JEE Advanced 2024 Paper 2 · Q11 · Integrated Rate Equations

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 $	imes 10^8$ 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 $4.5 	imes 10^9$ years; $\log_e 2 = 0.693$]

SolveKar AI · Accepted Answer

Mass ratio: $\frac{m_{Pb}}{m_U} = \frac{n_{Pb}\cdot 206}{n_U \cdot 238} = 7 \Rightarrow \frac{n_{Pb}}{n_U} = \frac{7\cdot 238}{206} = \frac{1666}{206}$. For first-order radioactive decay $N_U = N_0 e^{-\lambda t}$ with $\lambda = \frac{\ln 2}{t_{1/2}}$. So $\frac{n_{Pb}}{n_U} = e^{\lambda t} - 1$, giving $e^{\lambda t} = 1 + \frac{1666}{206} = \frac{1872}{206} = 9.087$. Thus $\lambda t = \ln(9.087) \approx 2.207$, hence $t = \frac{2.207}{0.693} 	imes 4.5 	imes 10^9 \approx 1.4265 	imes 10^{10}$ years $= 142.65 	imes 10^8$ years. So $P = 142.65$.

JEE Advanced 2024 Paper 2 · Q11 · Integrated Rate Equations

Solution