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

Vedclass · Accepted Answer

$75000$

Vedclass · Answer

(C) The activity of a radioactive sample is given by $R = R_0 e^{-\lambda t}$.Taking the natural logarithm on both sides,we get $\ln R = \ln R_0 - \lambda t$.Comparing this with the equation of a straight line $y = mx + c$,the slope of the graph is $m = -\lambda$.From the given graph,at $t = 0$,$\ln R_0 = 2$,which implies $R_0 = e^2 = 7.5$.The slope of the graph is $\lambda = -\frac{\Delta(\ln R)}{\Delta t} = -\frac{1 - 2}{10 	imes 10^3 - 0} = \frac{1}{10^4} = 10^{-4} 	ext{ s}^{-1}$.We know that the activity $R_0 = \lambda N_0$,where $N_0$ is the number of undecayed nuclei at $t = 0$.Therefore,$N_0 = \frac{R_0}{\lambda} = \frac{7.5}{10^{-4}} = 7.5 	imes 10^4 = 75000$.

The natural logarithm of the activity $R$ of a radioactive sample varies with time $t$ as shown. At $t=0$,there are $N_0$ undecayed nuclei. Then,$N_0$ is equal to [Take $e^2=7.5$].

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