Question

A hypothetical radioactive nucleus decays according to the following series
${}_{72}{A}^{180}\underrightarrow{\alpha }{A}_1\underrightarrow{{\beta }^{-}}{A}_2\underrightarrow{\alpha }{A}_3\underrightarrow{{\rm Y}}{A}_4$
If the mass number and atomic number of A are respectively $180$ and $72$. then the atomic number and mass number of A will respectively be:

• A. $69,171$
• B. $70,172$
• C. $68,172$
• D. $69,172$

Answer 1

The correct option is D $69,172$

Concept : $\alpha$ decay decreases the mass number of the nucleus be $4$ and decreases atomic number of nucleus by $2$.

$\beta$ decay increases the atomic number by $1$ whereas the mass number remains the same.

In $\gamma$ decay, the mass number and the atomic number remains the same.

$⟹$ The daughter nuclei are as follows :

${}_{70}^{176}{A}_1$, ${}_{71}^{176}{A}_2$, ${}_{69}^{172}{A}_3$, and ${}_{69}^{172}{A}_4$

Thus, the mass number and atomic number of $A_4$ will be $172$ and $69$ respectively.