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 remains undisintegrated. When the age of the sample is \( P \times 10^8 \) years, the ratio of the mass of Pb-206 to that of U-238 in the sample is found to be 7. The value of \( P \) is: \[ \text{[Given: Half-life of U-238 = \(4.5 \times 10^9\) years; \(\log_e2 = 0.693\)]} \]

Hint:

For radioactive decay problems, carefully calculate the ratio of masses and logarithmic values for precise age determination.

Answer 1

To solve the problem, we are asked to find the time it takes for the ratio of the mass of Pb-206 to U-238 to become 7, given that U-238 decays to Pb-206.

Key Concepts and Formulas:

• Radioactive Decay Law: $N(t) = N₀ \cdot e^{-\lambda t}$, where:
  • $N(t)$ is the number of radioactive nuclei at time $t$
  • $N₀$ is the initial number of radioactive nuclei
  • $\lambda$ is the decay constant
  • $t$ is the time
• Half-Life and Decay Constant: $T_{1/2} = \frac{\ln(2)}{\lambda}$, so $\lambda = \frac{\ln(2)}{T_{1/2}}$
• Mass and Number of Atoms: The number of atoms of an element is proportional to its mass, given a constant molar mass.

Steps to Solve:

1. Calculate the Decay Constant $\lambda$:

We are given the half-life of U-238 as $T_{1/2} = 4.5 \times 10^9$ years. Using the formula for $\lambda$, we get:

$$ \lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{4.5 \times 10^9 \, \text{years}} $$

2. Relating Masses to Number of Atoms:

Let the mass of U-238 at time $t$ be $m_U$ and the mass of Pb-206 at time $t$ be $m_Pb$. The number of U-238 atoms at time $t$ is $N_U = N₀ \cdot e^{-\lambda t}$, and the number of Pb-206 atoms formed at time $t$ is $N_Pb = N₀ (1 - e^{-\lambda t})$.

The ratio of their masses at time $t$ is:

$$ \frac{m_{Pb}}{m_U} = \frac{N_{Pb} \times \text{AtomicMass}_{Pb}}{N_U \times \text{AtomicMass}_{U}} = 7 $$

Using the atomic masses of U and Pb ($\text{AtomicMass}_U = 238$, $\text{AtomicMass}_{Pb} = 206$), we get:

$$ \frac{N_{Pb}}{N_U} = 7 \times \frac{238}{206} $$

Simplifying, we get:

$$ \frac{N_{Pb}}{N_U} = \frac{1666}{206} $$

3. Solve for $e^{-\lambda t}$:

Let $x = e^{-\lambda t}$, then:

$$ \frac{1666}{206} = \frac{1 - x}{x} $$

Multiplying both sides by $x$, we get:

$$ \frac{1666}{206} \cdot x = 1 - x $$

Rearranging the equation:

$$ \frac{1666}{206} \cdot x + x = 1 $$

Factoring out $x$:

$$ x \left( \frac{1666 + 206}{206} \right) = 1 $$

Thus:

$$ x = \frac{206}{1872} $$

4. Solve for Time $t$:

Taking the natural logarithm of both sides:

$$ \ln \left( \frac{206}{1872} \right) = -\lambda t $$

Substituting $\lambda = \frac{0.693}{4.5 \times 10^9}$, we get:

$$ t = \frac{-1}{\lambda} \ln \left( \frac{206}{1872} \right) $$

Substituting the values:

$$ t = \frac{-4.5 \times 10^9}{0.693} \ln(0.11) $$

$$ t = \frac{-4.5 \times 10^9}{0.693} \times (-2.207) $$

$$ t = \frac{4.5 \times 10^9}{0.693} \times 2.207 $$

$$ t = 14.334 \times 10^9 \, \text{years} $$

5. Convert the Time to the Desired Form:

We are asked to express the time as $P \times 10^8$ years:

$$ t = P \times 10^8 = 14.334 \times 10^9 $$

Thus:

$$ P = \frac{14.334 \times 10^9}{10^8} = 143.34 $$

The final answer is approximately 143.

Final Answer:
The correct answer is $\boxed{143}$.