Nuclear Physics: Recoil in gamma decay

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SUMMARY

The discussion focuses on calculating the difference between the gamma energy (Egamma) and the excitation energy of a nucleus during gamma decay, considering the recoil of the nucleus. The participant employs the linear elastic assumption and conservation of energy principles, leading to the equation Eexc = Egamma(1 + Egamma/c²m). The conclusion drawn is that the recoil energy is negligible, typically in the range of tens to hundreds of eV, which simplifies the calculations significantly. The participant questions whether this aligns with the Mössbauer effect, indicating a deeper exploration of nuclear recoil phenomena.

PREREQUISITES
  • Understanding of gamma decay and nuclear physics principles
  • Familiarity with conservation of momentum and energy
  • Knowledge of the Mössbauer effect in nuclear physics
  • Basic proficiency in using equations involving mass-energy equivalence
NEXT STEPS
  • Study the principles of the Mössbauer effect in detail
  • Learn about recoil corrections in nuclear reactions
  • Explore the implications of linear elastic collisions in nuclear physics
  • Investigate the relationship between energy levels and photon emission in nuclear decay
USEFUL FOR

This discussion is beneficial for students and researchers in nuclear physics, particularly those studying gamma decay, recoil effects, and the Mössbauer effect. It is also relevant for physicists involved in experimental nuclear research and applications.

Clouis523
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Homework Statement


Question wants you to

a) find the difference between Egamma and the excitation energy of the nucleus due to the fact that the nucleus recoils. (using approximation that Egamma is small relative to nucleus mass.

other parts are simple if I can get a)

Homework Equations



Unsure but obviously we have 931.494 MeV/u and I'm using a linear elastic assumption as well the assumption that energy is conserved.

The Attempt at a Solution



Starting from momentum = 0 I got the momentum of the photon must = - momentum of nucleus

p=0 => abs(mv)=abs(h/λ)

and since λ=ch/Egamma

→ abs(mv)=abs(Egamma/c)

→v=Egamma/cm

Now using Eexc=Egamma+Enucleus
Eexc=Egamma+0.5mv2
Eexc=Egamma+0.5m(Egamma/cm)2
Eexc=Egamma+0.5Egamma2/c2m

leading to the relation

E=Egamma(1+Egamma/c2m)

or

E=Egamma(1+Egamma/1862.988m)

if m is given in amu and Egamma in MeV

which if I apply the assumption given in the question that c2m >> Egamma I obtain

E=Egamma as the second term goes to zero.

Now my thought on this is that unless I'm missing something recoil energy is given in terms tens or hundreds max of eV which would be negligable. Am I missing something here? Because the follow up questions become moot as everything goes to a multiplicative factor of 1.
 
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Or this is the Mössbauer effect and I'm dumb
 

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