Question 115: Atoms & Nuclei

Question

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).

Accepted Answer

As the ${ }_5^{12} B$ is going under $\beta$-decay to form
$$_6^{12}C$$. We can write the balanced equation as follows
$$
{ }_5^{12} \mathrm{~B} ightarrow{ }_6^{12} \mathrm{C}+e^{-1}+\bar{
u}
$$
The $Q$-value of this reaction is given by
$$
\begin{aligned}
Q & =\left[m\left({ }_5^{12} \mathrm{~B}ight)-m\left({ }_6^{12} \mathrm{C}ight)ight] c^2 \\
& =[12.041-12.0] 	imes 931.5=13.041 \mathrm{MeV}
\end{aligned}
$$
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 $\left(K_\betaight)$ and the antineutrino $\left(K_{\bar{
u}}ight)$ is $K_\beta+K_{\bar{
u}}=13.041-4.041=9 \mathrm{MeV}$. In $\beta-$ decay, the kinetic energy of the $\bar{
u}$ 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 $K_{\bar{
u}}=0$ ).

Atoms & Nuclei

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