Question

During nuclear explosion, one of the products is $^{90}Sr$ with half-life of 28.1 years. If 1µg of $^{90}Sr$ 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.

Answer 1

$k = \frac {0.693}{t_{1/2}} = \frac {0.693}{28.1} \ Y^{-1}$
Here,
It is known that,

$t = \frac {2.303}{k} log \ \frac {[R]_0}{[R]}$

$10 = \frac {2.303}{0.693/28.1} log \ \frac {1}{R}$

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

$log \ [R] =\frac { 10\times 0.693}{2.303\times28.1}$

$[R] = antilog \ (-0.1071)$

$[R] = antilog (\bar1.8929)$

$[R] = 0.7814 \ µg$

Therefore, 0.7814 µg of $^{90}Sr$ will remain after 10 years.

$t = \frac {2.303} {k}log \ \frac {[R]_0}{[R]}$

$60 = \frac {2.303}{0.693/28.1} log \ \frac {1}{R}$

$60 = \frac {2.303}{0.693/28.1} (-log \ [R])$

$log \ [R] =\frac { 60\times 0.693}{2.303\times28.1}$

$[R] = antilog \ (-0.6425)$

$[R] = antilog \ (\bar1.3575)$

$[R] = 0.2278\ µg$

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