A radio-active substance has a half-life of h days, then its initial decay rate is given by (where…

A radio-active substance has a half-life of h days, then its initial decay rate is given by (where radio-active substance has initial mass $\mathrm{m}_0$ ).
  1. $\frac{m_0}{\mathrm{~h}}(\log 2)$
  2. $\left(\mathrm{m}_0 \mathrm{~h}\right)(\log 2)$
  3. $-\frac{\mathrm{m}_0}{\mathrm{~h}}(\log 2)$
  4. $\quad-\left(m_0 \mathrm{~h}\right)(\log 2)$

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

Asked in: MHT CET 2024 (10 May Shift 1)

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