
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.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

