Question

${}^{227}Ac$ has a half-life of 21.8 year with respect to radioactive decay. The decay follows two parallel paths, one leading the ${}^{227}Th$ and the other leading to ${}^{223}Fr$. The percentage yields of these two daughters nuclides are 1.2% and 98.8% respectively. The rate constant in $y{r}^{-1}$, for each of the separate paths is :

• A. $3.14\times {10}^{-2}y{r}^{-1}$
• B. $2.46\times {10}^{-2}y{r}^{-1}$
• C. $4.44\times {10}^{-2}y{r}^{-1}$
• D. None of these

Answer 1

The correct option is A $3.14\times {10}^{-2}y{r}^{-1}$

${\lambda }_{Ac}=\frac{0.693}{21.8}=3.18\times {10}^{-2}yea{r}^{-1}$
For the decay involving two parallel paths.
We have, ${\lambda }_{Ac}={\lambda }_{Th}+{\lambda }_{Fr}$
$Thus,fractionalyieldofTh=\frac{{\lambda }_{Th}}{{\lambda }_{Ac}}$
or ${\lambda }_{Th}=3.18\times {10}^{-2}\times \frac{1.2}{100}=3.82\times {10}^{-4}y{r}^{-1}$
$Also,fractionalyieldofFr=\frac{{\lambda }_{Fr}}{{\lambda }_{Ac}}$
$\therefore {\lambda }_{Fr}=3.18\times {10}^{-2}\times \frac{98.8}{100}=3.14\times {10}^{-2}y{r}^{-1}$