, affecting the photon's energy. The key realization is that the photon’s energy is not simply ΔMc² due to the atom's motion, leading to a redshift in the photon’s frequency. The energy must be expressed as hf', where f' accounts for the relativistic Doppler effect. This highlights the complexity of the problem, contrary to its initial appearance as straightforward.
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Homework Statement


An atom of mass M decays from an excited state to the ground state with a change in mass of ΔM<<M. In the decay process, the atom releases a photon. Use the laws of energy and momentum to determine the energy of the photon, assuming the atom decays from rest.


Homework Equations


The 4-momenta of the atom before the emission:

P=(Mc2,0,0,0)

After the emission:

P=(M-ΔM)c2, -pxc,0,0)

And the photon:

E=(M-ΔM)c2

p = hf

My main concern here is that this seems too straightforward. The book labels this problem as a challenging one. I think I'm missing a subtlety in the problem.
 
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Found it. The energy of the photon isn't delta(m)c^2 as the atom is moving. So in the atom's frame, the photon is redshifted. The energy of the photon (which is hf in a rest frame) is now hf', where f' is the shifted frequency due to the transverse relativistic doppler effect.
 
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