During the nuclear explosion, one of the products is 90Sr with half life of 6.93 years. If 1 $\mu$ g of 90Sr was absorbed in the bones of newly born baby in placed of Ca, how much time, in years, is required to reduce much time, in year, is required to reduce it by 90% if it not lost metabolically.
Answer (integer)
23
Solution
All nuclear decays follow first order kinetics
<br><br>t = ${1 \over k}\ln {{\left[ {{A_0}} \right]} \over {\left[ A \right]}}$
<br><br>= ${{\left( {{t_{1/2}}} \right)} \over {0.693}} \times 2.303{\log _{10}}10$
<br><br>= 10 × 2.303 × 1
<br><br>= 23.03 years
<br><br><b>Shortcut Method :</b>
<br><br>t<sub>90%</sub> = ${{10} \over 3} \times {t_{50\% }}$
<br><br>= ${{10} \over 3} \times 6.93$ = 23.1
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Chemical Kinetics are available — start with the harder ones if your accuracy is >70%.