Question

The age of a rock containing lead and uranium is equal to $1.5\times {{10}^{9}}$ yrs. The uranium is decaying into lead with half life equal to $4.5\times {{10}^{9}}$ yrs. If the ratio of lead to uranium present in the rock, assuming initially no lead was present in the rock is given by $259\times {{10}^{-k}}$. Then k will be given by (Given ${{2}^{{}^{1}/{}_{3}}}=1.259$).

Answer 1

Hint It should be known to us that the time taken for any radioactive substance for the radioactivity, especially by a specific isotope to fall to the half of its original value is known as the half life. This is the concept used to find the solution.

Complete step by step answer
Let us take the initial amount of Uranium as ${{N}_{0}}$.
Let the present amount of Uranium after $1.5\times {{10}^{9}}$ years be N.
So,
We can denote the amount of lead present as ${{N}_{0}}-N$
therefore The required ratio can be derived as $\dfrac{({{N}_{0}}-N)}{N}=\dfrac{(1-{{e}^{(-\lambda t)}})}{({{e}^{(-\lambda t)}})}$
Now,
$T_{1 / 2}=\dfrac{\operatorname{In} 2}{\lambda} \Rightarrow \lambda=\dfrac{\ln 2}{T_{1 / 2}}$
$\therefore \dfrac{N_{0}-N}{N}=\dfrac{1-e^{-\ln \dfrac{t}{T / 2}}}{e^{-\ln 2^{\dfrac{t}{T / 2}}}}=\dfrac{1-2^{-\dfrac{t}{T_{1 / 2}}}}{2^{-\dfrac{t}{T_{1 / 2}}}}=\dfrac{1-2^{-\dfrac{1}{3}}}{2^{-\dfrac{1}{3}}}=2^{1 / 3}-1=0.259$

Therefore, the correct answer is 0.259.

Note The main application of a half life is that it will give us an idea about the time taken by a radioactive substance to decay. The amount of time taken to react gives an idea about the reactivity. It should be known to us that half life is used by the scientists to determine the approximate age of the organic objects. This process is known as carbon-14. In the field of medical science radioactive isotopes are used as a treatment for medical tracers.
The radioisotopes can rage between a few microseconds to billions of years.