When \( {}^{210}_{82}\mathrm{Pb} \) decays to \( {}^{206}_{80}\mathrm{Pb} \), it emits

When 21082Pb decays to 20680Pb, it emits one alpha particle. This is because alpha decay involves the emission of an alpha particle, which is a helium nucleus consisting of two protons and two neutrons. During alpha decay, the atomic nucleus of the parent atom loses two protons and two neutrons, resulting in a new atom with a mass number that is four units less and an atomic number that is two units less than the parent atom. In this case, 21082Pb has a mass number of 210 and an atomic number of 82, while 20680Pb has a mass number of 206 and an atomic number of 80. Therefore, during the decay process, 21082Pb emits an alpha particle, which reduces the mass number by four to 206 and the atomic number by two to 80, resulting in the formation of 20680Pb.