Very Basic Nuclear Reaction Q Value Problem

[steeeeevo]

Hello,

I looked around before and could not find an answer to this question.

So given the reaction

neutron + X -> Y + gamma

and we assume that both initial particles are at rest.

Then using momentum balance we can find that the energy of the gamma is:

Egamma = -m_Y*c^2 + sqrt( (m_Y*c^2)^2 + 2Q*m_y*c^2).

Here is what I don't get. The question says to assume that m_Y*c^2 >> Q. Using the binomial theorem and rearranging the above you get that in the limit, E_gamma -> Q. Why does the energy of the gamma approach Q? I originally thought that the energy of the gamma would approach 0 since the increasing rest mass of Y would cause more energy to be needed as binding energy in Y, thus leaving less energy for the gamma.

Please help... ;)

Thanks.

 

[hamster143]

How do you define Q? As an excess energy left after you're done assembling Y from neutron and X? This energy is divided into kinetic energy of Y and energy of the photon. If Y is very heavy compared with Q, it needs very little energy to balance the momentum of the massless photon.

 

[sir-pinski]

Hello,

The reason why the energy of the gamma approaches Q in the limit is because Q represents the net energy released in the reaction. In other words, it is the difference between the initial and final energies of the particles involved. Since we are assuming that m_Y*c^2 >> Q, the energy released in the reaction is much smaller compared to the rest mass energy of Y. This means that the energy of the gamma will be very close to the energy released, which is Q.

As for your initial thought about the energy of the gamma approaching 0, it is important to remember that Q is a positive value and therefore, in the limit, the energy of the gamma cannot approach 0. It will always be equal to or greater than Q.

I hope this helps clarify your understanding. Let me know if you have any other questions.