![\[{}_{10}^{23}Ne\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e29ca3a0219f614cc29b3bae3e497cfb_l3.png "Rendered by QuickLaTeX.com")
![\[m({}_{10}^{23}Ne)=22.994466u\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d36ae4b63fd6ea291a0e94c2bc70425f_l3.png "Rendered by QuickLaTeX.com")
![\[m({}_{11}^{23}Na)=22.989770u\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9fdb5fa785250587b1dcfdd1ebc7f37a_l3.png "Rendered by QuickLaTeX.com")
Ans:
We know that the number of protons gets increased by 1, and one electron and an antineutrino is emitted from the parent nucleus during β− emission.
The reaction equation is given by
![\[{}_{10}^{23}Ne\to {}_{11}^{23}Na+{{e}^{-}}+\bar{v}+Q\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e8051586cd484c3ed674ab3732271766_l3.png "Rendered by QuickLaTeX.com")
We are given that –
Mass of an electron, me = 0.000548 u
Therefore, Q-value of the given reaction is given as:
![\[Q=[m({}_{10}^{23}Ne)-m({}_{11}^{23}Na)+{{m}_{e}}]{{c}^{2}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-46840ff220f02680e30bdc16c0dc45b1_l3.png "Rendered by QuickLaTeX.com")
There are 10 electrons and 11 electrons in given Ne and Na isotopes respectively. Hence, the mass of the electron gets cancelled out in the Q-value equation. We get –
Q = [ 22.994466 – 22.989770 ] c2
= (0.004696 c2) u
We know that 1 u = 931.5 MeV/c2. So,
Q = 0.004696 x 931.5 = 4.374 MeV
In comparison to e– and v_, the daughter nucleus is too heavy. As a result, it carries very little energy. The antineutrino’s kinetic energy is nearly zero. As a result, the released electrons’ maximal kinetic energy is nearly equal to the Q-value, i.e. 4.374 MeV.