The isotope ${}_{5}^{12}B$ having mass 12.014 u undergoes $\beta $-decay to ${}_{6}^{12}C,$ ${}_{6}^{12}C$ has an excited state of the nucleus (${}_{5}^{12}C*$ )at 4.041 MeV above its ground state. If ${}_{5}^{12}B$decays to ${}_{5}^{12}C*$, the maximum possible kinetic energy $\beta $-particle in units of MeV is (1u=$931.5MeV/{{c}^{2}}$, where c is the speed of light in vacuum)

Hint: In the above radioactive decay we can see that ${}_{5}^{12}B$ undergoes beta minus decay. It Is believed that the neutron gets converted to proton and hence the atomic number increases but the mass remains the same. It is given that the ${}_{5}^{12}C*$ has an excited state of the nucleus at 4.041 MeV. Hence some of the energy released during the beta decay will exist in the form of the excited state of the product formed during the decay. Therefore we can subtract the energy of excitation of ${}_{5}^{12}C*$ from the total energy of the $\beta $decay and obtain the maximum possible kinetic energy $\beta $-particle.

Complete answer:
The above beta minus decay can be represented by the equation,
${}_{5}^{12}B\to {}_{6}^{12}\text{C + 4}\text{.041MeV +}\beta \text{(particle)}$
It is given in the question that 1u=$931.5MeV/{{c}^{2}}$. Hence if we multiply this u to the change in mass of the decay we will obtain the net energy released during the decay is given by,
$\begin{align}
  & E=\Delta m\times 931.5MeV/{{c}^{2}} \\
 & \Rightarrow E=\left( m{}_{5}^{12}B-m{}_{6}^{12}C \right){{c}^{2}}\times 931.5MeV/{{c}^{2}} \\
 & \Rightarrow E=\left( 12.014-12 \right)\times 931.5MeV \\
 & \Rightarrow E=\left( 0.014 \right)\times 931.5MeV \\
 & \Rightarrow E=13.041MeV \\
\end{align}$
It is given in the question that ${}_{6}^{12}C$ has an excited state at 4.041 MeV. Hence subtracting the above energy from the energy of excitation we will get the maximum kinetic energy of the beta particle.
$\begin{align}
  & E=13.041-4.041 \\
 & \Rightarrow E=9MeV \\
\end{align}$
Hence the maximum possible kinetic energy $\beta $-particle in units of MeV is 9.

Note:
It is to be noted that the symbol * next the product i.e. ${}_{5}^{12}C*$ represents that the product is in its excitation form. It is to be noted that the 1u=$931.5MeV/{{c}^{2}}$ hence we could multiply this figure to the mass defect and hence we could obtain the required energy. This is because the figure was expressed in terms of energy and square the speed of light i.e. c.