Question

By successive decay ${ }_{92} U ^{238}$ changed to ${ }_{82} Pb ^{206}$, when a sample of uranium ore was analysed. It was found that it contains $1\, g$ of $U ^{238}$ and $0.1\, g$ of $Pb ^{206}$, considering that all the $Pb ^{206}$ 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. $0 \text{.} 1 1 5 5 \times 1 0^{8} \text{yrs}$
• B. $7 \text{.} 0 9 9 \times 1 0^{8} \text{yrs}$
• C. $0 \text{.} 1 5 4 \times 1 0^{- 9} \text{yrs}$
• D. $7 \text{.} 0 9 9 \times 1 0^{1 0} \text{yrs}$

Answer 1

Answer: (B)

Considering that the whole of $Pb ^{206}$ comes from $U ^{238}$, amount of $U ^{238}$ decayed for $=0.1\, g \,Pb^{206}$
$=\frac{0.1 \times 238}{206} \,g \,U ^{238} $
$=0.1155\, g\, U ^{238}$
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} yrs ^{-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} $
$=7.099 \times 10^8\, yrs$