JEE Advance - Physics (2016 - Paper 1 Offline - No. 16)
The isotope $$_5^{12}B$$ having a mass 12.014 u undergoes $$\beta $$-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 $$\beta$$-particle in units of MeV is (1u = 931.5 MeV/c2, where c is the speed of light in vacuum).
Answer
9
Explanation
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 $K_{\bar{\nu}}=0$ ).
$$ { }_5^{12} \mathrm{~B} \rightarrow{ }_6^{12} \mathrm{C}+e^{-1}+\bar{\nu} $$
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 $K_{\bar{\nu}}=0$ ).
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