Question

A positron is emitted from ${}_{11}^{23}\text{Na}$ . The ratio of the mass number and the 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 balanced reaction is:

${}_{11}^{23}\text{Na}{⟶}_{10}^{23}\text{Ne}{+}_{+1}^{0}e$

Resulting nuclide $={}_{23}^{10}Ne$

The ratio of mass number and the atomic number $=\frac{23}{10}$

Hence, option C is correct.