# Homework Help: Estimating the energy of an emitted photon in gamma decay

1. Apr 20, 2013

### dungonyu

1. The problem statement, all variables and given/known data

Consider a nucleus which is initially at rest and in an excited state with energy Ei. It then
decays to a lower energy state with energy Ef by emitting a gamma-ray photon.
Show that the energy of the photon is approximately given by

Eγ≈ΔE−(ΔE)^2/(2mc^2)

where ΔE = Ei - Ef and m is the mass of the nucleus.Note that ΔE is of the order MeV and
hence ΔE << mc^2

2. Relevant equations

Relativistic energy-momemtum formula:
E^2=(mc^2)^2+( pc)^2

3. The attempt at a solution

If I consider Ei=m*c^2 and Ef=m'*c^2, where m' is the mass of the nucleus after the emission,

After some calculations with the formula by applying conservation of energy and momentum,

1. momentum of the photon equals to momentum of the nucleus in magnitude, Eγ=pc,

2. mc^2= "square root of (m'c^2)^2+( pc)^2" + Eγ

I get Eγ= ΔE−(ΔE)^2/(2mc^2), which does not have any approximation

Am I wrong in the very first step by assuming

Ei=m*c^2 and Ef=m'*c^2 ? and applying the formula but forgetting some important approximation?

Last edited: Apr 21, 2013
2. Apr 21, 2013

### Staff: Mentor

Eγ= ΔE−(ΔE)^2/(2mc^2) gets negative for large ΔE, it has to be an approximation with limited validity.

You don't need the mass of the initial nucleus - it is at rest, it has zero momentum anyway.

3. Apr 21, 2013

### dungonyu

I dont understand. Shouldn't I approximate the energy of the emitted photon with the mass of the nucleus? And after the emission, both the photon and the nucleus carry non zero momentum.

I have enriched my steps in attempt. And I just wonder whether my first assumptions on Ei and Ef are correct

4. Apr 21, 2013

### Staff: Mentor

With the energy difference for the nucleus.
Sure, but you don't have the initial nucleus after the emission.

As ΔE << mc^2, you can use the nonrelativistic momentum of the nucleus.

There is still something missing, I think.

$$mc^2=\sqrt{(m'c^2)^2+(pc)^2} + Eγ$$
And what did you do afterwards?

5. Apr 21, 2013

### dungonyu

pc= Eγ. So (mc^2-Eγ)^2=(m'c^2)^2+(Eγ)^2

then
(mc^2)^2-2mc^2*(Eγ)+(Eγ)^2=(m'c^2)^2+(Eγ)^2
rearranging terms and cancelling (Eγ)^2 we get
2mc^2*(Eγ)= (mc^2)^2- (m'c^2)^2

if we assert Ei = mc^2 and Ef = m'c^2,
2mc^2*(Eγ)= (Ei)^2-(Ef)^2= (Ei-Ef)(Ei+Ef)=ΔE(2Ei-ΔE)

so by dividing 2mc^2, which is equal to 2Ei, the desired result follows. However is it awkward to assume that Ei = mc^2 because Ei is an excited energy level of the nucleus?

6. Apr 21, 2013

### Staff: Mentor

I think m should be the mass of the nucleus afterwards. Here you can use the assumption ΔE << mc^2 to fix that ;).