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(B) The radioactive decay law is given by $R = R_0 e^{-\lambda t}$.Taking the natural logarithm on both sides,we get $\ln R = \ln R_0 - \lambda t$.Thi

Vedclass · Accepted Answer

$4.62$

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(B) The radioactive decay law is given by $R = R_0 e^{-\lambda t}$.Taking the natural logarithm on both sides,we get $\ln R = \ln R_0 - \lambda t$.This is the equation of a straight line $y = mx + c$,where the slope $m = -\lambda$.From the graph,the line passes through $(0, 6)$ and $(40, 0)$.The slope is $	ext{slope} = \frac{0 - 6}{40 - 0} = -\frac{6}{40} = -0.15$.Thus,$-\lambda = -0.15$,which gives the decay constant $\lambda = 0.15 \, 	ext{sec}^{-1}$.The half-life $t_{1/2}$ is given by $t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.15} = 4.62 \, 	ext{sec}$.

For a certain radioactive process,the graph between $\ln R$ and $t \, (\text{sec})$ is obtained as shown in the figure. Then the value of the half-life for the unknown radioactive material is approximately $.... \, \text{sec}$.

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