1. Problem Analysis

We assume that at the time of the earth’s birth ($t=0$), the concentrations of $^{238}\text{U}$ and $^{235}\text{U}$ were identical. We need to find the time $t$ elapsed to reach current concentrations $n_1$ and $n_2$.

Parameters:

• Current concentrations: $n_1$ ($^{238}\text{U}$), $n_2$ ($^{235}\text{U}$)
• Half-lives: $\tau_1$, $\tau_2$
• Initial Condition: $N_{0,1} = N_{0,2} = N_0$

2. Mathematical Formulation

Using the radioactive decay law $N(t) = N_0 e^{-\lambda t}$, we can write the equations for both isotopes:

$$n_1 = N_0 e^{-\lambda_1 t} \quad \text{and} \quad n_2 = N_0 e^{-\lambda_2 t}$$

Divide the first equation by the second to eliminate $N_0$:

$$\frac{n_1}{n_2} = \frac{e^{-\lambda_1 t}}{e^{-\lambda_2 t}} = e^{(\lambda_2 – \lambda_1)t}$$

Take the natural logarithm of both sides:

$$\ln\left(\frac{n_1}{n_2}\right) = (\lambda_2 – \lambda_1)t$$ $$t = \frac{\ln(n_1/n_2)}{\lambda_2 – \lambda_1}$$

3. Substitution of Half-lives

We know that decay constant $\lambda = \frac{\ln 2}{\tau}$. Substituting this into the expression for $t$:

$$t = \frac{\ln(n_1/n_2)}{\frac{\ln 2}{\tau_2} – \frac{\ln 2}{\tau_1}}$$ $$t = \frac{\ln(n_1/n_2)}{\ln 2 \left( \frac{1}{\tau_2} – \frac{1}{\tau_1} \right)}$$ $$t = \frac{\ln(n_1/n_2)}{\ln 2 \left( \frac{\tau_1 – \tau_2}{\tau_1 \tau_2} \right)}$$