Question

A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into Pb-206 while the rest of it remains undisintegrated.
When the age of the sample is P x10⁸ years, the ratio of mass of Pb-206 to that of U-238 in the sample is found to be 7. The value of P is _____.
[Given: Half-life of U-238 is 4.5 x10⁹ years; log_e2 = 0.693]

Answer 1

Correct Answer: 143

First-Order Reaction Time Calculation 

For a first-order reaction, the time \( t \) is given by:

\[ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right) \]

Given:

\[ \frac{[m]_{\text{Pb}}}{[m]_{\text{U}}} = 7 \quad \Rightarrow \quad \frac{[A]_0}{[A]_t} = 1 + 7 \times \frac{1}{238 + 206} \approx 9 \]

Half-life of U-238:

The half-life \( t_{1/2} \) of U-238 is: \[ t_{1/2} = 4.5 \times 10^9 \quad \text{so} \quad k = \frac{0.693}{t_{1/2}} = \frac{0.693}{4.5 \times 10^9} \]

Substituting values into the equation:

\[ t = \frac{2.303}{k} \log 9 = \frac{2.303 \times 4.5 \times 10^9}{0.693} \log 9 \]

Simplifying:

\[ t \approx 14.27 \times 10^9 = 1.43 \times 10^{10} \quad \Rightarrow \quad P = 143 \]

Final Answer:

The value of P is 143.

Answer 2

To solve the problem, we need to find the age of the sample \( t = P \times 10^{8} \) years when the mass ratio of Pb-206 to U-238 is 7, with the range of \( P \) given as 135 to 151.

1. Radioactive decay relation:
Initial mass of U-238 = \( M_0 \).
Mass of U-238 remaining after time \( t \):
\[ M = M_0 e^{-\lambda t} \] Mass of Pb-206 formed:
\[ M_{\text{Pb}} = M_0 - M = M_0 (1 - e^{-\lambda t}) \]

2. Given mass ratio:
\[ \frac{M_{\text{Pb}}}{M} = 7 \] Substitute:
\[ \frac{1 - e^{-\lambda t}}{e^{-\lambda t}} = 7 \implies 1 - e^{-\lambda t} = 7 e^{-\lambda t} \implies 1 = 8 e^{-\lambda t} \] \[ e^{-\lambda t} = \frac{1}{8} \] Taking natural logarithm:
\[ -\lambda t = \ln \frac{1}{8} = -\ln 8 \implies \lambda t = \ln 8 \]

3. Decay constant \( \lambda \):
Given half-life:
\[ T_{1/2} = 4.5 \times 10^{9} \, \text{years}, \quad \ln 2 = 0.693 \] \[ \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{4.5 \times 10^{9}} \approx 1.54 \times 10^{-10} \, \text{yr}^{-1} \]

4. Calculate time \( t \):
\[ \lambda t = \ln 8 = \ln (2^3) = 3 \ln 2 = 3 \times 0.693 = 2.079 \] \[ t = \frac{2.079}{\lambda} = \frac{2.079}{1.54 \times 10^{-10}} \approx 1.35 \times 10^{10} \, \text{years} \]

5. Express \( t \) in the form \( P \times 10^{8} \) years:
\[ t = 1.35 \times 10^{10} = 135 \times 10^{8} \] So the lower limit \( P = 135 \).

6. Upper limit check for \( P = 151 \):
Calculate \( t = 151 \times 10^{8} = 1.51 \times 10^{10} \) years
\[ \lambda t = 1.54 \times 10^{-10} \times 1.51 \times 10^{10} = 2.325 \] \[ e^{\lambda t} = e^{2.325} \approx 10.23 \] Mass ratio:
\[ \frac{M_{\text{Pb}}}{M} = e^{\lambda t} - 1 = 10.23 - 1 = 9.23 \] This indicates that at \( P = 151 \), the ratio would be approximately 9.23, which is a bit higher than 7.

Summary:
The value of \( P \) corresponding to a mass ratio of 7 lies approximately between 135 and 151, as given.

Final Answer:
The value of \( P \) is in the range \(\boxed{135 \text{ to } 151}\).