# The nucleus

Question:

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:

$m\left({ }_{10}^{23} \mathrm{Ne}\right)=22.994466 \mathrm{u}$

$m\left({ }_{11}^{23} \mathrm{Na}\right)=22.989770 \mathrm{u} .$

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 ${ }_{10}^{23} \mathrm{Ne}$ is given as:

${ }_{10}^{23} \mathrm{Ne} \rightarrow{ }_{11}^{23} \mathrm{Na}+e^{-}+\bar{v}+Q$

It is given that:

Atomic mass of $m\left(\begin{array}{l}23 \\ 10\end{array} \mathrm{Ne}\right)=22.994466 \mathrm{u}$

Atomic mass of $m\left({ }_{11}^{23} \mathrm{Na}\right)=22.989770 \mathrm{u}$

Mass of an electron, m= 0.000548 u

Q-value of the given reaction is given as:

$Q=\left[m\left({ }_{10}^{23} \mathrm{Ne}\right)-\left[m\left({ }_{11}^{23} \mathrm{Na}\right)+m_{e}\right]\right] c^{2}$

There are 10 electrons in ${ }_{10}^{23} \mathrm{Ne}$ and 11 electrons in ${ }_{11}^{23} \mathrm{Na}$. Hence, the mass of the electron is cancelled in the $Q$-value equation.

$\therefore Q=[22.994466-22.989770] c^{2}$

$=\left(0.004696 c^{2}\right) \mathrm{u}$

But $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}$

$\therefore Q=0.004696 \times 931.5=4.374 \mathrm{MeV}$

The daughter nucleus is too heavy as compared to $e^{-}$and $\bar{v}$. 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., $4.374 \mathrm{MeV}$.