The decay rate of radium is proportional to the amount present at any time $t$. If initially 60 gms was present and half life period of radium is 1600 years, then the amount of radium present after 3200 years is

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Aicharya · Accepted Answer

Let m be the mass of substance at time t . Then $\begin{aligned} & \frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{km}, \text { where } \mathrm{k}\gt0 \\ & \Rightarrow \frac{\mathrm{dm}}{\mathrm{~m}}=-\mathrm{kdt} \end{aligned}$
Integrating on both sides, we get $\log m=-k t+c$ when $\mathrm{t}=0, \mathrm{~m}=60 \mathrm{gms}$ $\begin{array}{ll} \therefore \quad & \log 60=-k(0)+c \\ & \Rightarrow \mathrm{c}=\log 60 \end{array}$ $\begin{array}{ll} & \Rightarrow \mathrm{c}=\log 60 \\ \therefore \quad & \log \mathrm{~m}=-\mathrm{kt}+\log 60 \\ \therefore \quad & \text { when } \mathrm{t}=1600, \mathrm{~m}=\frac{60}{2}=30 \mathrm{gms} \\ \therefore \quad & \log 30=-1600 \mathrm{k}+\log 60 \\ & \Rightarrow 1600 \mathrm{k}=\log 2 \\ & \Rightarrow \mathrm{k}=\frac{1}{1600} \log 2 \end{array}$
Equation (i) becomes $\log m=\frac{-1}{1600}(\log 2) t+\log 60$ When $t=3200$ years $\begin{aligned} & \log \mathrm{m}=\frac{-1}{1600} \times \log 2 \times 3200+\log 60 \\ & \log \mathrm{~m}=-\log 4+\log 60 \\ & \Rightarrow \log \mathrm{~m}=\log \frac{60}{4} \\ & \Rightarrow \mathrm{~m}=15 \text { grams } \end{aligned}$

The decay rate of radium is proportional to the amount present at any time $t$. If initially 60 gms was…

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