Solving Gamma-Ray Photon Energy: 285 keV Electron-Positron Pair

[stacker]

Homework Statement

A gamma-ray photon produces an electron-positron pair, each with a kinetic energy of 285 keV.
What was the energy of the photon?

Homework Equations

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?

 

[AvroArrow]

Any help would be appreciated. The answer is 1.02 MeV. You can calculate this using the equation E = mc2, where m is the mass of the electron-positron pair. Since the kinetic energies of each particle is 285 keV, their total kinetic energy is 2*285 keV = 570 keV. This is equivalent to 570*1.6022*10^-16 J = 9.12*10^-14 J. The mass of the particle pair can be calculated from the equation E = mc2, which gives m = 9.12*10^-14/ (3.00*10^8)^2 = 9.11*10^-31 kg. Therefore, the total energy of the gamma ray photon is E = 2*9.11*10^-31*(3.00*10^8)^2 = 1.02 MeV.