# Nuclear Physics: Recoil in gamma decay

• Clouis523
In summary, the question asks for the difference between Egamma and the excitation energy of the nucleus, taking into account the recoil of the nucleus. The solution involves using the linear elastic assumption and conservation of energy to derive a relation between Egamma and the excitation energy. The result is that, under the assumption that Egamma is small relative to the nucleus mass, the difference between Egamma and the excitation energy is negligible.

## 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

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.

Or this is the Mössbauer effect and I'm dumb