Question

The isotope ${}_5^{12}B$ having a mass $12.014u$ undergoes $\beta$-decay to ${}_6^{12}C.{}_6^{12}C$ has an excited state of the nucleus $({}_6^{12}{C}^{\ast }$ ) 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

Answer: 9

${}_5^{12}B⟶{}_6^{12}C+{}_{-1}^0\beta$
Mass loss $=0.014u$
Energy produced $=13.041MeV\left(E=m{c}^2\right)$
$\therefore$ Maximum Kinetic energy of $\beta$-particle
$=13.041-4.041=9MeV$