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Q.

The isotope   512B having a mass 12.014 u undergoes  β-decay to  612C.  612C Has an excited state of the nucleus  612C*  at 4.041 MeV above its ground state. If   512B decays to  612C*, the maximum kinetic energy of the β  particle in units of MeV is (1 u=931.5 MeV/C2, where c is the speed of light in vacuum).

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answer is 9.

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Detailed Solution

Energy released E=[Δm]931.5=[12.014−12]931.5=13.041MeV Energy of β particle =13.041−4.041=9MeV
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