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The kinetic energy of alpha- particles ...

The kinetic energy of `alpha`- particles emiited in the decay of `._(88)Ra^(226)` into `._(86)Rn^(222)` is measured to be `4.78 MeV`. What is the total disintegration energy or the `'Q'`-value of this process ?

Text Solution

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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) = ((4-4)/(A))Q` , `Q = ((A)/(A-4)).KX_(a)`
`= ((226)/(226-4)) xx 4.78 MeV = (226)/(222) xx 4.78 MeV`
`Q = 4.865 MeV ~~ 4.87 MeV`
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