Question

Assuming that $^{\text{226}}\text{Ra}$ (${{\text{t}}_{\text{1}/\text{2}}}=\text{1}.\text{6}\times \text{1}{{0}^{\text{3}}}\text{yrs}$) is in secular equilibrium with $^{\text{238}}\text{U}$ (${{\text{t}}_{\text{1}/\text{2}}}=\text{4}.\text{5}\times \text{1}{{0}^{\text{9}}}\text{yrs}$) in a certain mineral, how many grams of radium will be present in for every gram of $^{\text{238}}\text{U}$in this mineral?
(A) $3.7\times {{10}^{-7}}$
(B) $3.7\times {{10}^{7}}$
(C) $3.4\times {{10}^{-7}}$
(D) $3.4\times {{10}^{7}}$

Answer 1

To calculate the grams of radium present, we should first find the relation between half-life and number of molecules of radium and uranium. We can write this relation as:
$\dfrac{Number\,of\,molecules\,of\,radium}{Number\,of\,molecules\,of\,uranium}=\dfrac{{{\left( {{t}_{\dfrac{1}{2}}} \right)}_{radium}}}{{{\left( {{t}_{\dfrac{1}{2}}} \right)}_{uranium}}}$

Complete step by step answer:
So, first we should know about radioactive decay.
- Radioactive decay is the spontaneous breakdown of the nucleus of the atom. This breakdown results in the release of energy and matter from the nucleus. From which the continuous breakdown occurs is known as radioisotopes.
- The half-life of a radioisotope describes how long it takes for half of the atoms in a given weight of atoms to decay.

We will use the following relationship between half-life and number of molecules.
$\dfrac{\text{Numner}\,\text{of molecules of radioactive element-1}}{\text{Number of molecules of radioactive element-2}}=\dfrac{{{\left( {{t}_{\dfrac{1}{2}}} \right)}_{1}}}{{{\left( {{t}_{\dfrac{1}{2}}} \right)}_{2}}}$

So, we can write that
$\dfrac{Number\,of\,molecules\,of\,radium}{Number\,of\,molecules\,of\,uranium}=\frac{{{\left( {{t}_{\dfrac{1}{2}}} \right)}_{radium}}}{{{\left( {{t}_{\dfrac{1}{2}}} \right)}_{uranium}}}$.........(1)

Let us now write the quantities that we should put in the above formula.
We are given that atomic weight of Radium=226$gmmo{{l}^{-1}}$
Let the weight of radium that has to be calculated=W$gmmo{{l}^{-1}}$
Atomic weight of uranium=238$gmmo{{l}^{-1}}$
Weight of uranium= 1 gram (because in question, it is give that we have to calculate weight of radium for every gram of uranium)
Now, to calculate N or number of molecules we know that:
Number of molecules= Moles $\times$ Avogadro's number

So, we can write that,
Number of molecules of Radium will be ${{N}_{radium}}=\dfrac{W}{226}\times {{N}_{A}}$
And number of molecules of Uranium will be ${{N}_{Uranium}}=\dfrac{1}{238}\times {{N}_{A}}$

We know that, for Radium half-life ${{\left( {{\text{t}}_{\frac{\text{1}}{\text{2}}}} \right)}_{Radium}}=1.6\times {{10}^{3\,}}years$ and for Uranium, half-life ${{\left( {{\text{t}}_{\dfrac{\text{1}}{\text{2}}}} \right)}_{Uranium}}=4.5\times {{10}^{9\,}}years$.

We will put these values in equation (1). So, we will get
$\dfrac{{{N}_{radium}}}{{{N}_{uranium}}}=\dfrac{{{\left( {{t}_{\dfrac{1}{2}}} \right)}_{radium}}}{{{\left( {{t}_{\frac{1}{2}}} \right)}_{uranium}}}$
$\dfrac{\dfrac{W}{226}\times {{N}_{A}}}{\dfrac{1}{238}\times {{N}_{A}}}=\dfrac{1.6\times {{10}^{3\,}}years}{4.5\times {{10}^{9\,}}years}$

So, we can write that,
$W=\dfrac{1.6\times {{10}^{3\,}}years}{4.5\times {{10}^{9\,}}years}\times \dfrac{226}{238}$
$W=3.4\times {{10}^{-7}}gm$
So, the correct answer is “Option C”.

Note: Note that we will get the answer in gram units because the molecular weight we have used contains the weight in grams as well. As we are directly taking the ratio of the half life of two elements, we need not to convert the half life into units other than years.