Hello all, I was recently assigned a problem in my modern physics class regarding finding the mean energy of Cosmic Microwave Background (CMB) photons. The problem reads as follows: The universe is permeated with primordial microwave radiation that has a mean wavelength of about 2.5 mm. (a) Find the mean energy (in eV) of the CMB photons, given that the energy of a photon is related to its frequency by the relationship E = hf (where f=frequency), where h is Planck's constant. (b) Consider a cosmic-ray proton that has total relativistic energy E in the CMB reference frame. Find E such that the Doppler shifted CMB photons striking the proton head-on have a mean energy of 100 MeV. I believe I have solved (a), with the solution as follows: E = hf = h(c/wavelength) E(mean) = h(c/mean wavelength) = (4.136x10^-15 eV*s)[(3.0x10^8 m/s)/2.5x10^-3 m] = 4.963 x 10^-4 eV The second part, (b), I have no idea how to find. Any help would be much appreciated!