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Question

(B) According to the law of radioactive decay,the activity $R$ at any time $t$ is given by $R = R_0 e^{-\lambda t}$,where $R_0$ is the initial activit

Vedclass · Accepted Answer

$R_1 = R_2 e^{-\lambda(t_1 - t_2)}$

Vedclass · Answer

(B) According to the law of radioactive decay,the activity $R$ at any time $t$ is given by $R = R_0 e^{-\lambda t}$,where $R_0$ is the initial activity at $t = 0$.At time $t_1$,the activity is $R_1 = R_0 e^{-\lambda t_1}$.At time $t_2$,the activity is $R_2 = R_0 e^{-\lambda t_2}$.Dividing the two equations:$\frac{R_1}{R_2} = \frac{R_0 e^{-\lambda t_1}}{R_0 e^{-\lambda t_2}}$$\frac{R_1}{R_2} = e^{-\lambda t_1} \cdot e^{\lambda t_2} = e^{-\lambda(t_1 - t_2)}$Therefore,$R_1 = R_2 e^{-\lambda(t_1 - t_2)}$.

The activity of a radioactive substance is $R_1$ at time $t_1$ and $R_2$ at time $t_2$. If $\lambda$ is the decay constant,then which of the following is correct?

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