Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:a)150b)360c)200d)400Correct answer is option 'C'. Can you explain this answer?

Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity

Open App

JEE Exam > JEE Questions > Using a nuclear counter the count rate of emi...

Join the Discussion on the App

Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:

a)
150
b)
360
c)
200
d)
400

Correct answer is option 'C'. Can you explain this answer?

FREE This question is part of

JEE Main 2019 January 10 Shift 1 Paper & Solutions

Attempt Test

Attempt this Test

Verified Answer

Using a nuclear counter the count rate of emitted particles from a rad...

at t = 0, A₀ = dN/dt = 1600 C/s
at t = 8s, A = 100 C/s

Therefor half life is t_1/2 = 2 sec
∴ Activity at t = 6 will be 1600 (1/2)³= 200 C/s

View all questions of this test

Most Upvoted Answer

Using a nuclear counter the count rate of emitted particles from a rad...

Given information:
- At t = 0, the count rate was 1600 counts per second.
- At t = 8 seconds, the count rate was 100 counts per second.

Objective:
To find the count rate at t = 6 seconds.

Understanding the problem:
The count rate of emitted particles from a radioactive source decreases with time. In this problem, we are given the count rates at t = 0 and t = 8 seconds, and we need to find the count rate at t = 6 seconds.

Solution:
To solve this problem, we can assume that the count rate follows an exponential decay model. The count rate at any given time (t) can be expressed as:

R(t) = R0 * e^(-λt)

Where:
- R(t) is the count rate at time t.
- R0 is the initial count rate at t = 0.
- λ is the decay constant.

Step 1: Finding the decay constant (λ):
We can find the decay constant (λ) using the given information. At t = 0, the count rate is 1600 counts per second. Substituting these values into the equation, we get:

1600 = R0 * e^(-λ * 0)
1600 = R0 * e^0
1600 = R0

So, R0 = 1600.

At t = 8 seconds, the count rate is 100 counts per second. Substituting these values into the equation, we get:

100 = 1600 * e^(-λ * 8)

To find the value of λ, we need to solve this equation.

Step 2: Solving the equation:
Dividing both sides of the equation by 1600:

100/1600 = e^(-λ * 8)

0.0625 = e^(-8λ)

Taking the natural logarithm (ln) of both sides:

ln(0.0625) = ln(e^(-8λ))

ln(0.0625) = -8λ

Now, we can solve this equation to find the value of λ.

Step 3: Calculating the count rate at t = 6 seconds:
Now that we have the value of λ, we can calculate the count rate at t = 6 seconds using the equation:

R(t) = R0 * e^(-λt)

Substituting the known values:

R(6) = 1600 * e^(-λ * 6)

Calculating this value will give us the count rate at t = 6 seconds.

Conclusion:
By following the steps outlined above, we can find that the count rate observed at t = 6 seconds is close to 200 counts per second (option C).

Answered by Infinity Academy · View profile

Attention JEE Students!

To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.

Physics for JEE Main & Advanced

Mock Tests for JEE Main and Advanced 2025

Chemistry for JEE Main & Advanced

Mathematics (Maths) for JEE Main & Advanced

View all answers Start learning for free

< Previous Question Next Question >

Explore Courses for JEE exam

Similar JEE Doubts

The French physicist Louis de-Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength of a material particle, its linear momentum p and planck constant h.The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they,(i) have lower velocities(ii) have no electrical and magnetic fields and(iii) are not emitted by the particle under consideration.The experimental confirmation of the deBroglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by deBroglie.Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron. He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also ture. This is summed up in what we now call the Heisenberg uncertainty principle : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of undertainty in the position, x and the uncertainity in the momentum (mv) must be greater than or equal to h/4. i.e.Q. The transition, so that the de - Broglie wavelength of electron becomes 3 times of its initial value in He+ ion will be

View answer

Can you explain the answer of this question below:The French physicist Louis de-Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength of a material particle, its linear momentum p and planck constant h.The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they,(i) have lower velocities(ii) have no electrical and magnetic fields and(iii) are not emitted by the particle under consideration.The experimental confirmation of the deBroglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by deBroglie.Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron. He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also ture. This is summed up in what we now call the Heisenberg uncertainty principle : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of undertainty in the position, x and the uncertainity in the momentum (mv) must be greater than or equal to h/4. i.e.Q.The correct order of wavelength of Hydrogen (1H1), Deuterium (1H2) and Tritium (1H3) moving with same kinetic energy is :A:H D TB:H = D = TC:H D TD:H D TThe answer is a.

View answer

The French physicist Louis de-Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength of a material particle, its linear momentum p and planck constant h.The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they,(i) have lower velocities(ii) have no electrical and magnetic fields and(iii) are not emitted by the particle under consideration.The experimental confirmation of the deBroglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by deBroglie.Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron. He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also ture. This is summed up in what we now call the Heisenberg uncertainty principle : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of undertainty in the position, x and the uncertainity in the momentum (mv) must be greater than or equal to h/4. i.e.Q. If the uncertainty in velocity position is same, then the uncertainty in momentum will be

View answer

We have two radioactive nucleiAandB. Both convert into a stable nucleusC. NucleusAconverts intoCafter emitting twoαα-particles and threeβ-particles. NucleusBconverts intoCafter emitting oneαα-particle and fiveβ-particles. At timet=0, nuclei ofAare 4N0and that ofBareN0. Half life ofA(into the conversion ofC) is 1min and that ofBis 2min. Initially number of nuclei ofCare zero.Q. What are number of nuclei ofC, when number of nuclei ofAandBare equal?

View answer

is known to undergo radioactive decay to formby emitting alpha and beta particles. A rock initially contained 68 × 10–6 g ofIf the number of alpha particles that it would emit during its radioactive decay ofin three half-lives is Z × 1018, then what is the value of Z? Correct answer is '1.20'. Can you explain this answer?

View answer

Top Courses for JEEView all

Physics for JEE Main & Advanced

Mock Tests for JEE Main and Advanced 2025

Chemistry for JEE Main & Advanced

Mathematics (Maths) for JEE Main & Advanced

Chapter-wise Tests for JEE Main & Advanced

Top Courses for JEE

View all

Share with a Friend

Answer this doubt

Question Description
Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:a)150b)360c)200d)400Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:a)150b)360c)200d)400Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:a)150b)360c)200d)400Correct answer is option 'C'. Can you explain this answer?.

Solutions for Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:a)150b)360c)200d)400Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:a)150b)360c)200d)400Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:a)150b)360c)200d)400Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:a)150b)360c)200d)400Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:a)150b)360c)200d)400Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:a)150b)360c)200d)400Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.

Explore Courses for JEE exam