The activities of three radioactive substance | vedclass

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Question

(B) The activity of a radioactive substance is given by $R = R_{0} e^{-\lambda t}$.Taking the natural logarithm on both sides,we get $\ln R = \ln R_{0

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$4: 3: 1$

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(B) The activity of a radioactive substance 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 slope $\lambda$ for each substance is:For $A$: $\lambda_{A} = \frac{6 - 0}{10 - 0} = 0.6 = \frac{3}{5}$.For $B$: $\lambda_{B} = \frac{4 - 0}{5 - 0} = 0.8 = \frac{4}{5}$.For $C$: $\lambda_{C} = \frac{2 - 0}{5 - 0} = 0.4 = \frac{2}{5}$.The half-life is given by $T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$.Thus,$T_{\frac{1}{2}}(A) : T_{\frac{1}{2}}(B) : T_{\frac{1}{2}}(C) = \frac{1}{\lambda_{A}} : \frac{1}{\lambda_{B}} : \frac{1}{\lambda_{C}} = \frac{5}{3} : \frac{5}{4} : \frac{5}{2}$.Multiplying by the $LCM$ of the denominators $(12)$,we get $20 : 15 : 30$,which simplifies to $4 : 3 : 6$.

The activities of three radioactive substances $A, B$ and $C$ are represented by the curves $A, B$ and $C$ in the figure. Then their half-lives $T_{\frac{1}{2}}(A) : T_{\frac{1}{2}}(B) : T_{\frac{1}{2}}(C)$ are in the ratio:

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