The rate of disintegration of a radioactive e | vedclass

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(D) Let $m$ be the mass of the radioactive element at time $t$. According to the problem,the rate of disintegration is proportional to its mass: $\fra

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$\log 2$

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(D) Let $m$ be the mass of the radioactive element at time $t$. According to the problem,the rate of disintegration is proportional to its mass: $\frac{dm}{dt} = -km$ (where $k > 0$ is the decay constant). Separating the variables,we get: $\frac{dm}{m} = -k \, dt$. Integrating both sides: $\int \frac{dm}{m} = -\int k \, dt \implies \ln(m) = -kt + C$. At $t = 0$,the initial mass $m = 6 	ext{ gm}$,so $\ln(6) = C$. Thus,$\ln(m) = -kt + \ln(6) \implies \ln(\frac{m}{6}) = -kt$. We want to find the time $t$ when the mass $m = 3 	ext{ gm}$: $\ln(\frac{3}{6}) = -kt \implies \ln(\frac{1}{2}) = -kt \implies -\ln(2) = -kt \implies t = \frac{\ln(2)}{k}$. Therefore,the time $t$ is proportional to $\log 2$.

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 $6 \text{ gm}$ will disintegrate into its mass of $3 \text{ gm}$ is proportional to

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