Question

An actinium ($\text{Ac}$) sample ($\text{A}:227;\text{Z}:89$) undergoes a series of $\alpha -\beta -\alpha -\gamma$ emissions to attain stability. What is the final daughter nucleus formed in the reaction?

• A. $22387\mathrm{Fr}$
• B. $23290\mathrm{Th}$
• C. $21986\mathrm{Rn}$
• D. $21985\mathrm{At}$

Answer 1

The correct option is C $21986\mathrm{Rn}$

An alpha emission decreases the atomic number ($\text{Z}$) by two and the mass number ($\text{A}$) by 4. A beta emission does not change the mass number, but increases the atomic number by 1. A gamma emission results in no change of the parent nucleus. Hence, the radioactive decay of the actinium nucleus is as follows:

$22789\mathrm{Ac}\stackrel{\alpha }{\to }22387\mathrm{Fr}\left(\mathrm{Francium}\right)$

$22387\mathrm{Fr}\stackrel{\beta }{\to }22388\mathrm{Ra}\left(\mathrm{Radium}\right)$

$22388\mathrm{Ra}\stackrel{\alpha }{\to }21986\mathrm{Rn}*\left(\mathrm{Radon}\right)$

$21986\mathrm{Rn}*\stackrel{\gamma }{\to }21986\mathrm{Rn}$

* indicates an energy state of the parent nucleus higher than that of the daughter nucleus.
Hence, the final product nucleus is $21986\mathrm{Rn}$.