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Question

The isotope 125B having a mass 12.014 u undergoes β -decay to 126C , 126C has an excited state of the nucleus (125C ) at 4.041 MeV above its ground state. If 125B decays to 126C, the maximum kinetic energy of the β-particle in units of MeV is
(1 u=931.5MeV/c2, where c is the speed of light in vacuum).

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Solution

125B126C+β+¯ν
For any radioactive particle decay, the energy released in the reaction will be given by Q=(mBmC)c2
Now mC=12 u and mB=12.041 u
Out of Q, 4.041 eV energy is used to excite 126C to its excited state 126C, then K.E. available to β particle and antineutrino is =(mBmC)c2Δmc2=0.014×931.54.041=9 MeV
The β particle will have maximum K.E when antineutrino will have minimum K.E i.e 0.
Therefore maximum K.E of the β particle is 9 MeV

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