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(D) The radioactive decay law for activity $A$ is given by $A = A_0 e^{-\lambda t}$,where $A_0 = \lambda N_0$ is the initial activity.Taking the natur

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$D$

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(D) The radioactive decay law for activity $A$ is given by $A = A_0 e^{-\lambda t}$,where $A_0 = \lambda N_0$ is the initial activity.Taking the natural logarithm on both sides:$\ln A = \ln(A_0 e^{-\lambda t})$$\ln A = \ln A_0 + \ln(e^{-\lambda t})$$\ln A = \ln A_0 - \lambda t$This equation is of the form $y = mx + c$,where $y = \ln A$,$x = t$,slope $m = -\lambda$,and intercept $c = \ln A_0$.Since the slope is negative $(-\lambda)$ and the intercept is positive $(\ln A_0)$,the graph of $\log A$ versus $t$ is a straight line with a negative slope. Looking at the provided figure,the line labeled $D$ represents this linear decrease.

The graph which represents the correct variation of logarithm of activity $(\log A)$ versus time $t$ is:

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