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Question
Find the energy liberated in the reaction
223Ra → 209Pb + 14C.
The atomic masses needed are as follows.
223Ra 209Pb 14C
22..018 u 208.981 u 14.003 u
Solution
Given :
Atomic mass of 223Ra, m(223Ra) = 223.018 u
Atomic mass of 209Pb, m(209Pb) = 208.981 u
Atomic mass of 14C, m(14C) = 14.003 u
Reaction :
`""^223"Ra" → ""^209"Pb" + ""^14"C"`
Energy , `E = [m(""^223Ra) - (m(""^209Pb) + m(""^14C))]c^2`
= `[223.018 "u" - (208.981 + 14.003) "u"] c^2`
= `0.034 xx 931 "MeV"`
= `31.65 "MeV"`
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