Question

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

• A. $4.5\times {10}^{9}year$
• B. $3.75\times {10}^{9}year$
• C. $6.25\times {10}^{9}year$
• D. None of these

Answer 1

The correct option is A $4.5\times {10}^{9}year$

The decay equation is as shown:

${}_{92}^{238}\text{U}{⟶}_{82}^{206}\text{Pb}+8{}_2^4\text{He}+6_{-1}^{0}e$

Total number of gram atoms of $Pb$ present $=\frac{20.6}{206}=0.1g-atom$. They correspond to the total number of gram atoms of U decayed.

Total number of gram atoms of $U$ present $=\frac{23.8}{238}=0.1g-atom$

Thus, $N=0.1g-atom$

$N_0=$ total number of g atoms of U present $+$ total number of gram atoms of U decayed

$N_0=0.1+0.1=0.2g-atom$

The integrated rate law expression for first-order decay process is :

$t=\frac{2.303}{\lambda }{\log }_{10}\frac{N_0}{N}=\frac{2.303\times \text{half life period}}{0.693}{\log }_{10}\frac{N_0}{N}$

$t=\frac{2.303\times 4.5\times {10}^9}{0.693}{\log }_{10}\frac{0.2}{0.1}$

Hence, the age of the ore is $t=4.5\times {10}^9$ years

Hence, option A is correct.