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(B) The activity $R$ of a radioactive sample is given by $R = R_0 e^{-\lambda t}$.Taking the logarithm on both sides,we get $\log R = \log R_0 - \lamb

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(B) The activity $R$ of a radioactive sample is given by $R = R_0 e^{-\lambda t}$.Taking the logarithm on both sides,we get $\log R = \log R_0 - \lambda t \log e$.Assuming the base of the logarithm is $e$,we have $\log R = \log R_0 - \lambda t$.This equation is in the form of a straight line $y = mx + c$,where the slope $m = -\lambda$.From the given graph,we can pick two points $(t_1, \log R_1) = (8, 8)$ and $(t_2, \log R_2) = (16, 6)$.The slope of the line is $m = \frac{\log R_2 - \log R_1}{t_2 - t_1} = \frac{6 - 8}{16 - 8} = \frac{-2}{8} = -0.25$.Since the slope $m = -\lambda$,we have $-\lambda = -0.25$,which gives the decay constant $\lambda = 0.25 	ext{ min}^{-1}$.The half-life $T_{1/2}$ is given by $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.25} = 2.772 	ext{ minutes}$.Rounding to the nearest value provided in the options,we get $T_{1/2} \approx 3.0 	ext{ minutes}$.

The graph shows the $\log$ of activity $\log R$ of a radioactive material as a function of time $t$ in minutes. The half-life (in minutes) for the decay is closest to

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