ExamBro
ExamBro
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.
Verified Solution

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

\(\therefore \frac{{N}_{0}}{3}={N}_{0} {e}^{-\lambda_{{mt}} {t}}\)

\(\Rightarrow \ell {n} \frac{1}{3}=-\lambda_{\text {net }} {t}\)

\(\Rightarrow 1.1 \frac{3 \times 0.693}{1400} {t} \Rightarrow {t} \approx 740 \text { year }\)

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app