How Does a Nucleus's Mass Change After Emitting a Gamma-Ray Photon?

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SUMMARY

The discussion focuses on the mass change of a nucleus after emitting a gamma-ray photon (Eγ) during its decay to the ground state. The key conclusion is that the change in mass (me - mg) can be approximated as Eγ/c² [1 + 1/2(v/c)] when the recoil speed (v) is much less than the speed of light (c). Participants emphasized the importance of conservation of 4-momentum and energy in deriving this relationship, while clarifying the correct application of signs in the equations. The discussion also highlighted the need for careful consideration of vectorial forms in momentum conservation.

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  • Understanding of relativistic energy and momentum conservation
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  • Basic grasp of special relativity principles
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Pedro de la Torre

Homework Statement


A stationary excited nucleus decays to its grpund state by emitting a γ-ray photon of energy Eγ. The grpund state nucleus recoils in the opposite direction at speed v. Show that when v<<c the change of mass of the nucleus is approximately:

me - mg = Eγ/c2 [1+ 1/2(v/c)]

Homework Equations


v/c=pc/E
PμPμ = constant = (mc2)2
E2 = (mc2)2 + (pc)2
Pγc = Eγ

The Attempt at a Solution


I used both, the cuadrimomentum conservation and the energy conservation. I think the solution should be directly found from the cuadrimomentum conservation, but I can not solve it.
 
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Pedro de la Torre said:

The Attempt at a Solution


I used both, the cuadrimomentum conservation and the energy conservation.
Conservation of 4-momentum implies conservation of both relativistic energy and relativistic 3-momentum.

I think the solution should be directly found from the cuadrimomentum conservation, but I can not solve it.
I'm not following what you mean here. It would be helpful if you showed some detail of your attempt to solve it this way.
 

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Your work looks good, except I don't understand why you suddenly put in a negative sign at the very end. ##\frac{E_{\gamma}}{E_g}## is positive. ##E_{\gamma} = |p_{\gamma}|c##.
So, ##1 + \frac{E_{\gamma}}{2 E_g} = 1+ \frac{v}{2c}##.

upload_2017-10-2_12-46-4.png
 
Last edited:
I did that because I considered there Pg = - Pγ. But I think I am realising that this is only for the vectorial form, Isn´t it?
 
Pedro de la Torre said:
I did that because I considered there Pg = - Pγ. But I think I am realising that this is only for the vectorial form, Isn´t it?
Yes.
 

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