Asked by Angie

A gamma ray (a high-energy photon of light) can produce a particle- antiparticle pair, each particle with mass 8.68 MeV/c2 when it enters the electric field of a heavy nucleus: γ ! λ+ + λ− .

What minimum γ-ray energy is required to accomplish this task?

I have searched but can't come up with a numerical answer in MeV units. Please Help
Answered by bobpursley
You know the mass created: 2*8.68Mev/c^2

so energy= masscreated*c^2=2*8.68Mev
Answered by Angie
thank you very, i did the calculations and it was simple math after your explanation..

i got answer as 17.36 MeV

=]
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