Question

By successive decay ${}_{92}{U}^{238}$ changed to ${}_{82}P{b}^{206}$, when a sample of uranium ore was analysed. It was found that it contains $1g$ of $U^{238}$ and $0.1g$ of $P{b}^{206}$, considering that all the $P{b}^{206}$ had accumulated due to the decay of $U^{238}$. Calculate the age of the ore. (Half-life of $U^{238}=4.5\times 10^{9}yrs$ ).

• A. $0\text{.}1155\times 10^8\text{yrs}$
• B. $7\text{.}099\times 10^8\text{yrs}$
• C. $0\text{.}154\times 10^{-9}\text{yrs}$
• D. $7\text{.}099\times 10^{10}\text{yrs}$

Answer 1

Answer: (B)

Considering that the whole of $P{b}^{206}$ comes from $U^{238}$, amount of $U^{238}$ decayed for $=0.1gP{b}^{206}$
$=\frac{0.1\times 238}{206}g{U}^{238}$
$=0.1155g{U}^{238}$
Hence, initial amount of $U^{238}=(1+0.1155)g{U}^{238}$ Thus, value of disintegration constant can be calculated as:
Now, $\lambda =\frac{0.693}{T_{1/2}}=\frac{0.693}{4.5\times 10^{9}yrs}$
$=0.154\times 10^{-9}yr{s}^{-1}$
$t=\frac{2.303}{\lambda }\log \frac{N_0}{N}$
$t=\frac{2.303}{0.154\times 10^{-9}}\log \frac{1.1155}{1}$
$=7.099\times 10^{8}yrs$