Question

The nuclide ${}_{21}^{50}\text{Sc}$ (mass 49.9516 amu) decays to form ${}_{22}^{50}\text{Ti}$ (mass 49.94479 amu). If emission gives $\beta$ - particles and neutrino and the kinetic energy of $\beta$-particles is 0.80 MeV, kinetic energy of neutrino in MeV is :
($1$ amu $=931.478$MeV).

(Answer should be greatest integer function of kinetic energy)

Answer 1

The mass defect is $49.9516amu-49.94479amu=6.81\times {10}^{-3}amu$. The energy that corresponds to mass defect is $6.81\times {10}^{-3}amu\times 931.478MeV=6.34MeV$. This corresponds to the sum of the K.E of beta particle and the K.E of neutrino.
Hence, the K.E of neutrino is $6.34MeV-0.80MeV=5.54MeV$