In a radioactive material,the activity at time $t_1$ is $R_1$ and at a later time $t_2$ it is $R_2$. If the decay constant of the material is $\lambda$,then:
- A$R_1 = R_2$
- B$R_1 = R_2 e^{-\lambda(t_1 - t_2)}$
- C$R_1 = R_2 e^{\lambda(t_1 - t_2)}$
- D$R_1 = R_2 \left( \frac{t_1}{t_2} \right)$
Explore More
Similar Questions
The half-life period of element $X$ is same as the mean life time of element $Y$. Assume initially $X$ and $Y$ have same number of atoms. Then
If $87.5 \%$ of atoms of a radioactive element decay in $6 \ days$, then the fraction of atoms of the element that decay in $8 \ days$ is
From $N$ atoms of a radioactive element,$n$ alpha particles are emitted per second. The half-life of the element is ......
Medium
View SolutionThe decay constant of a radioisotope is $\lambda$. If $A_1$ and $A_2$ are its activities at times $t_1$ and $t_2$ respectively,what is the number of nuclei that have decayed during the time interval $(t_1 - t_2)$?
The half-life of radium is $1620$ years and its atomic weight is $226 \ kg/kmol$. The number of atoms that will decay from its $1 \ g$ sample per second will be (Avogadro's number $N_A = 6.02 \times 10^{26} \ atoms/kmol$)
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo