The isotope $_{5}^{12} B$ having a mass 12.014 u undergoes $β$ -decay to $_{6}^{12} C .$ $_{6}^{12} C$ has an excited state of the nucleus $(_{6}^{12} C^{})$ at 4.041 MeV above its ground state. If $_{5}^{12} B$ decays to $_{6}^{12} C^{}$ , the maximum kinetic energy of the $β -$ particle in units of MeV is
(1u=931.5 MeVc^-2 , where c is the speed of light in vacuum)

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answer is 9.

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Detailed Solution

$Q - v a l u e = [12.014 u - (12 u + 4.041 \frac{M e V}{c^{2}})]$
$\Rightarrow Q = (0.014 u \times 931.5) M e V - (1.041) M e V = 9 M e V$
Hence, $β$ particle will have a maximum KE of 9 MeV

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