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## 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 = mc

^{2}

hf = KEmax +W

_{0}

## 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 +W

_{0}, and solve for KE? Does the 285 keV become converted to Joules and be substituted into the preceding equation as W?