91Ó°ÊÓ

The ${\mathrm{β}}^{+}$decay of Oxygen-$19$.

${}_7^{13}\text{N}{→}_6^{13}\text{C}{+}_{+1}^0{\mathrm{β}}^{+}+\mathrm{v}$

Atomic mass of ${}_7^{13}\text{N},{\mathrm{m}}_{\text{N}}=13.005738\text{u}$.

Atomic mass of ${}_6^{13}\mathrm{C},{\mathrm{m}}_{\mathrm{C}}=13.003355\text{u}$.

During electron capture the amount of energy released is dependent on the mass difference between parent and the daughter nucleus.

Which is, $\mathbf{Q}\mathbf{=}\mathbf{(}{\mathbf{m}}_{\mathbf{\text{parent}}}\mathbf{−}{\mathbf{m}}_{\mathbf{\text{daughter}}}\mathbf{)}{\mathbf{c}}^{\mathbf{2}}$.

Where,${\mathbf{m}}_{\mathbf{\text{parent}}}$is atomic mass of parent nucleus and${\mathbf{m}}_{\mathbf{\text{daughter}}}$ is atomic mass daughter nucleus.

Oxygen $19$ in beta decay will lose an electron and thus gain a charge, so one of the neutrons will become proton.

So it will become $\mathrm{F}-19$.

${}_8^{19}\text{O}{→}_{−1}^0\mathrm{c}{+}_9^{19}\text{F}$

So the daughter particle is $\mathrm{F}−19$.

The$\mathrm{Q}$ factor is:

$\begin{array}{c}Q=(m_{\text{parent}}−m_{\text{daughter}})c^2\\ =(m_0−m_F)(931.5\text{MeV}/u)\\ =(19.003577u−18.998403u)(931.5\text{MeV}/u)\\ =4.82\text{MeV}\end{array}$

The daughter particle is $F-19$.

The kinetic energy released is $\left|4.82\text{MeV}\right|$.