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.

Answer:

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.