Question: Find the energy that is released when a nucleus of lead 211 82 Pb (atomic mass = 210.988 735 u) undergoes decay to
Find the energy that is released when a nucleus of lead 21182Pb (atomic mass = 210.988 735 u) undergoes β– decay to become bismuth 21183Bi (atomic mass = 210.987 255 u).
ilpo 211 Pb
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The reaction and the atomic masses are When a 211 82 Pb nucleus decays into a bismuth 211 83 Bi nucl... View full answer
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