Question

A sample of $U^{238}$ (half-life $=4.5\times {10}^9$ years) ore is found to contain $23.8g$ of $U^{238}$ and $20.6g$ of $P{b}^{206}$. The age of the ore is:

• A. $4.50\times {10}^9$ years
• B. $4.14\times {10}^9$ years
• C. $4.65\times {10}^9$ years
• D. $Noneofthese$

Answer 1

Answer: (A)

$4.50\times {10}^9$ years

${}_{92}{U}^{238}\to {}_{82}P{b}^{206}+8{}_{2}H{e}^4+6{}_{-1}{e}^0$
$Pb$ present $\frac{20.6}{206}=0.1$ mol
= U decayed
= (X)
$\left[U_{present}\right(today\left)\right(a-x\left)\right]=\frac{23.8}{238}=0.1g$ atom or mol
$N_0=U_{present}+U_{decayed}$
$\left(a\right)=0.1+0.1$
$(a-x)+x$
$=0.2gatom$
= mol
$f=\frac{2.303}{K}log\frac{N_0}{N}$
$=\frac{2.303}{0.693/4.5\times {10}^{19}}log\frac{0.2}{0.1}$
$=4.5\times {10}^9$ years