
Two radioactive substances X and Y originally have \[{N_1}\]and\[{N_2}\]nuclei respectively. The half-life of X is half of the half-life of Y. After three half-lives of Y, the numbers of nuclei of both are equal. What is the ratio\[\dfrac{{{N_1}}}{{{N_2}}}\]?
A. \[\dfrac{8}{1}\]
B. \[\dfrac{1}{8}\]
C. \[\dfrac{3}{1}\]
D. \[\dfrac{1}{3}\]
Answer
232.8k+ views
Hint:Before we proceed with the problem, it is important to know about the half-life of a radioactive substance. The half-life of a radioactive substance is the amount of time it takes for one-half of the radioactive substance to decay. Now we are able to solve the problem step by step as follows.
Formula Used:
We know that from the question,
After n half-life, no of nuclei undecayed is,
\[N = \dfrac{{{N_0}}}{{{2^n}}}\]…….. (1)
Where, \[N\] is the number of nuclei decayed, \[{N_0}\] is the initial quantity of nuclei and $n$ is the number of half-lives.
Complete step by step solution:
Given, \[t_{(1/2)X} = \dfrac{t_{(1/2)Y}}{2}\]
So, 3 half-lives of Y = 6 half-lives of X. As the number of nuclei left are equal in the two cases.
\[{N_X} = {N_Y}\] after \[3t_{(1/2)Y}\]...............(By data)
\[\Rightarrow {N_X} = \dfrac{{{N_1}}}{{{2^n}}}\] and \[{N_Y} = \dfrac{{{N_2}}}{{{2^n}}}\].............(From the equation (1))
\[\Rightarrow \dfrac{{{N_1}}}{{{2^6}}} = \dfrac{{{N_2}}}{{{2^3}}}\]
After rearranging we get,
\[\dfrac{{{N_1}}}{{{N_2}}} = \dfrac{{{2^6}}}{{{2^3}}}\]
Upon further simplification,we get;
\[\Rightarrow \dfrac{{{N_1}}}{{{N_2}}} = \dfrac{{64}}{8}\]
\[ \therefore \dfrac{{{N_1}}}{{{N_2}}} = \dfrac{8}{1}\]
Therefore, the ratio of nuclei of radioactive substances is \[\dfrac{8}{1}\].
Hence, option A is the correct answer
Note: Radioactive decay happens when there is an emission of energy in the form of ionizing radiation. The ionizing radiation that is emitted may include alpha particles etc. In the radioactive decay process, each substance decays at its own rate. Therefore, the half-life of a particular substance is constant and is not affected by any physical conditions like temperature, pressure, etc that occur around it and only depends on the nature of the substance.
Formula Used:
We know that from the question,
After n half-life, no of nuclei undecayed is,
\[N = \dfrac{{{N_0}}}{{{2^n}}}\]…….. (1)
Where, \[N\] is the number of nuclei decayed, \[{N_0}\] is the initial quantity of nuclei and $n$ is the number of half-lives.
Complete step by step solution:
Given, \[t_{(1/2)X} = \dfrac{t_{(1/2)Y}}{2}\]
So, 3 half-lives of Y = 6 half-lives of X. As the number of nuclei left are equal in the two cases.
\[{N_X} = {N_Y}\] after \[3t_{(1/2)Y}\]...............(By data)
\[\Rightarrow {N_X} = \dfrac{{{N_1}}}{{{2^n}}}\] and \[{N_Y} = \dfrac{{{N_2}}}{{{2^n}}}\].............(From the equation (1))
\[\Rightarrow \dfrac{{{N_1}}}{{{2^6}}} = \dfrac{{{N_2}}}{{{2^3}}}\]
After rearranging we get,
\[\dfrac{{{N_1}}}{{{N_2}}} = \dfrac{{{2^6}}}{{{2^3}}}\]
Upon further simplification,we get;
\[\Rightarrow \dfrac{{{N_1}}}{{{N_2}}} = \dfrac{{64}}{8}\]
\[ \therefore \dfrac{{{N_1}}}{{{N_2}}} = \dfrac{8}{1}\]
Therefore, the ratio of nuclei of radioactive substances is \[\dfrac{8}{1}\].
Hence, option A is the correct answer
Note: Radioactive decay happens when there is an emission of energy in the form of ionizing radiation. The ionizing radiation that is emitted may include alpha particles etc. In the radioactive decay process, each substance decays at its own rate. Therefore, the half-life of a particular substance is constant and is not affected by any physical conditions like temperature, pressure, etc that occur around it and only depends on the nature of the substance.
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