
Two radioactive substances A and B have decay constants $5\lambda $ and $\lambda $ respectively. At $t = 0$ they have the same number of nuclei. The ratio of the number of nuclei of A to those of B will be ${\left( {\dfrac{1}{e}} \right)^2}$ after a time interval.
(A) $\dfrac{1}{{4\lambda }}$
(B) $4\lambda $
(C) $2\lambda $
(D) $\dfrac{1}{{2\lambda }}$
Answer
219.9k+ views
Hint: Radioactivity is the phenomenon of spontaneous disintegration of the atomic nucleus by the emission of highly penetrating radiations. The law of radioactive disintegration states that the rate of disintegration at any instant is directly proportional to the number of atoms of the element present at that instant.
Formula used
$N = {N_0}{e^{ - \lambda t}}$
Where, $N$ stands for the number of atoms at a given instant, ${N_0}$stands for the initial number of atoms, $\lambda $is called the decay constant or the disintegration constant and $t$ stands for the time
Complete step by step answer:
According to the law of radioactive disintegration, we can write the decay equation as
$N = {N_0}{e^{ - \lambda t}}$
Let the number of atoms of A be${N_A}$, its decay constant is given by $5\lambda $
Then we can write that the number of atoms of A is
${N_A} = {N_0}{e^{ - 5\lambda t}}$
Let the number of atoms of B be${N_B}$, its decay constant is given by $\lambda $
Then we can write the number of atoms of B as
${N_B} = {N_0}{e^{ - \lambda t}}$
Taking the ratio of ${N_A}$and${N_B}$, we get
$\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{{{N_0}{e^{ - 5\lambda t}}}}{{{N_0}{e^{ - \lambda t}}}} = {e^{ - 4\lambda t}}$
In the question, it is given that $\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{1}{{{e^2}}} = {e^{ - 2}}$
This means that, ${e^{ - 4\lambda t}} = {e^{ - 2}}$
$\Rightarrow 4\lambda t = 2$
$\Rightarrow t = \dfrac{2}{{4\lambda }} = \dfrac{1}{{2\lambda }} $
So, the ratio of number of nuclei of A to those of B will be ${\left( {\dfrac{1}{e}} \right)^2}$ after a time interval $\dfrac{1}{{2\lambda }}$
The answer is Option (D): $\dfrac{1}{{2\lambda }}$
Note
The disintegration constant represents the probability of an atom to disintegrate. The negative sign in the disintegration constant indicates that the number of atoms decreases with the increase in time. The number of un-disintegrated atoms of a radioactive substance decreases exponentially. $N$ and ${N_0}$ can be replaced by the mass of the material. The S.I. The unit of radioactivity is Becquerel (Bq).
Formula used
$N = {N_0}{e^{ - \lambda t}}$
Where, $N$ stands for the number of atoms at a given instant, ${N_0}$stands for the initial number of atoms, $\lambda $is called the decay constant or the disintegration constant and $t$ stands for the time
Complete step by step answer:
According to the law of radioactive disintegration, we can write the decay equation as
$N = {N_0}{e^{ - \lambda t}}$
Let the number of atoms of A be${N_A}$, its decay constant is given by $5\lambda $
Then we can write that the number of atoms of A is
${N_A} = {N_0}{e^{ - 5\lambda t}}$
Let the number of atoms of B be${N_B}$, its decay constant is given by $\lambda $
Then we can write the number of atoms of B as
${N_B} = {N_0}{e^{ - \lambda t}}$
Taking the ratio of ${N_A}$and${N_B}$, we get
$\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{{{N_0}{e^{ - 5\lambda t}}}}{{{N_0}{e^{ - \lambda t}}}} = {e^{ - 4\lambda t}}$
In the question, it is given that $\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{1}{{{e^2}}} = {e^{ - 2}}$
This means that, ${e^{ - 4\lambda t}} = {e^{ - 2}}$
$\Rightarrow 4\lambda t = 2$
$\Rightarrow t = \dfrac{2}{{4\lambda }} = \dfrac{1}{{2\lambda }} $
So, the ratio of number of nuclei of A to those of B will be ${\left( {\dfrac{1}{e}} \right)^2}$ after a time interval $\dfrac{1}{{2\lambda }}$
The answer is Option (D): $\dfrac{1}{{2\lambda }}$
Note
The disintegration constant represents the probability of an atom to disintegrate. The negative sign in the disintegration constant indicates that the number of atoms decreases with the increase in time. The number of un-disintegrated atoms of a radioactive substance decreases exponentially. $N$ and ${N_0}$ can be replaced by the mass of the material. The S.I. The unit of radioactivity is Becquerel (Bq).
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

States of Matter Chapter For JEE Main Chemistry

Trending doubts
Understanding Uniform Acceleration in Physics

Understanding Atomic Structure for Beginners

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Average and RMS Value in Electrical Circuits

Other Pages
Understanding Entropy Changes in Different Processes

Common Ion Effect: Concept, Applications, and Problem-Solving

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

JEE Main 2025-26 Mock Test: Ultimate Practice Guide for Aspirants

Understanding Excess Pressure Inside a Liquid Drop

CBSE Class 10 Sanskrit Set 4 52 Question Paper 2025 – PDF, Solutions & Analysis

