The isotopes and occur in nature in the atomic ratio 140:1. The half-life periods of and are and yea — Nuclear Chemistry and Radioactivity Chemistry Question
Question
The isotopes $\text{U}^{238}$ and $\text{U}^{235}$ occur in nature in the atomic ratio 140:1. The half-life periods of $\text{U}^{238}$ and $\text{U}^{235}$ are $4.5 \times 10^9$ and $7.2 \times 10^8$ years, respectively. Assuming that at the time of earth formation, they were present in equal numbers, the age of earth (in order of $10^9$ years) is ($\log 2 = 0.3$, $\log 140 = 2.1$)
💡 Solution & Explanation
Let initial numbers be $N_0$. Currently, $N_{238} = N_0 e^{-\lambda_1 t}$ and $N_{235} = N_0 e^{-\lambda_2 t}$. Their ratio is $\frac{N_{238}}{N_{235}} = e^{(\lambda_2 - \lambda_1) t} = 140$. Taking $\log_{10}$ on both sides: $\frac{(\lambda_2 - \lambda_1)t}{2.303} = 2.1$. Using $\lambda = \frac{2.303 \log 2}{t_{1/2}}$, we get $t \cdot \log 2 \left(\frac{1}{t_{1/2}(\text{U}^{235})} - \frac{1}{t_{1/2}(\text{U}^{238})}\right) = 2.1$. Substituting values: $t \cdot 0.3 \left(\frac{1}{0.72 \times 10^9} - \frac{1}{4.5 \times 10^9}\right) = 2.1$. This simplifies to $t \cdot 0.3 \left(\frac{3.78}{3.24 \times 10^9}\right) = 2.1$, which yields $t \cdot 10^{-9} = \frac{2.1 \times 10.8}{3.78} = 6$. So, $t = 6 \times 10^9 \text{ years}$. Therefore, correct answer is 6.