Question

A uranium mineral contains $^{238}U$ and $^{206}Pb$ in the ratio of $4:1$ by weight. Calculate the age of the mineral, ${t_{\dfrac{1}{2}}}$ of $^{238}U$= $4.5 \times {10^9}$years. Assume that all the lead present in the mineral is formed from disintegration of $^{238}U$ .

Answer 1

The relationship between the age of an element with its initial weight is given with a specific equation after integrating the equation for rate of radioactive disintegration and the value of decay constant can be substituted in this equation by its relationship with the half-life of element.
The initial weight of the mineral will be the sum of all of its components.

Formula Used: 1. $\lambda = \dfrac{{2.303}}{t}\log \dfrac{{{N_ \circ }}}{N}$
2. $\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$

Complete step by step answer:
Let us first understand the terms used in this question.
When we consider elements such as Uranium, these elements are radioactive which means they have an unstable nucleus and hence, they undergo spontaneous decay until they form a stable nucleus which can be either by emission of α particle , β particle or γ waves.
The half-life of a radioactive element is the amount of time which will be taken by the element to reduce its concentration to half its initial concentration due to decay.
There is a specific relation between decay constant and half life of an element.
$\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$
where, $\lambda$ = Decay Constant,${t_{\dfrac{1}{2}}}$ = half life

we have been given the value of half-life for $^{238}U$ as $4.5 \times {10^9}$
Substituting this value in the above equation we get;
$\lambda = \dfrac{{0.693}}{{4.5 \times {{10}^9}}}$
Let us keep this equation unsolved and as we will need to substitute these values further

The equation which relates the decay constant, age of mineral with its weight is
$\lambda = \dfrac{{2.303}}{t}\log \dfrac{{{N_ \circ }}}{N}$
where; $t = age\,of\,mi\,neral,{N_ \circ } = initial\,weight,N = \,weight\,at\,time\,t$

Substitute the value of decay constant we get;
$\dfrac{{0.693}}{{4.5 \times {{10}^9}}} = \dfrac{{2.303}}{t}\log \dfrac{{{N_ \circ }}}{N}$
It is given that; $^{238}U$ and $^{206}Pb$ occur in uranium mineral in the ratio of $4:1$
We can say that in the mineral, 4g of Uranium is present with every 1g of Lead.
Hence the initial weight the mineral will be;
Weight of Pb + weight of U
Substituting the values, we get:
$1 + 4 = 5g({N_ \circ })$
The weight at time t will be 4g. which will be $t$

Substituting these values in the above equation, we get:

$\Rightarrow \dfrac{{0.693}}{{4.5 \times {{10}^9}}} = \dfrac{{2.303}}{t}\log \dfrac{5}{4}$

Solving for $t$ we get :
$t = 1.45 \times {10^9}years$

So, $t = 1.45 \times {10^9}years$

Note: The half life on an element is also a measure of the stability of the nucleus . The shorter the half-life, the more unstable will be the nucleus.
Uranium is found along with Lead because Lead is a decay product of Uranium and is a stable Nuclei.
Radioactive decay leads to release of high amounts of energy and hence they are used in Nuclear Power Plants to generate energy.