The activity of a sample of radioactive mater | vedclass

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Question

(C) The activity $A$ of a radioactive sample at any time $t$ is given by $A = \lambda N = \lambda N_0 e^{-\lambda t}$.At time $t_1$, the activity is $

Vedclass · Accepted Answer

$A_2 = A_1 e^{(t_1 - t_2) / T}$

Vedclass · Answer

(C) The activity $A$ of a radioactive sample at any time $t$ is given by $A = \lambda N = \lambda N_0 e^{-\lambda t}$.At time $t_1$, the activity is $A_1 = \lambda N_0 e^{-\lambda t_1}$.At time $t_2$, the activity is $A_2 = \lambda N_0 e^{-\lambda t_2}$.Dividing the two equations, we get $\frac{A_2}{A_1} = \frac{\lambda N_0 e^{-\lambda t_2}}{\lambda N_0 e^{-\lambda t_1}} = e^{-\lambda(t_2 - t_1)} = e^{\lambda(t_1 - t_2)}$.Since the mean life $T = \frac{1}{\lambda}$, we can substitute $\lambda = \frac{1}{T}$.Therefore, $\frac{A_2}{A_1} = e^{(t_1 - t_2) / T}$, which implies $A_2 = A_1 e^{(t_1 - t_2) / T}$.

The activity of a sample of radioactive material is $A_1$ at time $t_1$ and $A_2$ at time $t_2$ $(t_2 > t_1)$. Its mean life is $T$.

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