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The kinetic energy of an alpha-particle ...

The kinetic energy of an `alpha`-particle which flies out of the nucleus of a `Ra^(226)` atom in radioactive disintergration is `4.78 MeV`. Find the total energy evolved during the eascape of the `alpha`-particle

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The standard relation between the kinetic energy of the `alpha` - particle `(T_(alpha))` and the Q - value (or total disintegration energy ) is
`T_(alpha) =((A-4)/(A))Q`
`Q=(A/(A-4)).T_(alpha)=(226/(226-4))xx4.78MeV`
`= (226)/(222) xx 4.78 MeV`
`:. Q = 4.86 MeV ~~ 4.87 MeV`
Notice that `T_(alpha)` is very close to (but smaller than) Q.
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