Question

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

• A. $\frac{22}{10}$
• B. $\frac{22}{11}$
• C. $\frac{23}{10}$
• D. $\frac{23}{12}$

Answer 1

The correct option is C $\frac{23}{10}$

The required nuclear reaction is
${}^{11}N{a}^{23}{\to }_{+1}{e}^0{+}_{10}N{a}^{23}$