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For the following nuclear reaction, what was the beginning radionuclide (X)? X → 86Rn222 + 2He4

Respuesta :

The original radioisotope underwent double-alpha decay, where the original nucleus lost a total of 4 protons and 4 neutrons.

The original radioisotope would have an atomic number of 86+4 (protons), and would have an atomic mass of 222+8 (protons + neutrons). This would make the element Thorium-230, or 90Th230.