The activity of a sample of a radioactive mat | vedclass

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(C) The activity of a radioactive sample at any time $t$ is given by the law of radioactive decay: $A = A_0 e^{-\lambda t}$.Since the mean life $T = 1

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The activity of a radioactive sample at any time $t$ is given by the law of radioactive decay: $A = A_0 e^{-\lambda t}$.Since the mean life $T = 1/\lambda$,we can write the equation as $A = A_0 e^{-t/T}$.At time $t_1$,the activity is $A_1 = A_0 e^{-t_1/T}$. From this,we find $A_0 = A_1 e^{t_1/T}$.At time $t_2$,the activity is $A_2 = A_0 e^{-t_2/T}$.Substituting the expression for $A_0$ into the equation for $A_2$,we get:$A_2 = (A_1 e^{t_1/T}) e^{-t_2/T} = A_1 e^{(t_1 - t_2)/T}$.Thus,the correct relation is $A_2 = A_1 e^{(t_1 - t_2)/T}$.

The activity of a sample of a radioactive material is ${A_1}$ at time ${t_1}$ and ${A_2}$ at time ${t_2}$ $({t_2} > {t_1})$. If its mean life is $T$,then:

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