JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2021
- Hard
- Numerical
A radioactive material decays by simultaneous emissions of two particles with half lives of \(1400\, years\) and \(700\, years\) respectively. What will be the time after which one third of the material remains? (Take In \(3=1.1\) ) (In \(years\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(740\)
Step-by-step Solution
Detailed explanation
\(\text { Given } \lambda_{1}=\frac{\ell {n} 2}{700} / \text { year }, \lambda_{2}=\frac{\ell {n} 2}{1400} / \text { year }\)
\(\therefore \lambda_{\text {net }}=\lambda_{1}+\lambda_{2}=\ell n 2\left[\frac{1}{700}+\frac{1}{1400}\right]=\frac{3 \ell {n} 2}{1400} / \text { year }\)
Now, Let initial no. of radioactive nuclei be No
\(\therefore \frac{{N}_{0}}{3}={N}_{0} {e}^{-\lambda_{{mt}} {t}}\)
\(\Rightarrow \ell {n} \frac{1}{3}=-\lambda_{\text {net }} {t}\)
\(\Rightarrow 1.1 \frac{3 \times 0.693}{1400} {t} \Rightarrow {t} \approx 740 \text { year }\)
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