Question

A K-40 nucleus decays by releasing a 1.46 MeV gamma ray. What is the equation for this reaction?

Answer 1

Potassium isotope K-40 decays somewhat in a different way. Thus, 89.3 % it, usually goes under β-decay, with the emission of an electron with the energy of 1.31 MeV. But for remaining 10.7 % its decay route is different. Here it goes via electron capture route and potassium is transferred to argon. This nucleus is in an excited state and emits γ-rays with an energy of 1.46 MeV. This mode of disintegration includes also some characteristic x-rays – because one of the orbital electrons is captured by the nucleus and leaves a hole that is filled with other electrons.

40-K + e− →  40-Ar + νe(electron neutrino).