JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2018
- Hard
- MCQ
At some instant, a radioactive sample \(S_1\) having an activity \(5\,\mu Ci\) has twice the number of nuclei as another sample \(S_2\) which has an activity of \(10\,\mu Ci.\) The halflives of \(S_1\) and \(S_2\) are
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Answer & Solution
Correct Answer
(B) \(5\) years and \(20\) years, respectively
Step-by-step Solution
Detailed explanation
Given \(: \mathrm{N}_{1}=2 \mathrm{N}_{2}\)
Activity of radioactive substance \(=\lambda \mathrm{N}\)
Half life period \(\mathrm{t}=\frac{\ln 2}{\lambda}\) or, \(\lambda=\frac{\ln 2}{\mathrm{t}}\)
\(\lambda_{1} \mathrm{N}_{1}=\frac{\ln 2}{\mathrm{t}_{1}} \times \mathrm{N}_{1}=5 \mu \mathrm{C}_{\mathrm{i}}\) ...... \((i)\)
\(\lambda_{2} \mathrm{N}_{2}=\frac{\ln 2}{\mathrm{t}_{2}} \times \mathrm{N}_{2}=10 \mu \mathrm{C}_{\mathrm{i}}\) ...... \((ii)\)
Dividing equation \((ii)\) by \((i)\)
\(\frac{t_{2}}{t_{1}} \times \frac{N_{1}}{N_{2}}=\frac{1}{2}\)
\(\frac{t_{2}}{t_{1}}=\frac{1}{4} \Rightarrow t_{1}=4 t_{2}\)
i.e., Half life of \(S_{1}\) is four times of sample \(S_{2}\) Hence \(5\) years and \(20\) years.
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