Lead 212/82Pb (atomic mass = 211.991871 u) is converted into bismuth 212/83Bi (atomic mass = 211.991255 u) via β- decay. What is the energy (in MeV) released in this process?

Answer 1

Q = (211.991871 – 211.991255) •1.66•10^-19•(3•10^8) = 9.2•10^-14 J =
=5.75•10^5 eV = 0.575 MeV