During the nuclear explosion, one of the products is ^{90}Sr  with a half-life of 6.93 years. If \mathrm{1\ \mu g\: of\: ^{90}Sr} was absorbed in the bones of a newly born baby in place of Ca, how much time, in the year, is required to reduce it by 90% if it is not lost metabolically-----------
Option: 1 23.03
Option: 2 46.06
Option: 369.06
Option: 422.66
 

Answers (1)

We know this formula:- 

\\\mathrm{t\: =\: \frac{2.303}{k}log\frac{[R_{o}]}{[R]}}

For 90% completion, the required time will be -

\\\\\mathrm{t\: =\: \frac{2.303}{k}log\frac{[R_{o}]}{[0.1R_{o}]}}

\\\mathrm{t\: =\: \frac{2.303}{k}log10}

\\\mathrm{t\: =\: \frac{2.303}{k}}

 

Half-life is given as 6.93 years. So, 
\\\mathrm{k\: =\: \frac{0.693}{t_{\frac{1}{2}}}}

\\\mathrm{k\: =\: \frac{0.693}{6.93}\: =\: 0.1}

Thus, t is given as fro 90% completion:

\\\mathrm{t\: =\: \frac{2.303}{k}}

\\\mathrm{t\: =\: \frac{2.303}{0.1}}

Thus, t = 23.03 years

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