1.

The kinetic energy of alpha - particles emitted 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 ?

Answer»

Solution :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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