• # question_answer The nucleus $_{10}^{23}$ Ne decays by ${{\beta }^{-}}-emission.$ Write down the $\beta -decay$ equation and determine the maximum kinetic energy of the electrons emitted. Given that, $m(_{10}^{23}Ne)=22.994466\,u,$ $m(_{11}^{23}Na)=22.989770\,u.$ Or With the help of an example, explain, how the neutron to proton ratio changes during $\alpha \text{-decay}$ of a nucleus.

The ${{\beta }^{-}}-decay$ equation of  $_{10}^{23}Ne$ is $_{10}^{23}Ne\xrightarrow{-\beta }_{11}^{23}Na+-{{1}^{{{e}^{0}}}}+\overline{v}+Q$ Mass defect, $\Delta \,m=m(_{10}^{23}Ne)-m(_{11}^{23}Na)$ $=22.994466-22.989770=0.004696\,u$ $Q=\Delta \,m\times 931=4.372\,MeV$ The maximum energy released in ${{\beta }^{-}}-decay$ is equal to the Q-value.             ${{E}_{e}}=Q=4.37\,MeV.$ Since, $_{10}^{23}Na$ nucleus is much heavier than electron and anti-neutrino, practically whole of the energy released is carried by electron-anti-neutrino pair. Because anti-neutrino gets zero energy, the electron will carry the maximum energy. So, the maximum kinetic energy of the electron is 4.37 MeV. Or Consider the $\alpha -decay$ $_{92}^{238}U\xrightarrow{\alpha -decay}_{90}^{234}Th+_{2}^{4}He$ Neutron-proton ratio before $\alpha -decay$             $=\frac{238-92}{92}=\frac{146}{92}=1.58$ Neutron-proton ratio after $\alpha -decay$             $=\frac{234-90}{90}=\frac{144}{90}=1.6$ Thus, the ratio increases.