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(C) The activity $A$ of a radioactive substance at time $t$ is given by the law of radioactive decay: $A = A_0 e^{-\lambda t}$.Taking the natural loga

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(C) The activity $A$ of a radioactive substance at time $t$ is given by the law of radioactive decay: $A = A_0 e^{-\lambda t}$.Taking the natural logarithm on both sides,we get: $\ln A = \ln A_0 - \lambda t$.This equation represents a straight line with a negative slope $(-\lambda)$ and a $y$-intercept of $\ln A_0$.When more gas is injected at time $T$,the activity $A$ increases abruptly,causing a vertical jump in the value of $\ln A$ at $t=T$.After the injection,the decay continues with the same decay constant $\lambda$,so the slope of the $\ln A$ versus $t$ graph remains the same (i.e.,$-\lambda$).Therefore,the graph should consist of two parallel straight-line segments with a negative slope,separated by a vertical jump at $t=T$. This corresponds to graph $C$.

At time $t=0$ some radioactive gas is injected into a sealed vessel. At time $T$ some more of the gas is injected into the vessel. Which one of the following graphs best represents the logarithm of the activity $A$ of the gas with time $t$?

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