Question

In an ore containing uranium, the ratio of $\text{U}-238$ to $\text{Pb}-206$ is $3.$ Assuming that all the lead present in the ore is the final stable product of $\text{U}-238.$ If age of the ore is $1.868\times {10}^n$ years. The value of the $n$ is (integer only).
(Take the half life of $\text{U}-238$ to be $4.5\times {10}^9$ years. $(\ln \frac{4}{3}=0.2876)$

Answer 1

Given, ${\left(t_{\frac{1}{2}}\right)}_{\text{U}}=4.5\times {10}^9\text{years}\frac{N_{\text{U}}}{N_{\text{Pb}}}=3$

Let, the age of the ore be $\left(t_0\right),$ let us assume that initially the ore had only $\text{U}-238$, the no. of nuclei present in the $\text{U}-238$ be $N_0$, and after time $t_0$ it becomes,

$N_U=N_0{e}^{-\lambda t_0}.......\left(1\right)$

The no. of $\text{Pb}$ nuclei created in the ore, in time $t_0$ will be,

$N_{Pb}=N_0(1-e^{-\lambda t_0}).......\left(2\right)$

From (1) and (2) we get,

$\frac{N_{Pb}}{N_U}=\frac{N_0(1-e^{-\lambda t_0})}{N_0{e}^{-\lambda t_0}}$

$\frac{1}{3}=\frac{(1-e^{-\lambda t_0})}{e^{-\lambda t_0}}$

$e^{-\lambda t_0}=3-3e^{-\lambda t_0}$

$4e^{-\lambda t_0}=3$

$e^{\lambda t_0}=\frac{4}{3}$

Taking log on both sides we get,

$\lambda t_0=\ln \frac{4}{3}$

$\therefore t_0=\frac{\ln \frac{4}{3}}{\lambda }=\frac{\ln \frac{4}{3}}{\frac{\ln 2}{t_{1/2}}}$

$\therefore t_0=1.86\times {10}^9\text{years.}$

$\therefore n=9$