Bismuth-208, which has an atomic mass of 208u and 83 protons in the nucleus, decays through the emission of 2 alpha particles and a beta-positive particle. What isotope results from this decay?
20883Bi ==> xyZ + 42a + 42a + 01B x = 208 - 8 = 200 y = 83 - 5 = 78 Proton number = 78 therefore Z = Pt Z= 20078Pt
