Question

A sample of pitch blend is found to contain 50% uranium and 2.425% lead. Of this lead only 93% was $P{b}^{206}$ isotope. If the disintegration constant is $1.52\times {10}^{-10}y{r}^{-1}$, the age of pitch blend deposits could be :

• A. $3.3\times {10}^{8}year$
• B. $5.6\times {10}^{8}year$
• C. $4.5\times {10}^{8}year$
• D. None of these

Answer 1

The correct option is A $3.3\times {10}^{8}year$

Assume 1 g of the sample is present.
Uranium present in the sample $=\frac{50}{100}g=\frac{0.50}{238}g$ - $atom$
$=2.10\times {10}^{-3}g$ - $atom$
Note: mass of U is divided by its molar mass to obtain number of gram atoms.
Pb present in the sample $=\frac{2.425}{100}g=\frac{2.425}{100\times 206}g$ - $atom$
Pb formed from uranium decay $=\frac{2.425\times 93}{100\times 206\times 100}$
$=0.109\times {10}^{-3}g$ - $atom$
Thus, $N=2.10\times {10}^{-3}g$ - $atom$
$N_0=(2.10+0.109)\times {10}^{-3}g$ - $atom$
$=2.209\times {10}^{-3}g$ - $atom$
Now, the integrated rate law expression for the nuclear decay process is $t=\frac{2.303}{\lambda }{\log }_{10}\frac{N_0}{N}$
$=\frac{2.303}{1.52\times {10}^{-10}}{\log }_{10}\frac{2.209\times {10}^{-3}}{2.10\times {10}^{-3}}$
Hence, the age of pitch blend deposits could be $t=3.3\times {10}^{8}year$