Question

During nuclear explosion one of the products is ${}^{90}Sr$ with half-life of 28.1 yr. If $1\mu g$ of ${}^{90}Sr$ was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 yr and 60 yr if it is not lost metabolically?

Answer 1

$t_{\frac{1}{2}}=28.1yr;k=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{28.1}y{r}^{-1}$
For 1st order reaction
$t=\frac{2.303}{k}log\frac{a}{(a-x)}$
Case I $\alpha =1\mu g;t=10yr;k=\frac{0.693}{28.1}y{r}^{-1}$
$10yr=\frac{2.303}{0.693}\times 28.1yrlog\frac{a}{(a-x)}log\frac{a}{(a-x)}=\frac{10yr\times 0.693}{2.303\times 28.1yr}=0.107\frac{a}{(a-x)}=Antilog\left(0.107\right)=1.279(a-x)=\frac{a}{1.279}=\frac{\left(1\mu g\right)}{1.279}=0.7819\mu g$
Amount left after 10 yr $=0.7819\mu g$

Case II: $\alpha =1\mu ,t=60yr;k=\frac{0.693}{28.1}y{r}^{-1}$
$60yr=\frac{2.303}{\frac{0.693}{28.1/y{r}^{-1}}}\times log\frac{a}{(a-x)}log\frac{a}{(a-x)}=\frac{\left(60yr\right)\times \frac{0.693}{28.1y{r}^{-1}}}{2.303}=0.642\frac{a}{(a-x)}=Antilog0.642=4.385(a-x)=\frac{a}{4.385}=\frac{\left(1\mu g\right)}{4.385}=0.2280\mu g$
Amount left after 60 yr $=0.2280\mu g$