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## Homework Statement

Question wants you to

a) find the difference between E

_{gamma}and the excitation energy of the nucleus due to the fact that the nucleus recoils. (using approximation that E

_{gamma}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/E

_{gamma}

→ abs(mv)=abs(E

_{gamma}/c)

→v=E

_{gamma}/cm

Now using E

_{exc}=E

_{gamma}+E

_{nucleus}

E

_{exc}=E

_{gamma}+0.5mv

^{2}

E

_{exc}=E

_{gamma}+0.5m(E

_{gamma}/cm)

^{2}

E

_{exc}=E

_{gamma}+0.5E

_{gamma}

^{2}/c

^{2}m

leading to the relation

E=E

_{gamma}(1+E

_{gamma}/c

^{2}m)

or

E=E

_{gamma}(1+E

_{gamma}/1862.988m)

if m is given in amu and E

_{gamma}in MeV

which if I apply the assumption given in the question that c

^{2}m >> E

_{gamma}I obtain

E=E

_{gamma}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.