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

For a first-order reaction, the time \( t \) is given by: \[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t} \] Given: \[ \frac{[m]_{\text{Pb}}}{[m]_U} = 7 \implies \frac{[A]_0}{[A]_t} = 1 + 7 \times \frac{1}{238 + 206} \approx 9 \] Half-life of U-238: \[ t_{1/2} = 4.5 \times 10^9 \quad \text{so,} \quad k = \frac{0.693}{t_{1/2}} = \frac{0.693}{4.5 \times 10^9} \] Substituting: \[ t = \frac{2.303}{k} \log 9 = 2.303 \times \frac{4.5 \times 10^9}{0.693} \log 9 \] Simplify: \[ t \approx 14.27 \times 10^9 = 143 \times 10^8 \quad \implies \quad P = 143. \]