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

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SUMMARY

The energy of a photon released during atomic decay is determined by the change in mass (ΔM) of the atom, which decays from an excited state to the ground state. The initial 4-momentum of the atom is represented as P=(Mc², 0, 0, 0), while after decay, it is P=(M-ΔM)c², -pxc, 0, 0). The energy of the emitted photon is given by E=(M-ΔM)c². It is crucial to account for the relativistic Doppler effect, as the photon is redshifted in the atom's frame, resulting in a frequency shift from hf to hf', where f' is the shifted frequency.

PREREQUISITES
  • Understanding of atomic decay processes
  • Familiarity with 4-momentum in relativistic physics
  • Knowledge of the relativistic Doppler effect
  • Basic principles of energy conservation in physics
NEXT STEPS
  • Study the principles of 4-momentum in special relativity
  • Learn about the relativistic Doppler effect and its applications
  • Explore energy conservation in atomic decay processes
  • Investigate photon energy calculations in different reference frames
USEFUL FOR

Students in physics, particularly those studying atomic physics and relativity, as well as educators seeking to clarify concepts related to photon energy and atomic decay.

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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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