problem 55 page 1390 asks for "the upper limit that theobservation sets on the mass of a neutrino". Obs. was a supernova 170 000 ly away. 8 neutrinos in 6s, (energies betw. 8 MeV and 40 MeV) and 11 n's in 13 s ( 10 MeV n. needs 10 s more than photon totravel this distance) The expert solution does not give this ans. It only converts light years to seconds! The book gives the ans. as " 19 eV/(c squared)" Please explain this ans. Brian. problem 55 page 1390 asks for "the upper limit that theobservation sets on the mass of a neutrino". Obs. was a supernova 170 000 ly away. 8 neutrinos in 6s, (energies betw. 8 MeV and 40 MeV) and 11 n's in 13 s ( 10 MeV n. needs 10 s more than photon totravel this distance) The expert solution does not give this ans. It only converts light years to seconds! The book gives the ans. as " 19 eV/(c squared)" Please explain this ans. Brian.
Explanation / Answer
19eV is energy in electron volts. 1 electron voltis 1.6e-19 Joules, so for very small energies it is more convenientto use the electron volt rather than Joules. From Einstein's equation e=mc*c, the solution you gave hasjust rearranged this to m=e/c*c Was this what you needed clarification on?