Question

A neutron breaks into a proton, an electron (ß-particle) and antineutrino $\left.{ }_{0} n^{1} \longrightarrow { }_{1} H ^{1}+{ }_{-1} \beta^{0}+\bar{v}\right)$. The energy released in the process in $MeV$ is

• A. 731 MeV
• B. 7.31 MeV
• C. 0.731 MeV
• D. 1.17 MeV

Answer 1

Answer: (C)

From Einstein's mass energy relation
$\Delta E=(\Delta m) c^{2}$
Mass of proton $=1.6725 \times 10^{-27} \,kg$
Mass of electron $=0.0009 \times 10^{-27}\, kg$
Their sum $=1.6734 \times 10^{-27} \,kg$
Mass of neutron $=1.6747 \times 10^{-27} \,kg$
Their difference $\Delta m=0.0013 \times 10^{-27} \,kg$
$\therefore \Delta E=\left(0.0013 \times 10^{-27}\right)\left(3 \times 10^{8}\right)^{2}$
$=1.17 \times 10^{-13} \,J$
Also, $ 1.6 \times 10^{-19} J =1 eV$
$\therefore \Delta E=\frac{1.17 \times 10^{-13}}{1.6 \times 10^{-19}}$
$=0.731 \times 10^{6} eV$
$=0.731 \,MeV .$