JEE MAIN - Chemistry (2020 - 7th January Morning Slot - No. 4)
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
23to23.03
Explanation
All nuclear decays follow first order kinetics
t = $${1 \over k}\ln {{\left[ {{A_0}} \right]} \over {\left[ A \right]}}$$
= $${{\left( {{t_{1/2}}} \right)} \over {0.693}} \times 2.303{\log _{10}}10$$
= 10 × 2.303 × 1
= 23.03 years
Shortcut Method :
t90% = $${{10} \over 3} \times {t_{50\% }}$$
= $${{10} \over 3} \times 6.93$$ = 23.1
t = $${1 \over k}\ln {{\left[ {{A_0}} \right]} \over {\left[ A \right]}}$$
= $${{\left( {{t_{1/2}}} \right)} \over {0.693}} \times 2.303{\log _{10}}10$$
= 10 × 2.303 × 1
= 23.03 years
Shortcut Method :
t90% = $${{10} \over 3} \times {t_{50\% }}$$
= $${{10} \over 3} \times 6.93$$ = 23.1
Comments (0)
