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 pitchblende 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

Hint: Pitchblende is a mineral containing uranium in 560 % and lead in 2.425 % and it is radioactive in nature. Radioactive means pitch blend is not stable. Due to its radioactivity pitch blend contains oxides of lead and minimum amount of helium in it.

Complete step by step solution:
- In the question it is given that the pitch blend contains 50% uranium and 2.425% lead.
- That means the amount of uranium is 50.
- It is also given that 2.425 % of lead in pitch blend contains 93 % of $P{{b}^{206}}$ isotope.
- Therefore the amount of $P{{b}^{206}}$ in pitch blend is
\[\begin{align}
& =2.425\times \dfrac{93}{100} \\
& =2.255 \\
\end{align}\]
- There is a formula to calculate the age of the pitch blend deposited and it is as follows.
\[t=\dfrac{2.303}{\lambda }\log \dfrac{{{N}_{o}}}{N}\]
Here, $t$ = age of the chemical
$\lambda $ = disintegration constant
${{N}_{o}}$ = Total amount of the chemical
$N$ = amount of chemical does not contain radioisotope
- We know that ${{N}_{o}}$ $= 50 + 2.255$ $= 52.255$
$N = 50$
- Substitute all the known values in the above to get the age of the pitch blend.
\[\begin{align}
& t=\dfrac{2.303}{\lambda }\log \dfrac{{{N}_{o}}}{N} \\
& =\dfrac{2.303}{1.52\times {{10}^{-10}}}\log \dfrac{52.255}{50} \\
& =3.3\times {{10}^{8}}year \\
\end{align}\]
- Therefore the age of pitchblende deposits is $3.3\times {{10}^{8}}year$ .

So, the correct option is A.

Note: The total amount of lead present in the pitch blend contains 93 % of $P{{b}^{206}}$ and 7 % percent of $P{{b}^{207}}$ . With uranium a small amount of radium is also present in the pitch blend. Because radium is the decay product formed from uranium.