A positron is emitted from ${ }_{11}^{23}\mathrm{Na}$. The ratio of the atomic mass and atomic number of the resulting nuclide is

A positron is emitted from ${ }_{11}^{23}\mathrm{Na}$. The ratio of the atomic mass and atomic number of the resulting nuclide is

${ }_1^1\mathrm{H}{\to }_0^1\mathrm{n}{+}_{+1}^0\mathrm{e}$

A proton is transformed to a neutron during positron emission, resulting in a one-unit fall in atomic number but no change in atomic mass.

${ }_{11}^{23}\mathrm{Na}{⟶}_{+1}^0\mathrm{P}{+}_{10}^{23}\mathrm{E}$

$\therefore \frac{\mathrm{n}}{\mathrm{p}}\text{ ratio }=\frac{23}{10}$