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Polonium-210 decays by alpha decay to Lead-206 ...

84210Po --> 82206Pb + 24He2+ + 15.9531 MeV

84220Po (210.0485 amu) --> 82206Pb (206.0388 amu) + 24He (4.003874 amu) + (0.005826 amu)

1 amu = 931.162 MeV

thus 0.005826 amu = 5.4249 MeV

Most of this (5.305 MeV) ends up in the Alpha particle, leaving the Lead recoiling with 0.1199 MeV. Rarely a 0.8 MeV Gamma is emitted (0.0012%), changing the distribution of energy and momentum.

Nucleonics Fundamentals, 1959 David B. Hoisington

The Radiochemical Manual, 1962

The 15.9531 MeV is some sort of "ground state" energy (not sure what), the released decay energy is calculated from the "mass defect".

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Q: What is the properly balanced alpha decay equation indicating the resulting products Po 84210 right arrow plus?
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