Question

A mixture consists of two radioactive materials \[{A_1}\] and \[{A_2}\]with half-lives of \[20\,{\text{s}}\] and \[10\,{\text{s}}\] respectively. Initially the mixture has \[40\,{\text{g}}\] of \[{A_1}\] and \[160\,{\text{g}}\] of \[{A_2}\]. The active amount of the two in the mixture will becomes equal after:
A. \[20\,{\text{s}}\]
B. \[40\,{\text{s}}\]
C. \[60\,{\text{s}}\]
D. \[80\,{\text{s}}\]

Answer 1

Hint: Use the formula for the population of radioactive material at time t. This formula gives the relation between initial population of radioactive material, time and half-life of the radioactive material. Rewrite this equation for both the materials in the mixture, equate these equations and determine the time t.

Formula used:
The population \[N\] of the radioactive material at time \[t\] is given by
\[ \Rightarrow N = {N_0}{\left( {\dfrac{1}{2}} \right)^{t/T}}\] …… (1)
Here, \[{N_0}\] is the initial population of the radioactive material and \[T\] is the half-life of the radioactive material.

Complete step by step solution:
We have given that the half-life of material \[{A_1}\] is \[20\,{\text{s}}\] and the initial population of radioactive material \[{A_1}\] is \[40\,{\text{g}}\].
\[ \Rightarrow{T_1} = 20\,{\text{s}}\]
\[ \Rightarrow{N_{01}} = 40\,{\text{g}}\]
We have given that the half-life of material \[{A_2}\] is \[10\,{\text{s}}\] and the initial population of radioactive material \[{A_2}\] is \[160\,{\text{g}}\].
\[ \Rightarrow{T_2} = 10\,{\text{s}}\]
\[ \Rightarrow{N_{02}} = 160\,{\text{g}}\]
Rewrite equation (1) for radioactive material \[{A_1}\].
\[ \Rightarrow{N_1} = {N_{01}}{\left( {\dfrac{1}{2}} \right)^{t/{T_1}}}\]
Here, \[{N_1}\] is the population of the radioactive material \[{A_1}\] at time \[t\].
Substitute \[40\,{\text{g}}\] for \[{N_{01}}\] and \[20\,{\text{s}}\] for \[{T_1}\] in the above equation.
\[ \Rightarrow{N_1} = \left( {40\,{\text{g}}} \right){\left( {\dfrac{1}{2}} \right)^{t/\left( {20\,{\text{s}}} \right)}}\]
Rewrite equation (1) for radioactive material \[{A_2}\].
\[ \Rightarrow{N_2} = {N_{02}}{\left( {\dfrac{1}{2}} \right)^{t/{T_2}}}\]
Here, \[{N_2}\] is the population of the radioactive material \[{A_2}\] at time \[t\].
Substitute \[160\,{\text{g}}\] for \[{N_{02}}\] and \[10\,{\text{s}}\] for \[{T_2}\] in the above equation.
\[ \Rightarrow{N_2} = \left( {160\,{\text{g}}} \right){\left( {\dfrac{1}{2}} \right)^{t/\left( {10\,{\text{s}}} \right)}}\]
Let us assume that the active amounts of the radioactive materials \[{N_1}\] and \[{N_2}\] in the mixture is the same at time \[t\].
\[ \Rightarrow{N_1} = {N_2}\]
Substitute \[\left( {40\,{\text{g}}} \right){\left( {\dfrac{1}{2}} \right)^{t/\left( {20\,{\text{s}}} \right)}}\] for \[{N_1}\] and \[\left( {160\,{\text{g}}} \right){\left( {\dfrac{1}{2}} \right)^{t/\left( {10\,{\text{s}}} \right)}}\] for \[{N_2}\] in the above equation.
\[ \Rightarrow \left( {40\,{\text{g}}} \right){\left( {\dfrac{1}{2}} \right)^{t/\left( {20\,{\text{s}}} \right)}} = \left( {160\,{\text{g}}} \right){\left( {\dfrac{1}{2}} \right)^{t/\left( {10\,{\text{s}}} \right)}}\]
\[ \Rightarrow {\left( {\dfrac{1}{2}} \right)^{t/\left( {20\,{\text{s}}} \right)}} = 4{\left( {\dfrac{1}{2}} \right)^{t/\left( {10\,{\text{s}}} \right)}}\]
\[ \Rightarrow \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^{t/20}}}}{{{{\left( {\dfrac{1}{2}} \right)}^{t/10}}}} = 4\]
\[ \Rightarrow {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{20}} - \dfrac{t}{{10}}}} = 4\]
\[ \Rightarrow {\left( {\dfrac{1}{2}} \right)^{\dfrac{{20t - 10t}}{{200}}}} = 4\]
\[ \Rightarrow {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{20}}}} = 4\]
Take a log on both sides of the above equation.
\[ \Rightarrow \log {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{20}}}} = \log 4\]
\[ \Rightarrow \dfrac{t}{{20}}\log \left( {\dfrac{1}{2}} \right) = \log {2^2}\]
\[ \Rightarrow \dfrac{t}{{20}}\left( {\log 1 - log2} \right) = 2\log 2\]
\[ \Rightarrow \dfrac{t}{{20}}\left( {0 - log2} \right) = 2\log 2\]
\[ \Rightarrow t = - 40\,{\text{s}}\]
\[ \therefore t = 40\,{\text{s}}\]
Therefore, the active amount of radioactive material in the mixture will be equal at time \[40\,{\text{s}}\].

Hence, the correct option is B.

Note: As the time cannot be negative, the negative sign is neglected in the calculation. One can also solve the same question in another way using the formula for decay constant and the decay rate equation for the radioactive material. First determine the decay constants for both the materials and write the decay rate equations for these materials. Equate these two equations and solve it for time t.