Question

A sample of $U^{238}$ (half life $=4.5\times {10}^9$year), ore is found to contain $23.8\text{g}$ of $U^{238}$ and $20.6\text{g}$ of $P{b}^{206}$ The age of the ore is $X\times {10}_9$ year. Calculate the value of $2X$ :

Answer 1

The corresponding equation is,
${}_{92}{U}^{238}{\to }_{82}P{b}^{206}+8_{2}H{e}^4+6_{-1}{e}^{\circ }$
Pb present =$\frac{20.6}{206}$=0.1 g atom = U decayed.
U present =$\frac{23.8}{238}$= 0.1 g atom
Thus initial amount of $U^{238}=(0.1+0.1)\text{g atom}=0.2\text{g atom}$
Since half of the initial amount are degrading, the age of the ore will be equal to the half life.
Thus $X==4.5\times {10}^9$year
Hence $2X=9\times {10}^9$ year