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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 `(KE_alpha)` and the Q-value (or total disintegration energy ) is
`KE_a=((A-4)/A).Q`
`Q=(A/(A-4)).KE_a`.
`=(226/(226-4))xx4.78` MeV= `226/222xx4.78` MeV
Q=4.865 MeV `approx` 4.87 MeV
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AAKASH SERIES-NUCLEI-Practice Exercise
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