# Can somebody explain why the momentum of a gamma ray can be ignored relative to carbon's momentum in the following reaction: alpha + Be --> C + gamma?

Before the discovery of the neutron, it was proposed that the penetrating radiation produced when beryllium was bombarded with alpha particles consisted of high-energy $\gamma$ rays (up to $50\,\mathrm{MeV}$) produced in reactions such as $\alpha + ^9\mathrm{Be} \to ^{13}\mathrm{C} + \gamma$

a.) Calculate the $Q$ value for this reaction.

b.) If 5-MeV alpha particles are incident on $^9\mathrm{Be}$, calculate the energy of the $^{13}\mathrm{C}$ nucleus and, hence, determine the energy of gamma radiation assuming it is emitted as a single photon. Hint: You may neglect the momentum of the γ ray relative to the 13C nucleus. Masses: m(4He) = 4.0026u, m(9Be) = 9.0122u, m(13C) = 13.0034u

My main gripe with this problem is that i dont understand why we can neglect the momentum of the gamma ray when conserving momentum. I did the problem with the assumption but can the problem still be done without it?

• You would not know the direction relative to the direction of the alpha particle, so this would become angle-dependent.
– user137289
Commented Sep 17, 2017 at 19:29

In the lab frame, and near threshold the fused nucleus will end with most of the momentum of the alpha because there is relatively little energy to give the photon and it's momentum being $p_\gamma = E_\gamma/c$ will be quite small. This choice would also have the advantage of giving you an energy for the photon that is what an instrument would measure if you set up the experiment in the easiest way—and the way described in the problem text.
• You were right about the problem statement; I have retracted my answer and given you the upvote because the rest is just a comment: @Elvis you can prove it because the total energy in the center-of-mass frame will be approximately $E=m~c^2 + p^2/(2m) + p~c$ where $m$ is the mass of the carbon atom. This is a quadratic in $p$ that you can solve, finding that the answer is approximately $p=(E-mc^2)/c.$ for $E-mc^2\ll mc^2.$ This in turn means $p/(2m)\ll c$ and the lion's share of the energy is in the $p~c$ not the $p^2/(2m),$ hence we can pretend that the carbon atom's momentum is zero. Commented Sep 17, 2017 at 21:18