Solving Nuclear Problem: Minimum Photon Energy

[matt222]

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


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

what do you think guys i am doing well

 

[nrqed]

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


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

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

 

[matt222]

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]

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:

https://www.physicsforums.com/showthread.php?t=235644

My post there indicated that your binding energy formula was wrong; the 2.22 MeV in this problem is the difference between the rest mass of the proton and neutron when they are separated and when they are combined.

In the last post in this thread, you have written down some energy and momentum conservation conditions for nonrelativistic speeds. That's the right procedure to follow, however, I think there are two problems: First, you have neglected the photon's momentum and energy. For example, the momentum that the proton and neutron have after the collision is equal to the momentum that the photon had before the collision.

Second, I think the speeds involved are relativistic speeds; if that's true you'll need the relativistic energy and momentum formulas.