During nuclear explosion, one of the products is


During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.


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

It is known that,

$t=\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$

$\Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{[R]}$

$\Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}}(-\log [R])$

$\Rightarrow \log [R]=-\frac{10 \times 0.693}{2.303 \times 28.1}$

$\Rightarrow[\mathrm{R}]=$ antilog $(-0.1071)$

$=$ antilog $(\overline{1} .8929)$

$=0.7814 \mu \mathrm{g}$

Therefore, 0.7814 μg of 90Sr will remain after 10 years.


$t=\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$

$\Rightarrow 60=\frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{[R]}$

$\Rightarrow \log [\mathrm{R}]=-\frac{60 \times 0.693}{2.303 \times 28.1}$


$=\operatorname{antilog}(\overline{1} .3575)$

$=0.2278 \mu \mathrm{g}$

Therefore, 0.2278 μg of 90Sr will remain after 60 years.



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