Question

Given the text from Hindi to be translated:


$$^{218} \mathrm{Po}$$ ($t_{1/2}=183$ seconds) decays via $\alpha$-emission to $$^{214}\mathrm{Pb}$$ ($t_{1/2}=161$ seconds), while $$^{214}\mathrm{Pb}$$ is a $\beta$-emitter. An experiment started with 1 mole of pure $$^{218}\mathrm{Po}$$. After how much time will the number of nuclei of $$^{214}\mathrm{Pb}$$ be the highest?

  • A 147.5
  • B 247.5
  • C 182
  • D 304

✨ Free AI Tools for You

Powered by Chatterbot AI

Answer

To solve the problem, we need to determine the time at which the number of $_{82}^{214}\mathrm{~Pb}$ nuclei will be at its maximum. Given data includes:

  • Half-life of $^{218}\mathrm{Po}$ ($t_{1/2}$) = 183 seconds
  • Half-life of $^{214}\mathrm{~Pb}$ ($t_{1/2}$) = 161 seconds

We start by understanding the decay constants for each:

$$ \lambda_{\alpha} = \frac{\ln(2)}{183 \text{ seconds}} $$

$$ \lambda_{\beta} = \frac{\ln(2)}{161 \text{ seconds}} $$

When we have a situation where an isotope $A$ decays to $B$, and $B$ in turn decays to $C$, the time at which the intermediate product $B$ reaches its maximum concentration is given by the formula:

$$ t_{\text{max}} = \frac{\ln\left(\frac{\lambda_{\beta}}{\lambda_{\alpha}}\right)}{\lambda_{\alpha} - \lambda_{\beta}} $$

Inserting the values we have calculated:

$$ t_{\text{max}} = \frac{\ln\left(\frac{\frac{\ln(2)}{161}}{\frac{\ln(2)}{183}}\right)}{\frac{\ln(2)}{183} - \frac{\ln(2)}{161}} $$

Simplifying the expression inside the logarithm:

$$ t_{\text{max}} = \frac{\ln\left(\frac{183}{161}\right)}{\frac{\ln(2)(183 - 161)}{183 \cdot 161}} $$

$$ = \frac{\ln\left(\frac{183}{161}\right)}{\ln(2) \cdot \left(\frac{22}{183 \cdot 161}\right)} $$

$$ = \frac{\ln\left(1.136\right)}{\ln(2)} \cdot \frac{183 \cdot 161}{22} $$

$$ t_{\text{max}} \approx \frac{0.1278}{0.693} \cdot \frac{29463}{22} $$

$$ \approx 0.1843 \cdot 1339.23 $$

$$ \approx 247.5 \text{ seconds} $$

Thus, the time at which the number of $_{82}^{214}\mathrm{~Pb}$ nuclei reaches its maximum is (\boldsymbol{247.5}) seconds.

Therefore, the correct answer is:

B
$$ 247.5 $$


Was this helpful?

India's 1st AI Doubt Solver for CBSE, JEE, and NEET

Ask a Question for Free

and then it's just ₹212 a month

Learn how Chatterbot AI can help you succeed

Cover Image of the article Computer Science Class 11 CBSE - The Ultimate Guide with Notes, Solutions and AI

Computer Science Class 11 CBSE - The Ultimate Guide with Notes, Solutions and AI

This ultimate guide for CBSE Computer Science class 11 has detailed notes, NCERT solutions, cheat sheets, and our free AI-powered doubt-solving assistant, Chatterbot AI.

Cover Image of the article JEE Advanced 2024 Exam Date Announced: Complete Guide to Eligibility, Syllabus, and Preparation Tips

JEE Advanced 2024 Exam Date Announced: Complete Guide to Eligibility, Syllabus, and Preparation Tips

JEE Advanced 2024 on May 26! Get exam schedule, syllabus, prep tips & more in this guide. Ace India's top engineering test with topper strategies.

Cover Image of the article How to Crack NEET: The Ultimate Blueprint to Outsmart the Exam and Unlock Your Medical Dreams

How to Crack NEET: The Ultimate Blueprint to Outsmart the Exam and Unlock Your Medical Dreams

Ace NEET with expert strategies: Discover effective exam strategies, time management, core concepts mastery, problem-solving techniques, revision tips, and AI-assisted doubt clearing with Chatterbot AI.

Cover Image of the article How to Crack IIT: Smart Self-Study Strategies and AI Tools for Success

How to Crack IIT: Smart Self-Study Strategies and AI Tools for Success

Ace IIT JEE in 6 months without coaching. Discover expert self-study strategies for Physics, Chemistry, and Math. Master time management, mock tests, and leverage AI tools like Chatterbot AI for personalized doubt-solving.