A radio active substance has half-life of $h$ days, then its initial decay rate is given by Note that at…

A radio active substance has half-life of $h$ days, then its initial decay rate is given by Note that at $\mathrm{t}=0, \mathrm{M}=\mathrm{m}_{\mathrm{o}}$
  1. $\frac{\mathrm{m}_{\mathrm{o}}}{\mathrm{h}}(\log 2)$
  2. $\left(\mathrm{m}_0 \mathrm{~h}\right)(\log 2)$
  3. $-\frac{\mathrm{m}_{\mathrm{o}}}{\mathrm{h}}(\log 2)$
  4. $\left(-\mathrm{m}_0 \mathrm{~h}\right)(\log 2)$

Solution

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 $\mathrm{t}=0, \mathrm{M}=\mathrm{m}_0$ $\begin{array}{ll} \therefore \quad & \log \mathrm{m}_0=0+c \\ & \Rightarrow \mathrm{c}=\log \mathrm{m}_0 \\ \therefore \quad & \log \mathrm{M}=-\mathrm{kt}+\log \mathrm{m}_0 \\ & \Rightarrow \log \frac{M}{\mathrm{~m}_0}=-\mathrm{kt} \end{array}$ $\begin{gathered} \text { When } \mathrm{t}=\mathrm{h}, \mathrm{M}=\frac{1}{2} \mathrm{~m}_0 \\ \therefore \quad \log \left(\frac{\frac{1}{2} \mathrm{~m}_0}{\mathrm{~m}_0}\right)=-\mathrm{kh} \\ \Rightarrow \log \frac{1}{2}=-\mathrm{kh} \\ \Rightarrow \log 2=\mathrm{kh} \\ \Rightarrow \mathrm{k}=\frac{\log 2}{\mathrm{~h}} ...(i)\\ \mathrm{Initial} \text { decay rate, } \\ \frac{\mathrm{dM}}{\mathrm{dt}}=-\mathrm{km} \mathrm{~m}_0 \\ =\frac{-\mathrm{m}_0}{\mathrm{~h}} \log 2...[From(i)] \end{gathered}$

Asked in: MHT CET 2024 (15 May Shift 1)

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