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…

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:
  1. \(R_1=R_2\)
  2. $R_{1}=R_{2} e^{-\lambda\left(t_{1}-t_{2}\right)}$
  3. $R_{1}=R_{2} e^{\lambda\left(t_{1}-t_{2}\right)}$
  4. $R_1 = R_2 \left( \frac{t_2}{t_1} \right)$

Solution

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

Asked in: NEET 2006

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