To calculate the binding energy released when alpha decay converts radium 226, with atomic mass of 226.025, into radon 222, with atomic mass 222.01757, I understand that we subtract the atomic mass of radium 226 from the sum of the atomic masses of the radon 222 and the alpha particle, which is the helium nucleus (atomic mass 4.002603). My question involves the two electrons that are "left over" and I assume are contained in the radon atom after the transmutation. In other words, during the alpha decay, two protons and two neutrons are emitted as an He 4 nucleus. We are then left with 86 protons and 136 neutrons and to get radon 222. Where are the 88 electrons that radium 226 started with? Since a helium nucleus is emitted, I assume the electrons are in the radon, producing a negatively charged atom. So I would think that to calculate the binding energy, I would have to include the mass of the two extra electrons in addition to the atomic mass of radon. But this seems to give me an incorrect answer. Or is the book mistaken? or is 4.002603 the mass of the helium atom, rather than that of the alpha particle as reported. I am confused about these two electrons! How are they accounted for?