Two different radioactive elements with half-lives $T_1$ and $T_2$ have undecayed atoms $N_1$ and $N_2$ respectively present at a given instant. The ratio of their activities at that instant is

  • A
    $\frac{N_1 T_1}{N_2 T_2}$
  • B
    $\frac{N_2 T_2}{N_1 T_1}$
  • C
    $\frac{N_1 T_2}{N_2 T_1}$
  • D
    $\frac{N_1 N_2}{T_1 T_2}$

Explore More

Similar Questions

After $40 \, days$,the $1/16$th part of a radioactive element remains undecayed. What is its half-life (in $, days$)?

In a radioactive disintegration,the ratio of the initial number of atoms to the number of atoms present at an instant of time equal to its mean life is:

Radon $({Rn})$ decays into Polonium $({Po})$ by emitting an $\alpha$-particle with a half-life of $4\, days$. $A$ sample contains $6.4 \times 10^{10}$ atoms of $Rn$. After $12\, days$,the number of atoms of $Rn$ left in the sample will be:

If a radioactive element having a half-life of $30 \ min$ is undergoing beta decay,the fraction of the radioactive element that remains undecayed after $90 \ min$ will be:

If the decay or disintegration constant of a radioactive substance is $\lambda$,then its half-life and mean life are respectively:

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