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The kinetic energy of `alpha`-particle emitted in the `alpha`-decay of `""_(88)Ra^(266)` is [given, mass number of Ra = 222 u]
A. `5.201` Me V
B. `3.301` Me V
C. `6.023` Me V
D. `4.871` Me V

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Correct Answer - D
(d) the nuclear should be `""_(88)Ra^(226) to ""_(86)Rn ^(222)+ alpha `
Energy corresponding mass defect
`Q =[{:("(mass number of Ra )"),("-(Mass number of Ru )"),("-m (4He )"):}]Uxx931.5Me V = 4.871MeV `

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