Question

$A{c}^{227}$ has a half-life of 22 years. The decay follows two parallel paths, one leading the $T{h}^{227}$ and the other leading of $F{r}^{223}$, the percentage yields of these two daughters nuclides are 2% and 98% respectively. What is the rate constant in $y{r}^1$, for each of the separate paths?

• A. $6.30\times {10}^{-4}y{r}^{-1},3.087\times {10}^{-2}y{r}^{-1}$
• B. $6.30\times {10}^{-3}y{r}^{-1},3.087\times {10}^{-2}y{r}^{-1}$
• C. $63.0\times {10}^{-3}y{r}^{-1},30.87\times {10}^{-3}y{r}^{-1}$
• D. None of these

Answer 1

Answer: (A)

$6.30\times {10}^{-4}y{r}^{-1},3.087\times {10}^{-2}y{r}^{-1}$

This is an example of parallel reaction.

$T{h}^{227}\leftarrow A{c}^{227}\to F{r}^{223}$

$A{c}^{227}\to F{r}^{223}$ (98%) $K_2$

$A{c}^{227}\to T{h}^{227}$ (2%) $K_1$

$K_1$ + $K_2$ $=K$

This radio active decays follow 1st order reaction mechanism.

$K=\frac{0.693}{22}$ $Y{r}^{-1}$

Yield Ratio = $T{h}^{227}$ : $F{r}^{223}$ $=1:48$

i.e. $K_1$ : $K_2$ $=1:48$

$K_1$= $K\ast \frac{1}{50}=\frac{\left(\frac{0.693}{22}\right)}{50}=6.3\ast$ ${10}^{-4}$ $Y{r}^{-1}$

$K_2$ $=K\ast \frac{48}{50}$ $\frac{0.693}{22}$ $\ast \frac{48}{50}$ $=3.087\ast$ ${10}^{-2}$ $Y{r}^{-1}$