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Question

A stationary U238 nucleus undergoes α- decay. If the kinetic energy of product nucleus is E, the total energy released in the process is-

A
E
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B
58.5E
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C
59.5E
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D
60.5E
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Solution

The correct option is C 59.5E
mass of alpha particle ma=4m, where m is mass of a proton.
mass of Uranium nucleus is mu=238m after emission the residue will have mass mr=2384=234m

We know KE=momentum2/2m
so for residue nucleus, it should be like this E=p2r/2mr

or momentum of residue nucleus is pr=22mrE=2468mE
But during emission the alpha particle should get same momentum in opposite direction to make net momentum as zero
before and after emission so it should also be pa=2468mE

If kinetic energy of alpha particle is Ea then Ea=p2a/2ma=468mE2×4m=58.5E

So net energy released in the emission is the sum of above two i.e. 59.5E


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