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(B) The activity of a radioactive source is given by $A(t) = A_0 e^{-\lambda t}$,where $\lambda = \frac{1}{	au}$ is the decay constant.For source $P$

Vedclass · Accepted Answer

$2$

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(B) The activity of a radioactive source is given by $A(t) = A_0 e^{-\lambda t}$,where $\lambda = \frac{1}{	au}$ is the decay constant.For source $P$,$\lambda_P = \frac{1}{	au}$. For source $Q$,$\lambda_Q = \frac{1}{2	au}$.The rate of change of activity is $R = -\frac{dA}{dt} = -\frac{d}{dt}(A_0 e^{-\lambda t}) = A_0 \lambda e^{-\lambda t}$.Given that $A_0$ is the same for both at $t = 0$,we have $R_P(t) = A_0 \lambda_P e^{-\lambda_P t}$ and $R_Q(t) = A_0 \lambda_Q e^{-\lambda_Q t}$.At $t = 2	au$:$R_P = A_0 (\frac{1}{	au}) e^{-(\frac{1}{	au})(2	au)} = \frac{A_0}{	au} e^{-2}$.$R_Q = A_0 (\frac{1}{2	au}) e^{-(\frac{1}{2	au})(2	au)} = \frac{A_0}{2	au} e^{-1}$.Taking the ratio: $\frac{R_P}{R_Q} = \frac{\frac{A_0}{	au} e^{-2}}{\frac{A_0}{2	au} e^{-1}} = 2 \cdot \frac{e^{-2}}{e^{-1}} = 2 e^{-1} = \frac{2}{e}$.Comparing this with $\frac{n}{e}$,we get $n = 2$.

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