(A) The radioactive decay follows the law $N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$, where $T_{1/2}$ is the half-life.Given $N(0) = 1600$ an

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The radioactive decay follows the law $N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$, where $T_{1/2}$ is the half-life.Given $N(0) = 1600$ and $N(8) = 100$.$100 = 1600 \left( \frac{1}{2} \right)^{8/T_{1/2}}$$\frac{1}{16} = \left( \frac{1}{2} \right)^{8/T_{1/2}}$$\left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^{8/T_{1/2}}$$4 = \frac{8}{T_{1/2}} \implies T_{1/2} = 2 \, s$.Now, at $t = 6 \, s$, the number of half-lives elapsed is $n = \frac{6}{2} = 3$.The count rate is $N(6) = 1600 \times \left( \frac{1}{2} \right)^3 = \frac{1600}{8} = 200 \, \text{counts per second}$.

Class 12 Physics Nuclei Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

Medium

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 at $t = 8 \, s$, it was $100$ counts per second. The count rate observed, in counts per second, at $t = 6 \, s$ is close to:

JEE MAIN 2019 All JEE MAIN

AIEEE 2012 All AIEEE

A
$200$
B
$150$
C
$400$
D
$360$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Nuclei

Subtopic

Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

Previous Year

JEE MAIN 2019 Paper All JEE MAIN years →

Previous Year

AIEEE 2012 Paper All AIEEE years →

The half-life period of a radioactive element $X$ is same as the mean life time of another radioactive element $Y$. Initially,they have the same number of atoms. Then:

Difficult

View Solution

For a substance,the average life for $\alpha$-emission is $1620 \ years$ and for $\beta$-emission is $405 \ years$. After how much time will $\frac{1}{4}$ of the material remain due to simultaneous emission?

Difficult

View Solution

The half-life of ${}^{198} {Au}$ is $3 \, \text{days}$. If the atomic weight of ${}^{198} {Au}$ is $198 \, \text{g/mol}$, then the activity of $2 \, \text{mg}$ of ${}^{198} {Au}$ is ..... $\times 10^{12} \, \text{disintegration/second}$.

DifficultJEE MAIN 2021

View Solution

The initial activity of a certain radioactive isotope was measured as $16000 \ counts \ min^{-1}$. Given that the only activity measured was due to this isotope and that its activity after $12 \ h$ was $2000 \ counts \ min^{-1}$,its half-life,in hours,is nearest to

Medium

View Solution

The half-life of a radioactive isotope $X$ is $1.37 \times 10^9 \, y$. It decays to a stable isotope $Y$. In a rock sample from the moon,the ratio of $X$ to $Y$ is $1:7$. The age of the rock is:

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

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 at $t = 8 \, s$, it was $100$ counts per second. The count rate observed, in counts per second, at $t = 6 \, s$ is close to:

Similar Questions

The half-life period of a radioactive element $X$ is same as the mean life time of another radioactive element $Y$. Initially,they have the same number of atoms. Then:

For a substance,the average life for $\alpha$-emission is $1620 \ years$ and for $\beta$-emission is $405 \ years$. After how much time will $\frac{1}{4}$ of the material remain due to simultaneous emission?

The half-life of ${}^{198} {Au}$ is $3 \, \text{days}$. If the atomic weight of ${}^{198} {Au}$ is $198 \, \text{g/mol}$, then the activity of $2 \, \text{mg}$ of ${}^{198} {Au}$ is ..... $\times 10^{12} \, \text{disintegration/second}$.

The initial activity of a certain radioactive isotope was measured as $16000 \ counts \ min^{-1}$. Given that the only activity measured was due to this isotope and that its activity after $12 \ h$ was $2000 \ counts \ min^{-1}$,its half-life,in hours,is nearest to

The half-life of a radioactive isotope $X$ is $1.37 \times 10^9 \, y$. It decays to a stable isotope $Y$. In a rock sample from the moon,the ratio of $X$ to $Y$ is $1:7$. The age of the rock is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module