For an α-decay, the mass number of the nucleus is always reduced by 4 and the charge number reduced by 2 for the given nucleus.
The energy released in the reaction:
For an α - decay, we have the decay process for 212/84Po as:
84Po212 → 2He4 + 82Pb208 + ΔE
The following nuclei are observed to decay by emitting an α particle: 212/84Po
Write out the decay process for 212/84Po.
Determine the energy released in this reaction. Be sure to take into account the mass of the electrons associated with the neutral atoms.
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