QJEEAdv16Ph07ab: JEE Advanced 2016 Physics - Class 12 JEEAdv16Ph07ab Solution: The isotope having a mass 12.014 u undergoes -decay to . has an …

As the ${ }_5^{12} B$ is going under $\beta$ -decay to form $_6^{12}C$ . We can write the balanced equation as follows

${ }_5^{12} \mathrm{~B} \rightarrow{ }_6^{12} \mathrm{C}+e^{-1}+\bar{\nu}$

The $Q$ -value of this reaction is given by

$\begin{aligned}Q & =\left[m\left({ }_5^{12} \mathrm{~B}\right)-m\left({ }_6^{12} \mathrm{C}\right)\right] c^2 \\\\& =[12.041-12.0] \times 931.5=13.041 \mathrm{MeV}\end{aligned}$

The energy $Q=13.041 \mathrm{MeV}$ is released in the reaction. Out of this energy, 4.041 $\mathrm{MeV}$ is used to excite ${ }_6^{12} \mathrm{C}$ to its excited state ${ }_6^{12} \mathrm{C}^*$ . Thus, the kinetic energy available to the $\beta$ -particle $\left(K_\beta\right)$ and the antineutrino $\left(K_{\bar{\nu}}\right)$ is $K_\beta+K_{\bar{\nu}}=13.041-4.041=9 \mathrm{MeV}$ . In $\beta-$ decay, the kinetic energy of the $\bar{\nu}$ can vary from zero to a maximum value. Hence, the maximum kinetic energy of the $\beta$ -particle is $K_{\beta, \max }=9 \mathrm{MeV}$ (when $K_{\bar{\nu}}=0$ ).