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)

$0.228\mu g$

Half-life
$t_{1/2}=28.1yr$
$k=\frac{0.693}{t_{1/2}}=\frac{0.692}{28.1}y{r}^{-1}$
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}$60\ yr = \dfrac {2.303}{0.693/28.1} \times \log \dfrac {a}{(a - x)}\log \dfrac {a}{(a - x)} = \dfrac {(60\ yr)\times 0.693/ 28.1 yr^{-1})}{2.303} = 0.642\dfrac {a}{(a - x)} = antilog\ 0.642 = 4.385(a - x) = \dfrac {a}{4.385} = \dfrac {(1\mu g}{4.385} = 0.2280\mu g$Amountleftafter$60\ yr = 0.2280\mu g$.