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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

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Question : 17 of 39
Marks: +1, -0
During nuclear explosion, one of the products is 90^{90}Sr with half-life of 28.1 years. If 1 µg of 90^{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 ?
Solution:  
As radioactive disintegration follows first order kinetics. Hence Decay constant of 90 Sr, (λ)=0.693t1/2=0.69328.190\ \mathrm{Sr},\ (\lambda)=\frac{0.693}{t_{1/2}}=\frac{0.693}{28.1} =2.466×10−2 yr−1=2.466 \times 10^{-2}\ \mathrm{yr}^{-1}
To calculate the amount left after 10 years
Given, [R0]=1 μg,[R_{0}]=1\ \mu\mathrm{g},
t=10t=10 years,
k=2.466×10−2 yr−1, [R]=?k=2.466 \times 10^{-2}\ \mathrm{yr}^{-1},\ [R]=?
Using formula, λ=2.303tlog⁡[R0][R]\lambda=\frac{2.303}{t}\log\frac{[R_{0}]}{[R]}
or 2.466×10−2=2.30310log⁡1[R]2.466 \times 10^{-2}=\frac{2.303}{10}\log\frac{1}{[R]}
or, 2.466×10−2×102.303=−log⁡[R]\frac{2.466 \times 10^{-2} \times 10}{2.303}=-\log[R]
or, log⁡[R]=−0.1071\log[R]=-0.1071
or, [R]=[R]= Antilog (−0.1071)=0.7814 μg(-0.1071)=0.7814\ \mu\text{g}
To calculate the amount left after 60 years, t = 60 years, [R0][R_0] = 1 µg, [R] = ?
or, 2.466×10−2=2.30360log⁡1[R]2.466 \times 10^{-2}=\frac{2.303}{60}\log\frac{1}{[R]}
or, 2.466×10−2×602.303=−log⁡[R]\frac{2.466 \times10^{-2} \times 60}{2.303}=-\log[R]
or,log⁡[R]=−0.6425\log[R]=-0.6425
or, [R]=[R]= Antilog (−0.6425)=0.2278 μg(-0.6425)=0.2278\ \mu\text{g}
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