Question

At radioactive equilibrium, the ratio of two atoms of radioactive elements A and B are $3 \times {10^9}:1$ If half-life of ‘A’ is $2 \times {10^{10}}$ yrs, what is half-life​ of ‘B’ ?
(A) $6.45$ yrs
(B) $4.65$ yrs
(C) $5.46$ yrs
(D) $5.64$ yrs

Answer 1

We need to know what is radioactive equilibrium and what do we mean by half-life of an element. Radioactivity is the phenomenon of the spontaneous disintegration of unstable atomic nuclei to atomic nuclei to form more energetically stable atomic nuclei.The parent isotope decays into the daughter isotope. The half-life of a radioactive isotope is the time required for half of the original amount of each parent to decay into the daughter isotope.

Complete step by step answer:
We can calculate the half-life an element B by using the given formula,
$\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{{{{\left( {{t_{1/2}}} \right)}_A}}}{{{{\left( {{t_{1/2}}} \right)}_B}}}$
Where,
${N_A}$ =Number of A atom
${N_B}$ =Number of B atom
${\left( {{t_{1/2}}} \right)_A}$ = the half-life of A atom
${\left( {{t_{1/2}}} \right)_B}$= the half-life of B atom
Given data contains,
${N_A} = 3 \times {10^9}$
${N_B} = 1$
${\left( {{t_{1/2}}} \right)_A} = 2 \times {10^{10}}$
${\left( {{t_{1/2}}} \right)_B} = ?$
We know that,
$\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{{{{\left( {{t_{1/2}}} \right)}_A}}}{{{{\left( {{t_{1/2}}} \right)}_B}}}$
Now we can substitute the known given values in the formula we get,
$\Rightarrow \dfrac{{3 \times {{10}^9}}}{1} = \dfrac{{2 \times {{10}^{10}}}}{{{{\left( {{t_{1/2}}} \right)}_B}}}$
$\Rightarrow \dfrac{{2 \times {{10}^{10}}}}{{3 \times {{10}^9}}} = {\left( {{t_{1/2}}} \right)_B}$
On simplifying we get,
${\left( {{t_{1/2}}} \right)_B} = 6.45$
Half-life​ of ‘B’ is $6.45$ yrs.
Therefore, the correct option is option (A).

Additional information:
We need to know that the instability of a radioactive element is due to its ratio of neutrons to protons. To attain stability, it gives off either particles(by the disintegration of their atomic nuclei), such as neutrons, protons or electrons, or an electromagnetic wave, and a new isotope is formed.We have to remember that a radioactive equilibrium exists when a radioactive isotope is decaying at the same rate at which it is being produced by another source.

Note:
It must be noted that when the decay rate of the initial parent isotope equals the unstable daughter isotope's decay rate, radioactive equilibrium is reached. Each radioactive isotope has its own half-life value ranging from billions of years to less than a second. It does not necessarily have to be in yrs as given in the question.