The age of a rock containing lead and uranium is equal to $1.5\times {10}^9$ years. The uranium is decaying into lead with half life equal to $4.5\times {10}^9$ years. Find the ratio of lead to uranium present in the rock, assuming that initially no lead was present in the rock. Given that $2^{\frac{1}{3}}=1.259$

Let $N_0$ be the initial amount of uranium in the rock and let $N$ be the present amount of uranium after $1.5\times {10}^9$ years

So the amount of lead present ion the rock is $(N_0-N)$

So the required ratio is $\frac{N_0-N}{N}=\frac{1-e^{-\lambda t}}{e^{-\lambda t}}$

Since $T_{1/2}=\frac{\ln \left(2\right)}{\lambda }$

$$\begin{gathered} \Rightarrow \lambda =\frac{\ln 2}{T_{1/2}} \\ \Rightarrow e^{-\lambda t}=e^{-\left(\frac{{\log }_{e}2}{T_{1/2}}\right)t}=e^{-\log \left(2\right)\frac{t}{T_{1/2}}}=2^{-\frac{t}{T_{1/2}}} \\ \frac{N_0-N}{N}=\frac{1-e^{-\left(\frac{{\log }_{e}2}{T_{1/2}}\right)t}}{e^{-\left(\frac{{\log }_{e}2}{T_{1/2}}\right)t}}=\frac{1-2^{-\frac{t}{T_{1/2}}}}{2^{-\frac{t}{T_{1/2}}}}=\frac{1-2^{-{\displaystyle \frac{1}{3}}}}{2^{-{\displaystyle \frac{1}{3}}}} \\ \Rightarrow \frac{N_0-N}{N}=2^{\frac{1}{3}}-1=0.259 \end{gathered}$$