Question

How to calculate the maximum Beta energy $(0.634MeV)$ of fluorine-18 ?

Answer 1

Hint: We will first find out how a fluorine-18 atom disintegrates under normal conditions. Since, fluorine-18 is a lighter isotope of fluorine-19, it is unstable at normal pressure and temperature and has a half-life of about $109.7\min $. This results in the disintegration of the fluorine-18 atom into an oxygen-18 atom and a positron releasing Beta energy.

Complete answer:
The equation for decay of fluorine-18 by positron emission can be given as:
$_{9}^{18}F\to _{8}^{18}O+_{+1}^{0}\beta +energy$
Now, the absolute mass of $_{9}^{18}F$ and $_{8}^{18}O$ atoms are:
$\begin{align}
  & _{9}^{18}F:18.0009380 \\
 & _{8}^{18}O:17.9991610 \\
\end{align}$
Thus, the difference in mass can be calculated to be:
$\begin{align}
  & \Rightarrow \vartriangle m=18.0009380-17.9991610 \\
 & \Rightarrow \vartriangle m=0.001777amu \\
\end{align}$
Therefore, net energy released will be equal to:
$\begin{align}
  & \Rightarrow E=931.5\times 0.001777MeV \\
 & \Rightarrow E=1.6552755MeV \\
\end{align}$
Here, $931.5MeV$ is the nuclear binding energy of 1amu.
When a positron is ejected from the parent nucleus, the daughter $(Z-1)$ must also shed an orbital electron to balance its charge. Therefore, the mass of two electrons is ejected, not one. An energy equivalent of two electrons (one positron and one electron) is released.
We know,
$\Rightarrow {{m}_{e}}C=0.511MeV$
Where, ${{m}_{e}}$ is the mass of the electron and $C$ is the speed of light.
Therefore, total energy released by electron and positron emission is equal to:
$\begin{align}
  & =2\times 0.511MeV \\
 & =1.022MeV \\
\end{align}$
Thus, the maximum Beta energy can be calculated as:
$\begin{align}
  & =1.6552755-1.022MeV \\
 & =0.6332755Me \\
 & \approx 0.633MeV \\
\end{align}$
Hence, the maximum Beta energy is calculated to be $0.633MeV$ .

Note:
We should know the decay reaction of some of the common radioactive elements or isotopes. In this question the key was to understand the fact that the daughter nucleus will also emit an electron to balance the charge in the equation. If we missed out on this fact then our answer would have been different and wrong. Also, we have to be very careful while performing these lengthy calculations.