- #1

Rahulrj

- 107

- 0

## Homework Statement

An Fe nucleus (A=57) decays from an excited stated to the ground state by emitting a gamma ray. The energy of the photon is 14.4 KeV when the nucleus is held fixed. If the nucleus is free to recoil then the energy of the photon emitted will be?

## Homework Equations

## E = mc^2##

## E^2 = (pc)^2 + E_{0}^2##

##KE (T) = p^2/2m##

## The Attempt at a Solution

So writing down the energy conservation

(E* - energy of Fe before decay, E' - energy of the lower state nucleus, T_{recoil} - KE from the recoiling of the nucleus)

Case 1: when nucleus is fixed ##T_{recoil} = 0##

therefore

##E* = E'+14.4##

Case 2: When nucleus is allowed to recoil, ##p_{photon} = p_{recoil}##

##E* = E'+E_{photon}+T_{recoil}##

Now ##T_{recoil} = p_{recoil}^2/2m## ( the KE of recoil is small enough that non relativistic equation can be used)

Substituting the equation in case 1

## E'+14.4= E'+E_{photon}+T_{recoil}##

## E_{photon}+T_{recoil} = 14.4##

I am stuck here as I do not know how to figure out T_{recoil}, Can somebody please help?