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${ }_{10}^{23} Ne$ का नाभिक, $\beta^{-}$ उत्सर्जन के साथ क्षयित होता है। इस $\beta^{-}$ -क्षय के लिए समीकरण लिखिए और उत्सर्जित इलेक्ट्रॉनों की अधिकतम गतिज ऊर्जा ज्ञात कीजिए।
$m\left({ }_{10}^{23} Ne \right)=22.994466 \,u$ $u ; m\left({ }_{11}^{23} Na \right)=22.089770\, u$
Solution
In $\beta^{-}$ emission, the number of protons increases by $1,$ and one electron and an antineutrino are emitted from the parent nucleus. $\beta^{-}$ emission of the nucleus
$_{10}^{23} Ne _{10}^{23} Ne \rightarrow_{11}^{23} Na +e^{-}+\bar{v}+Q$
It is given that:
Atomic mass $m\left(_{10}^{23} Ne \right) o f=22.994466 u$
Atomic mass $m\left(_{11}^{23} N a\right) o f=22.989770 u$
Mass of an electron, $m_{e}=0.000548 u$
$Q$ – value of the given reaction is given as:
$Q=\left[m\left(^{23}_{10} N e\right)-\left[m\left(_{11}^{23} Na\right)+m_{e}\right]\right] c^{2}$
There are 10 electrons in $_{10} N e^{23}$ and 11 electrons in $_{11}^{23} N a$. Hence, the mass of the electron is cancelled in the $Q$ – value equation. $\therefore Q=[22.994466-22.9897770] c^{2}$
$=\left(0.004696 c^{2}\right) u$
But $1 u=931.5 MeV / c ^{2}$
$\therefore Q=0.004696 \times 931.5=4.374 MeV$
The daughter nucleus is too heavy as compared to $e^{-}$ and $v .$ Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., $4.374 MeV$.