Question

The age of a rock containing lead and uranium is equal to $1.5\times {10}^{9}yrs.$ The uranium is decaying into lead with half life equal to $4.5\times {10}^{9}yrs$. If the ratio of lead to uranium present in the rock, assuming initially no lead was present in the rock s given by $259\times {10}^{-k}$. Then $k$ will be given by (Given $2^{1/3}=1.259)$.

Answer 1

$\text{Let initially there be}{N}_0\text{uranium atoms.}\text{Uranium converts to leaal}$.

$\begin{array}{c}\text{after time}t=1.5\times {10}^{9}yrs=\frac{4.5\times {10}^9}{3}yrs=\frac{t_{1/2}}{3}\\ \text{Converted atoms of Uranium be}{N}_{0}N\\ \text{Left uranium atoms be}N\end{array}$

$\begin{array}{c}n=\text{number of half lives}=1/3\\ N=\frac{N_0}{2^n}\\ N=\frac{N_0}{2^{1/3}}=\frac{M_0}{1.259}\end{array}$

$\text{lead atoms now = 18.}{N}_0-N$

$\text{uranium atoms now =}N$

$\text{ratio}=\frac{N_0-N_1}{N_1}=\frac{N_0}{N_1}-1=1.259-1=0.259$

$\begin{array}{cc}\text{ratio}=& 259\times {10}^{-k}=0.259\\ \frac{259}{0.259}& ={10}^k\\ {10}^3& =k^k\\ k& =3⟶\text{Ans.}\end{array}$