Question

In an old rock, the ratio of nuclei of uranium and lead is \[1:1\]. Half of the life of Uranium is \[45 \times {10^9}\;{\text{yrs}}\]. Initially it contains only uranium nuclei. How old is the rock?

Answer 1

Hint: In this question, the concept of the decay constant is used to calculate the constant ratio for the number of atoms of radio nuclei that convert in a given period of time. Calculate the value the value of\[\lambda \], where \[\lambda \] is the decay constant

Complete step by step answer:
Consider the given question, we have given the ratio of nuclei of the uranium and the ratio of nuclei of the lead in a old rock as
\[ \Rightarrow \dfrac{{{\text{Uranium}}}}{{{\text{Lead}}}} = \dfrac{1}{1}\]

And let us represent the uranium by \[U\] and represent the lead by \[Pb\].
\[ \Rightarrow \dfrac{U}{{Pb}} = \dfrac{1}{1}\]
We have also given the half life of the uranium as
\[U = 4.5 \times {10^9}\;yrs\]

The old rock initially contains the uranium nuclei. Let us represent the time by\[t\], represent the initial uranium in the rock by \[{U_0}\]. And the uranium after the time \[t\] represented by \[{U_t}\] and let represent the lead after time \[t\] by\[P{b_t}\].
As we know that the initial uranium present in the old rock can be written as,
\[ \Rightarrow {U_0} = {U_t} + P{b_t}\]

We have \[{U_0} = 2\] that is the initial uranium form. and the uranium after time\[t\],
\[{U_t} = \left( {{U_t} + P{b_t}} \right){e^{ - \lambda t}}......\left( 1 \right)\]

As we know that the constant ratio for the number of atoms of radio nuclei is used to convert in a given period of time. Where\[\lambda \] is decay constant.
\[ \Rightarrow \lambda = \dfrac{{0.693}}{{{t_{1/2}}U}}\]

Now we substitute the values in the above expression and obtain
\[ \Rightarrow \lambda = 0.1551 \times {10^{ - 9}}\;{\text{yr}}{{\text{s}}^{ - 1}}\]
Now we substitute the values in equation \[\left( 1 \right)\] as
\[ \Rightarrow 1 = \left( {1 + 1} \right){e^{\left( { - 0.1551 \times {{10}^{ - 9}}} \right)t}}\]
Now we take the log both sides,
\[ \Rightarrow \log \left( {\dfrac{1}{2}} \right) = \log \left( {{e^{\left( { - 0.1551 \times {{10}^{ - 9}}} \right).t}}} \right)\]
After simplification we get,
\[ \Rightarrow 0.1551 \times {10^{ - 9}}t = 0.693\]

Now we solve the equation further for $t$.
\[\therefore t = 4.46 \times {10^9}\;{\text{yrs}}\]

Therefore, the rock is old \[t = 4.46 \times {10^9}\;{\text{yrs}}\].

Note:As we know that the half-life means that the time taken to decay the material to its half quantity. Generally, the decay of the radioactive material obtains the logarithmic way that is it is difficult to obtain the life of the material, so we calculate the half-life of these materials.