(C) The number of days from $January \, 1^{st}$ to $January \, 24^{th}$ is $23 \, days$.The number of half-lives $n$ is calculated as $n = \frac{23}{8

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

(C) The number of days from $January \, 1^{st}$ to $January \, 24^{th}$ is $23 \, days$.The number of half-lives $n$ is calculated as $n = \frac{23}{8.04} \approx 2.86$.Since $n After $3$ half-lives,the activity would be $A = A_0 \times (1/2)^3 = 600 \times (1/8) = 75 \, Bq$.Because the elapsed time $(23 \, days)$ is less than $3$ half-lives $(3 \times 8.04 = 24.12 \, days)$,the remaining activity must be greater than $75 \, Bq$.

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

Medium

The nuclide $^{133}I$ is radioactive,with a half-life of $8.04 \, days$. At noon on $January \, 1$,the activity of a certain sample is $600 \, Bq$. The activity at noon on $January \, 24$ will be

A
$75 \, Bq$
B
Less than $75 \, Bq$
C
More than $75 \, Bq$
D
$150 \, Bq$

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

The counting rate observed from a radioactive source at $t=0$ second was $1600$ counts per second and at $t=8$ seconds it was $100$ counts per second. The counting rate observed,as counts per second at $t=6$ seconds,will be

Difficult

View Solution

In a radioactive sample,there are $1.414 \times 10^6$ active nuclei. If they reduce to $10^6$ within $10 \text{ min}$,then the half-life of this sample will be ....... $\text{min}$.

Medium

View Solution

The activity of a radioactive sample is $1.6 \, curie$ and its half-life is $2.5 \, days$. Its activity after $10 \, days$ will be .......... $curie$.

Medium

View Solution

The variation of decay rate with the number of active nuclei is correctly shown in which graph?

EasyMHT CET 2020

View Solution

The decay constant of radium is $4.28 \times 10^{-4}$ per year. Its half-life will be .......... $years$.

Easy

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

The nuclide $^{133}I$ is radioactive,with a half-life of $8.04 \, days$. At noon on $January \, 1$,the activity of a certain sample is $600 \, Bq$. The activity at noon on $January \, 24$ will be

Similar Questions

The counting rate observed from a radioactive source at $t=0$ second was $1600$ counts per second and at $t=8$ seconds it was $100$ counts per second. The counting rate observed,as counts per second at $t=6$ seconds,will be

In a radioactive sample,there are $1.414 \times 10^6$ active nuclei. If they reduce to $10^6$ within $10 \text{ min}$,then the half-life of this sample will be ....... $\text{min}$.

The activity of a radioactive sample is $1.6 \, curie$ and its half-life is $2.5 \, days$. Its activity after $10 \, days$ will be .......... $curie$.

The variation of decay rate with the number of active nuclei is correctly shown in which graph?

The decay constant of radium is $4.28 \times 10^{-4}$ per year. Its half-life will be .......... $years$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module