- #1

- 132

- 0

## 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

## Homework Equations

## The Attempt at a Solution

my answer for 1:

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