# Nuclear problem

Nuclear problem,,,

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

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

what do you think guys i am doing well

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nrqed
Homework Helper
Gold Member
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

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

2.22=939+938-E(MINIMUM)

E(MINIMUM)=1874.78Mev
No, you are doing as if the neutron and proton were remaining at rest. You must take into account that they must move so you must consider both conservation of energy and of momentum. In addition, you don't take into account that the proton and neutron's rest mass energies are there both before and after the interaction!! So it's completely off.
Write the total energy before and after. Write the total momentum before and after
i

Energy before colission is and after collsion is:

m1v1/2+m2v2/2=m1u1/2+m1u2/2

total momentum:

m1v1+m2v2=m1u1+m2u2

so if both proton and neautron have equal velocities so total energy would equal to:

m1+m2=m1+m2

was that true

alphysicist
Homework Helper
Hi matt222,

I replied to another thread from you with this same problem about 5 days ago; did you not see it? It was at: