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Nuclear Physics: Recoil in gamma decay

  • Thread starter Clouis523
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
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Or this is the Mössbauer effect and I'm dumb
 

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