You know the mass created: 2*8.68Mev/c^2
so energy= masscreated*c^2=2*8.68Mev
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
2 answers
thank you very, i did the calculations and it was simple math after your explanation..
i got answer as 17.36 MeV
=]
i got answer as 17.36 MeV
=]