The decay constant of a radioactive element is $3 \times 10^{-6} \ min^{-1}$. Its half-life is

  • A
    $2.31 \times 10^5 \ min$
  • B
    $2.31 \times 10^6 \ min$
  • C
    $2.31 \times 10^{-6} \ min$
  • D
    $2.31 \times 10^{-7} \ min$

Explore More

Similar Questions

The decay of $_{92}U^{235}$ is a reaction of .......... order.

The $C^{14}$ to $C^{12}$ ratio in a wooden article is $13\%$ that of the fresh wood. Calculate the age of the wooden article. Given that the half-life of $C^{14}$ is $5770 \ years$.

Difficult
View Solution

If $n_t$ number of radioatoms are present at time $t$, the following expression will be a constant:

The half-life of a radioactive substance is $120$ days. After $480$ days,the amount of $4 \ g$ of the substance remaining will be: (in $g$)

Half-life is the time in which $50\%$ of a radioactive element disintegrates. Carbon-$14$ disintegrates $50\%$ in $5770$ years. Find the half-life of carbon-$14$ in years.

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