Question

Two radioactive sources \[X\] and \[Y\] of half lives \[1\,{\text{h}}\] and \[2\,{\text{h}}\] respectively initially contain the same number of radioactive atoms. At the end of \[2\,{\text{h}}\], their rates of disintegration are in the ratio of:
(A) \[4:3\]
(B) \[3:4\]
(C) \[1:2\]
(D) \[2:1\]

Answer 1

Hint:First, use the formula to find the relation between the initial and final concentration after half-life, for each source. Find the rate of decay for both the sources and substitute the values of concentration in that equation, to find the ratio.

Complete step by step answer:
In the given question, we are given that in both the radioactive sources, there contains an equal number of atoms, initially.The half life of the first radioactive source is \[1\] hour while the half life of the second radioactive source is \[2\] hours. We are asked to find the ratio of their rates of disintegration after a time duration of \[2\] hours.
The two radioactive sources are given as \[X\] and \[Y\] .
Half-life of \[X\],
\[{T_1} = 1\,{\text{h}}\]
Half-life of \[Y\],
\[{T_2} = 2\,{\text{h}}\]
Let the initial amount of the substance be \[{N_0}\] ; the amount left after disintegration be \[{N_1}\] for the first source and the amount left after disintegration be \[{N_2}\] for the second source.
We have a relation which relates the initial concentration and the final concentration:
\[N\left( t \right) = {N_0} \times {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{t_{1/2}}}}}}\] …… (1)
Where,
\[N\left( t \right)\] indicates the concentration after time \[t\] .
\[{N_0}\] indicates the initial concentration.
\[{t_{1/2}}\] indicates the half life time.
For the first source:
Substitute the values, \[{t_{1/2}} = 1\,{\text{h}}\] and \[t = 2\,{\text{h}}\] in equation (1), we get:
\[
{N_1} = {N_0} \times {\left( {\dfrac{1}{2}} \right)^{\dfrac{2}{1}}} \\
\Rightarrow{N_1} = {N_0} \times {\left( {\dfrac{1}{2}} \right)^2} \\
\Rightarrow{N_1} = \dfrac{1}{4}{N_0} \\
\]
For the second source:
Substitute the values, \[{t_{1/2}} = 2\,{\text{h}}\] and \[t = 2\,{\text{h}}\] in equation (1), we get:
\[
{N_2} = {N_0} \times {\left( {d\dfrac{1}{2}} \right)^{\dfrac{2}{2}}} \\
\Rightarrow{N_2} = {N_0} \times {\left( {\dfrac{1}{2}} \right)^1} \\
\Rightarrow{N_2} = \dfrac{1}{2}{N_0} \\
 \]
The expression which gives decay rate is:
\[R = \lambda {\rm N}\] …… (2)
Where,
\[R\] indicates the rate of decay.
\[\lambda \] indicates decay constant.
\[{\rm N}\] indicates the amount of substance left after decay.
Again, we have another formula for half-life:
\[\lambda = \dfrac{{0.693}}{{{t_{1/2}}}}\] …… (3)
Using the equation (3) in equation (2), we get:
\[R = \dfrac{{0.693}}{{{t_{1/2}}}} \times {\rm N}\]
So, the rate of decay for the first source \[X\] is:
\[{R_1} = \dfrac{{0.693}}{{{{\left( {{t_{1/2}}} \right)}_1}}} \times {{\rm N}_1}\] …… (4)
So, the rate of decay for the second source \[Y\] is:
\[{R_2} = \dfrac{{0.693}}{{{{\left( {{t_{1/2}}} \right)}_2}}} \times {{\rm N}_2}\] …… (5)
Dividing, equation (4) by equation (5), we get:
\[ \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\dfrac{{0.693}}{{{{\left( {{t_{1/2}}} \right)}_1}}} \times {{\rm N}_1}}}{{\dfrac{{0.693}}{{{{\left( {{t_{1/2}}} \right)}_2}}} \times {{\rm N}_2}}} \\
\Rightarrow\dfrac{{{R_1}}}{{{R_2}}}= \dfrac{{{{\left( {{t_{1/2}}} \right)}_2}}}{{{{\left( {{t_{1/2}}} \right)}_1}}} \times \dfrac{{{{\rm N}_1}}}{{{{\rm N}_2}}} \\
\Rightarrow\dfrac{{{R_1}}}{{{R_2}}}= \dfrac{2}{1} \times
\dfrac{{\dfrac{{{N_0}}}{4}}}{{\dfrac{{{N_0}}}{2}}} \\
\therefore \dfrac{{{R_1}}}{{{R_2}}}= 1 \\
\]
Hence, it can be concluded that the ratio is \[1:1\].

Note: While solving the problem, you must know that final concentration is a function of time, whereas the initial concentration is an independent factor. To get the answer, you must substitute the formula of decay constant in both the ratio equations, to make it disappear.