Modern Physics Ques 203
- The isotopes ${ } _5^{12}$ B having a mass $12.014 $ $u$ undergoes $\beta$-decay to ${ } _6^{12} C$. ${ } _6^{12} C$ has an excited state of the nucleus $\left({ } _6^{12} C^{*}\right)$ at $4.041$ $MeV$ above its ground state. If ${ } _5^{12} B$ decays to ${ } _5^{12} C$, the maximum kinetic energy of the $\beta$-particle in units of $MeV$ is ( $1 u=931.5 $ $MeV / c^{2}$, where $c$ is the speed of light in vacuum)
(2016 Adv.)
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Answer:
Correct Answer: 203.$(9)$
Solution:
Formula:
- ${ } _5^{12} B \rightarrow{ } _6^{13} C+{ } _{-1}^{0} e+\bar{\nu}$
Mass of ${ } _6^{12} C=12.000 u$ (by definition of $1$ $ a.m.u.$)
$Q$-value of reaction,
$ \begin{aligned} Q= & \left(M _B-M _C\right) \times c^{2}=(12.014-12.000) \times 931.5 \\ = & 13.041 MeV \\ & 4.041 MeV \text { of energy is taken by }{ } _{12}^{65} C^{*} \\ \Rightarrow & \text { Maximum KE of } \beta \text {-particle is }(13.041-4.041)=9.0 MeV \end{aligned} $
BETA
