Question

A radioactive isotope X with half life \[1.5 \times {10^9}\]years decays into a stable nucleus Y. A rock sample contains both elements X and Y in ratio 1:15. Find the age of the rock.

Answer 1

Hint: To answer this question we must understand the concept of half life. We should also know how the concentration of reactants and rate of reaction affects the half life. We can put in the values in the following equation to get our desired result.
\[{N_{(t)}} = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{t_{\dfrac{1}{2}}}}}}}\]

Complete step by step solution:
During natural radioactive decay, not all atoms of an element are instantaneously changed to atoms of another element instead the decay process takes a long time. Sometimes the reaction never reaches completion. Here, it is important to note the concept of half life. It is the time in which the initial concentration is decayed and reduced to half.
From the question we can see that the half life is \[1.5 \times {10^9}\]
And X and Y are present in the ratio 1:15
Thus, we can write Y=15X

Let us assume Z to be the amount of radioactive isotope X initially present.
Therefore, X+Y=Z
$\Rightarrow $ X + 15X = Z
$\Rightarrow $ 16X = Z
Hence, \[\dfrac{Z}{X} = 16\]
We know that for second order reactions,
\[\lambda t = 2.303\log \dfrac{Z}{X}\]; where t is the age of the rock and lambda represents the decay constant which is the natural logarithmic value of 2 =0.693.
Substituting the values we have obtained so far,
\[\dfrac{{0.693}}{{1.5 \times {{10}^9}}}t = 2.303\log 16\]
Or, \[\dfrac{{0.693}}{{1.5 \times {{10}^9}}}t = 2.303 \times 1.204\]
Therefore,
\[t = 6 \times {10^9}years\]
Hence, the answer is \[6 \times {10^9}years\].

Note: Radioactive dating utilizes the concept of half life and radioactive decay. It is a process by which the approximate age of an object is determined through the use of certain radioactive nuclides.