Question

The number of ${238}_U$ atoms in an ancient rock equals $\frac{1}{3}$ times number of ${206}_{Pb}$ atoms. The half life of decay of ${238}_U$ is $4.5\times {10}^{9}y$. Assuming that all ${206}_{Pb}$ atoms are formed the decay of ${238}_U$, the estimate age of the rock is

• A. $2.25\times {10}^{9}Y$
• B. $4.5\times {10}^{9}Y$
• C. $9.0\times {10}^{9}Y$
• D. $1.35\times {10}^{9}Y$

Answer 1

The correct option is C $9.0\times {10}^{9}Y$

$\begin{array}{c}\text{it is given Pb is formed only by Uranium}\end{array}$

$\begin{array}{c}\text{so,, initially Let us say U have}{N}_0\text{atom}\\ \text{then Pb should have zero}\end{array}$

$\text{and given that after time (t)}$

$\text{Number of}{U}^{238}=\frac{1}{3}\text{Number of Pb}$

$\begin{array}{c}\text{we can say that amount of}{U}^{238}\text{left will be}\\ 25\%.\end{array}$

$\begin{array}{ccccccc}& & & & & & U^{238}\to P{b}^{206}\end{array}$

$\begin{array}{ccc}t=0& 1& 0\\ t=t& 1-x& x\end{array}$ $\begin{array}{cc}\because 1-x& =\frac{x}{3}\\ x& =\frac{3}{4}\text{or}75\%.\text{(decayed)}\end{array}$

$\text{Now}$

$\begin{array}{cc}\frac{N_0}{4}& =N_0{e}^{-\lambda t}\\ & \because \lambda =\frac{\ln 2}{T_{1/2}}\end{array}$

$\begin{array}{cc}\ln 4& =\frac{\ln 2}{T_{1/2}}\times t\\ t& =4.5\times {10}^9\times 2\text{year}\\ & =9\times {10}^9\text{year}\end{array}$