1. The problem statement, all variables and given/known data Consider the following reaction that is possible when a proton (p+) collides with a photon (γ): p+ + γ → ∆+. Suppose that the photon is a cosmic microwave-background (CMB) photon of energy 2.3 × 10−4eV. Calculate the minimum energy that the proton must have for this reaction to take place. m∆ = 1232eV/c2. Hints: • The minimum energy will be for a head-on collision. Be careful with the signs of the momenta when doing the calculation. • The answer is a very large energy, so you will be able to make the approximation that the the proton energy is much bigger than it’s rest energy. • You might find it interesting to look up Greisen-Zatsepin-Kuzmin limit. (is this relevant?) 2. Relevant equations K = E − mc2 E2 −(pc)2 =(mc2)2 3. The attempt at a solution (See attached table) I am thinking that because I'm after the minimum energy of the proton for a head on collision, would this mean that the delta baryon is going to have zero kinetic energy and momentum thus a total energy 1232 MeV? I can then say that the proton has momentum pc = -2.3*10-10MeV. But that just makes the energy of the proton also equal to 938.3 MeV within the precision of my calculator. That's not right, I should get Eproton>>mprotonc2. Where to now? Does the change in mass of 293.7 MeV have anything to do with this? Thanks!