Q136 / 200 · NEET 2006 · Physics

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
    \(R_1=R_2\)
  • Correct answer
    B
    \(R_1=R_2 e^{-\lambda\left(t_1-t_2\right)}\)
  • C
    \(R_1=R_2 e^{\lambda\left(t_1-t_2\right)}\)
  • D
    \(\mathrm{R}_1=\mathrm{R}_2\left(\mathrm{t}_2 / \mathrm{t}_1\right)\)

Solution

$R_1=R_0 e^{-\lambda_4}$ and $R_2=R_0 e^{-\lambda \lambda_2}$ $$ \begin{aligned} \Rightarrow \quad \frac{R_1}{R,} & =\frac{e^{-\lambda \lambda_1}}{e^{-\lambda \lambda_2}} \\ & =e^{-\lambda\left(t_1-t_2\right)} \\ \Rightarrow \quad R_1 & =R_2 e^{-\lambda\left(t_1-t_2\right)} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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