Question

At radioactive equilibrium, the ratio between two atoms of radioactive elements A and B is $3.1 \times {10^9}:1$ respectively. If ${t_{50}}$ of element A is $2 \times {10^{10}}$ years , then ${t_{50}}$ of elements B is:
A.$6.2 \times {10^9}$ years
B.$6.45$ years
C.$2 \times {10^{10}}$ years
D.$3.1 \times {10^9}$ years

Answer 1

By equating the relationship of half life with number of atoms for element A with the relationship of half life with the number of atoms for element B, we can find the half life of element B since we already know the ratio of the number of atoms in the two elements rather than the actual number of atoms itself.
Formula used:
1. $\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{{{t_{50A}}}}{{{t_{50B}}}}$

Complete step by step answer:
Since, the question mentions, there is a radioactive equilibrium,
which means that the rate of decay of element A will be equal to the rate of decay of element B.
Therefore, we can say that:
$- (\dfrac{{d{N_A}}}{{dt}}) = - (\dfrac{{d{N_B}}}{{dt}})$
${N_A} = Number\,of\,atoms\,in\,A,{N_B} = Number\,of\,atoms\,in\,B$
Using Bateman’s Equation, we known that :
$- (\dfrac{{d{N_A}}}{{dt}}) = {K_A}{N_A}$
$- (\dfrac{{d{N_B}}}{{dt}}) = {K_B}{N_B}$
Hence, Substituting these values in the equilibrium equation we get,
${K_A}{N_A} = {K_B}{N_B}$
We know that the relationship between the decay constant and half life of an element is given by the formula
$K = \dfrac{{0.693}}{{{t_{50}}}}$
Where, $K = decay\,constant,{t_{50}} = Half\,life$
So, The value of decay constant for Element A and Element B will be:
Element A : ${K_A} = \dfrac{{0.693}}{{{t_{50A}}}}$
Elements B: ${K_B} = \dfrac{{0.693}}{{{t_{50B}}}}$
Substituting these values for decay constant in the equilibrium equation we get,
$\dfrac{{0.693}}{{{t_{50A}}}}{N_A} = \dfrac{{0.693}}{{{t_{50B}}}}{N_B}$
Rearranging the equation, we get:
$\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{{{t_{50A}}}}{{{t_{50B}}}}$
In the question the ratio of the number of atoms is given to us which is nothing but $\dfrac{{{N_A}}}{{{N_B}}}$ value.
From the question, we know that: $\dfrac{{{N_A}}}{{{N_B}}}$ = $3.1 \times {10^9}:1$ and ${t_{50A}}$= $2 \times {10^{10}}$ years
Substituting equations in the above equation, we get:
$3.1 \times {10^9} = \dfrac{{2 \times {{10}^{10}}}}{{{t_{50B}}}}$
Solving this equation for the half-life value of Element B
we get; ${t_{50B}} = 6.45$ years
Hence, Option B is correct.

Note:
Radioactive elements have unstable nuclei which will undergo spontaneous decay and form a new nucleus, called the daughter nuclei. This decay process continues till the nucleus of the new formed species is not stable. However, the mass of the radioactive species never actually reaches zero , although it is supposed to theoretically. So a sample that once contained a radioactive element will never be free from radiation, but the radiation level might reach a safe level which does not harm human tissue.