A gamma-ray photon produces an electron-positron pair, each with a kinetic energy of 285 keV.
What was the energy of the photon?
E = mc2
hf = KEmax +W0
The Attempt at a Solution
E = 2(9.11*10^-31 kg)(3.0*10^8 m/s)^2 = 1.64*10^-13 J = 1.02 MeV
This isn't the answer but a similar problem in my textbook arrived at this answer. And where does 285 keV fit in all this? I know I can convert this to joules.
Should I be using a different equation possibly hf = KEmax +W0, and solve for KE? Does the 285 keV become converted to Joules and be substituted into the preceding equation as W?