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JEE Advanced 2016
Paper-1 2016
Question
The isotope ${ }_5^{12} \mathrm{~B}$ having a mass 12.014 u undergoues $\beta$ - decay to ${ }_6^{12} \mathrm{C}_{16}^{12} \mathrm{C}$ has an excited state of the nucleus $\left({ }_6^{12} \mathrm{C}^*\right)$ at 4.041 MeV above its ground state. If ${ }_5^{12} \mathrm{~B}$ decays to ${ }_6^{12} \mathrm{C}^*$, the maximum kinetic energy of the $\beta$ - particle in units of MeV is
$\left(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^2\right.$, where c is the speed of light in vacuum $)$.
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Question Tags
JEE Advance
Physics
Easy
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