Question

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

• A. $0.184\mu g$
• B. $0.025\mu g$
• C. $0.262\mu g$
• D. $0.228\mu g$

Answer 1

Answer: (D)

Half-life

$t_{1/2}=28.1yr$

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{28.1}y{r}^{-1}$

For first order reaction,

$t=\frac{2.303}{k}\log \frac{a}{(a-x)}$

$a=1\mu ,t=60yr,k=\frac{0.693}{28.1}y{r}^{-1}$

$60yr=\frac{2.303}{0.693/28.1}\times \log \frac{a}{(a-x)}$

$\log \frac{a}{(a-x)}=\frac{(60yr)\times 0.693/28.1y{r}^{-1})}{2.303}=0.642$

$\frac{a}{(a-x)}$ =antilog $0.642=4.385$

$(a-x)=\frac{a}{4.385}=\frac{(1\mu g)}{4.385}=0.2280\mu g$

Amount left after $60yr=0.2280\mu g$