The decay rate of radium is proportional to t | vedclass

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(B) Let $m$ be the mass of the substance at time $t$. Given that the decay rate is proportional to the amount present, we have $\frac{dm}{dt} = -km$,

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) Let $m$ be the mass of the substance at time $t$. Given that the decay rate is proportional to the amount present, we have $\frac{dm}{dt} = -km$, where $k > 0$. Separating the variables, we get $\frac{dm}{m} = -k dt$. Integrating both sides, we obtain $\ln m = -kt + c$. At $t = 0$, $m = 60$, so $\ln 60 = c$. Thus, $\ln m = -kt + \ln 60$. Given the half-life $T_{1/2} = 1600 	ext{ years}$, at $t = 1600$, $m = \frac{60}{2} = 30$. Substituting these values: $\ln 30 = -1600k + \ln 60$, which gives $1600k = \ln 60 - \ln 30 = \ln 2$. So, $k = \frac{\ln 2}{1600}$. Now, for $t = 3200$, the amount $m$ is given by $\ln m = -\left(\frac{\ln 2}{1600}ight)(3200) + \ln 60$. $\ln m = -2 \ln 2 + \ln 60 = -\ln 4 + \ln 60 = \ln \left(\frac{60}{4}ight) = \ln 15$. Therefore, $m = 15 	ext{ grams}$.

The decay rate of radium is proportional to the amount present at any time $t$. If initially $60 \text{ gms}$ was present and half-life period of radium is $1600 \text{ years}$, then the amount of radium present after $3200 \text{ years}$ is (in $\text{ grams}$)

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