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Nuclei — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsNuclei3 Marks
Question
The nucleus $^{23}_{10}\text{Ne}$ decays by $\beta$ emission. Write down the $\beta$-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
$\text{m}(^{23}_{10}\text{Ne})=22.994466\text{ u}.$
$\text{m}(^{23}_{10}\text{Na})=22.089770\text{ u}.$
✓

Answer

The $\beta$-decay of $^{23}_{10}\text{Ne}$ may be represented as:
$^{23}_{10}\text{Ne}\rightarrow\ ^{23}_{11}\text{Na }\ -\ ^{0}_{-1}\text{e }\ +\ \overline{\text{v}}\ +\ \text{Q}$
Ignoring the rest mass of antineutrino $(\overline{\text{v}})$ and electron, we get Mass defect,
$\Delta\text{m}=\text{m}(^{23}_{10}\text{Ne})-\text{m}(^{23}_{11}\text{Na})$
= 22.994466 - 22.989770
= 0.004696 u
$\therefore\ \text{Q}=0.004696\times931\text{ MeV}=4.372\text{ MeV}.$
This energy of 4.3792 MeV, is shared by e- and $\overline{\text{v}}$ pair because, $^{23}_{11}\text{Na}$ is very massive.
The maximum K.E. of e- = 4.372 MeV, when energy carried by $\overline{\text{v}}$ is zero.

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