Given the text from Hindi to be translated: ^218Po (t_(1/2)=183 seconds) decays via alpha-emission to ^214Pb (t_(1/2)=161 seconds), while ^214Pb is a beta-emitter. An experiment started with 1 mole of pure ^218Po. After how much time will the number of nuclei of ^214Pb be the highest? A 147.5 B 247.5 C 182 D 304
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
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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
$$
247.5
$$
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