`._(84)^(218)Po(t_(1//2)=183` sec) decay to `._(82)Pb(t_(1//2)=161` sec) by `alpha`-emission, while `Pb^(214)` is a `beta`-emitter. In an experiment starting with 1 mole of pure `Po^(218)`, how many time would be required for the number of nuclei of `._(82)^(214)Pb` to reach maximum ?

Options

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

Answer

To solve the problem, we need to determine the time required for the number of nuclei of \( _{82}^{214}Pb \) to reach its maximum after starting with 1 mole of pure \( _{84}^{218}Po \). The decay process involves two steps: the alpha decay of polonium and the beta decay of lead.

### Step-by-Step Solution:

1. **Determine the decay constants (\( \lambda \)) for both isotopes:**
- For \( _{84}^{218}Po \):
\[ \lambda_1 = \frac{0.693}{t_{1/2}} = \frac{0.693}{183 \, \text{sec}} \approx 3.786 \times 10^{-3} \, \text{s}^{-1} \]
- For \( _{82}^{214}Pb \):
\[ \lambda_2 = \frac{0.693}{t_{1/2}} = \frac{0.693}{161 \, \text{sec}} \approx 4.304 \times 10^{-3} \, \text{s}^{-1} \]

2. **Use the formula for the time to reach maximum concentration of \( _{82}^{214}Pb \):**
\[ t_{max} = \frac{2.303}{\lambda_1 - \lambda_2} \log\left(\frac{\lambda_1}{\lambda_2}\right) \]

3. **Substitute the values of \( \lambda_1 \) and \( \lambda_2 \) into the formula:**
- Calculate \( \lambda_1 - \lambda_2 \):
\[ \lambda_1 - \lambda_2 = 3.786 \times 10^{-3} - 4.304 \times 10^{-3} = -5.183 \times 10^{-4} \, \text{s}^{-1} \]
- Calculate the logarithm:
\[ \log\left(\frac{\lambda_1}{\lambda_2}\right) = \log\left(\frac{3.786 \times 10^{-3}}{4.304 \times 10^{-3}}\right) \approx \log(0.878) \approx -0.0556 \]

4. **Calculate \( t_{max} \):**
\[ t_{max} = \frac{2.303}{-5.183 \times 10^{-4}} \times (-0.0556) \]
- This simplifies to:
\[ t_{max} \approx \frac{2.303 \times 0.0556}{5.183 \times 10^{-4}} \approx 247.5 \, \text{sec} \]

5. **Final Result:**
The time required for the number of nuclei of \( _{82}^{214}Pb \) to reach maximum is approximately **247.5 seconds**.