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
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
Relativistic energy-momemtum formula:
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?