Ncert Exercises & Solutions

4.17 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.

$K = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{28.1} y^{- 1}$
$H e r e,$
$I t i s k n o w n t h a t,$
$t = \frac{2.303}{K} \log \frac{{[R]}_{0}}{[R]}$
$\Rightarrow 10 = \frac{2.303}{0.693} \log \frac{1}{[R]}$
$\Rightarrow \frac{2.303}{0.693} (- \log [R])$
$28.1$
$\Rightarrow \log [R] = - \frac{10 \times 0.693}{2.303 \times 28.1}$
$\Rightarrow [R] = a n t i \log) - 0.1071)$
$= a n t i \log (1.8929)$
$= 0 . 7814_{μ g}$

Therefore, 0.7814 µg of ${}_{90}S r$ will remain after 10 years.
Again,

$t = \frac{2.303}{K} \log \frac{{[R]}_{0}}{[R]}$
$\Rightarrow 60 = \frac{2.303}{0.693} \log \frac{1}{[R]}$
$\log [R] = - \frac{60 \times 0.693}{2.303 \times 28.1}$
$\Rightarrow [R] = a n t i \log (- 0.6425)$
$= a n t i \log (1.3575)$
$= 0.2278 μ g$

Therefore, 0.2278 µg of $^{90} S r$ will remain after 60 years.

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