
A radioactive substance of half-life 6 min is placed near a Geiger counter which is found to register 1024 particles per minute. How many particles per minute will be registered after 42 min?
A) 4/min
B) 8/min
C) 5/min
D) 7/min
Answer
233.1k+ views
Hint: In this question radioactive decay is happening and the numbers of particles are being registered in Geiger counter. We are going to use the formula of the amount of substance left after a radioactive decay. First we will find decay constant $\lambda$. Then we will find the number of particles registered by Geiger counter after 42 min.
Complete step by step solution:
Given:
Half life $\tau = 6\min $
Number of particles registered per minute ${N_o} = 1024$
Time $t = 42\min $
Let the number of particles left after 42 minutes is N
Decay constant $\lambda $ is given by
$\lambda = \dfrac{{\ln 2}}{\tau }$
Putting the value of half life $\tau = 6\min $
$\Rightarrow \lambda = \dfrac{{0.693}}{6}$
$\Rightarrow \lambda = 0.1155$
Now we will find the number of particles left after 42 minutes using following formula,
$\Rightarrow N = {N_o}{e^{ - \lambda t}}$
Putting the values of ${N_o},\lambda {\text{ and t}}$
$\Rightarrow N = 1024{e^{ - 0.1155 \times 42}}$
$\Rightarrow N = 1024{e^{ - 4.851}}$
$\Rightarrow N = 1024 \times 0.00782$
$\Rightarrow N = 8.008$
The number of particles left after 42 minutes is $N = 8.008$.
Note: Geiger counter is a device used in nuclear physics to detect or count the number of particles being emitted from a radioactive source. It is also known as the Geiger Muller counter. It works on the basis of ionization radiation. It means that the particles inside the Geiger Muller tube get ionized and counter detects the particles. The Geiger Muller tube is filled with an inert gas such as Helium, Argon or Neon. The Geiger counter is the best device to detect radiation.
In the given question we have used the formula of radioactive decay. This is a first order equation of radioactive decay. We have found decay constant in the question is a very important variable because it is the probability of decay of substance per minute or it can be defined by the rate at which the population of particles of a radioactive material decreases due to radioactive decay.
Complete step by step solution:
Given:
Half life $\tau = 6\min $
Number of particles registered per minute ${N_o} = 1024$
Time $t = 42\min $
Let the number of particles left after 42 minutes is N
Decay constant $\lambda $ is given by
$\lambda = \dfrac{{\ln 2}}{\tau }$
Putting the value of half life $\tau = 6\min $
$\Rightarrow \lambda = \dfrac{{0.693}}{6}$
$\Rightarrow \lambda = 0.1155$
Now we will find the number of particles left after 42 minutes using following formula,
$\Rightarrow N = {N_o}{e^{ - \lambda t}}$
Putting the values of ${N_o},\lambda {\text{ and t}}$
$\Rightarrow N = 1024{e^{ - 0.1155 \times 42}}$
$\Rightarrow N = 1024{e^{ - 4.851}}$
$\Rightarrow N = 1024 \times 0.00782$
$\Rightarrow N = 8.008$
The number of particles left after 42 minutes is $N = 8.008$.
Note: Geiger counter is a device used in nuclear physics to detect or count the number of particles being emitted from a radioactive source. It is also known as the Geiger Muller counter. It works on the basis of ionization radiation. It means that the particles inside the Geiger Muller tube get ionized and counter detects the particles. The Geiger Muller tube is filled with an inert gas such as Helium, Argon or Neon. The Geiger counter is the best device to detect radiation.
In the given question we have used the formula of radioactive decay. This is a first order equation of radioactive decay. We have found decay constant in the question is a very important variable because it is the probability of decay of substance per minute or it can be defined by the rate at which the population of particles of a radioactive material decreases due to radioactive decay.
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

