The rate of disintegration of a radioactive e | vedclass

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Question

(C) Let $m$ be the mass of the radioactive element at time $t$. The rate of disintegration is $\frac{dm}{dt}$,which is proportional to $m$. $\frac{dm}

Vedclass · Accepted Answer

$\log 3$

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(C) Let $m$ be the mass of the radioactive element at time $t$. The rate of disintegration is $\frac{dm}{dt}$,which is proportional to $m$. $\frac{dm}{dt} = -km$,where $k > 0$. Separating variables,we get $\frac{dm}{m} = -k \, dt$. Integrating both sides,$\int \frac{1}{m} \, dm = -k \int dt + C$,which gives $\log m = -kt + C$. Initially,at $t = 0$,$m = 1.5 = \frac{3}{2}$. So,$\log \left(\frac{3}{2}ight) = -k(0) + C$,which implies $C = \log \left(\frac{3}{2}ight)$. The equation becomes $\log m = -kt + \log \left(\frac{3}{2}ight)$,or $\log \left(\frac{m}{3/2}ight) = -kt$,which simplifies to $\log \left(\frac{2m}{3}ight) = -kt$. When $m = 0.5 = \frac{1}{2}$,we have $\log \left(\frac{2 	imes (1/2)}{3}ight) = -kt$. $\log \left(\frac{1}{3}ight) = -kt$. $-\log 3 = -kt$,so $t = \frac{1}{k} \log 3$. Thus,the required time is proportional to $\log 3$.

The rate of disintegration of a radioactive element at time $t$ is proportional to its mass at that time. Then the time during which the original mass of $1.5 \text{ g}$ will disintegrate into its mass of $0.5 \text{ g}$ is proportional to

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