Find the energy liberated in the reaction: 223Ra→209Pb+14C The atomi – InterviewSolution

1.	Find the energy liberated in the reaction: 223Ra→209Pb+14C The atomic masses needed are as follows: 223Ra223.018u 209Pb208.981u 14C14.003u
Answer» Find the energy liberated in the reaction: $^{223} R a \to^{209} P b +^{14} C$ The atomic masses needed are as follows: $^{223} R a 223.018 u$ $^{209} P b 208.981 u$ $^{14} C 14.003 u$

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