# Gamma photon momentum

## Homework Statement

A Gamma photon is used in order to dissociate deuterium into a proton and neutron. the binding energy is 2.22Mev anf the rest energies for the proton and neutron are 938Mev and 939Mev respectively. determine the minmum photon energy to achieve this. consider two cases:
1- both proton and neutron acquire collinear equal velocities parallel to the photon momentum
2-the neutron stays staionary after the collission
3- why non-collinear not considered in this analysis

## The Attempt at a Solution

Binding energy= (total number of proton and neutron)- minimum photon energy

2.22=939+938-E(MINIMUM)

E(MINIMUM)=1874.78Mev

for part 2:

since the neutron is stationary so this mean we have a zero energy for neutron and we will left only with proton energy and the minimum photon energy in this case would be:

E(MINIMUM)=939-2.22=936.78Mev

for part 3:

for non-collinear this is because the momentum is not conserved

no one have an idea for my Q if i am on the wr8 track o not

59 viewers no one can help me

alphysicist
Homework Helper
Hi matt222,

I believe the binding energy is equal to the difference in rest mass between the combined proton+neutron and the separated proton and neutron. An energy of 2.22 MeV is required to separate the proton and neutron. If the energy added were exactly equal to the binding energy, then the proton and neutron would end up with zero kinetic energy.

However, in this case, with a high energy photon causing the separation, more than just energy needs to be conserved. What does that mean the minimum photon energy needs to be?

binding energy can also be the minimum energy required to decompose a molecule, an atom, or a nucleus into its components. So I agree with alphysicist.

alphysicist
Homework Helper
matt222,