Question

${}^{222}Ac$ has a half-life of $22$ years with respect to radioactive decay. The decay follows two parallel paths: ${}^{227}Ac{\to }^{227}Th$ and ${}^{227}Ac{\to }^{223}Fr$. If the percentage of the two daughter nuclides are $2.0$ and $98.0$, respectively, the decay constant (in $yea{r}^{-1}$) for ${}^{227}Ac{\to }^{227}Th$ path is closest to

• A. $6.3\times {10}^{-2}$
• B. $6.3\times {10}^{-3}$
• C. $6.3\times {10}^{-1}$
• D. $6.3\times {10}^{-4}$

Answer 1

The correct option is D $6.3\times {10}^{-4}$

The half-life of ${}^{222}Ac$ is 22 years
$t_{1/2}=22$ years

The overall decay constant $\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{22yr}=0.0315/yr$

The fractional yield of ${}^{227}Ac{\to }^{227}Th$ path
$=\frac{2.0}{2.0+98.0}=0.02$

The decay constant for ${}^{227}Ac{\to }^{227}Th$ path

${\lambda }^{\prime }=\text{fractional yield}\times \lambda$

${\lambda }^{\prime }=0.02\times 0.0315/yr$

${\lambda }^{\prime }=6.3\times {10}^{-4}/yr$