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 (${}_5^{12}{C}^{\ast }$ ) at $4.041MeV$ above its ground state. If ${}_5^{12}B$ decays to ${}_6^{12}{C}^{\ast }$, the maximum kinetic energy of the $\beta$-particle in units of $MeV$ is
($1u=931.5MeV/c^2$, where c is the speed of light in vacuum).

Answer 1

${}_5^{12}B{\to }_6^{12}C+{\beta }^{-}+\overline{\nu }$
For any radioactive particle decay, the energy released in the reaction will be given by $Q=(m_B-m_C)c^2$
Now $m_C=12u$ and $m_B=12.041u$
Out of $Q$, 4.041 eV energy is used to excite ${}_6^{12}\text{C}$ to its excited state ${}_6^{12}{C}^{\ast }$, then K.E. available to $\beta$ particle and antineutrino is $=(m_B-m_C)c^2-\Delta m{c}^2=0.014\times 931.5-4.041=9MeV$
The $\beta$ particle will have maximum K.E when antineutrino will have minimum K.E i.e 0.
Therefore maximum K.E of the $\beta$ particle is $9MeV$