A radioactive substance has a half-life of h | vedclass

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(C) Let $M$ be the mass of the substance at time $t$. The rate of decay is given by the differential equation: $\frac{dM}{dt} = -kM$,where $k > 0$. In

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$-\frac{m_0}{h} \ln 2$

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(C) Let $M$ be the mass of the substance at time $t$. The rate of decay is given by the differential equation: $\frac{dM}{dt} = -kM$,where $k > 0$. Integrating this,we get $\ln M = -kt + c$. At $t = 0, M = m_0$,so $\ln m_0 = c$. Thus,$\ln M = -kt + \ln m_0$,which simplifies to $\ln(\frac{M}{m_0}) = -kt$. Given the half-life is $h$,at $t = h$,$M = \frac{m_0}{2}$. Substituting these values: $\ln(\frac{1}{2}) = -kh \Rightarrow -\ln 2 = -kh \Rightarrow k = \frac{\ln 2}{h}$. The initial decay rate is the value of $\frac{dM}{dt}$ at $t = 0$. $\frac{dM}{dt} = -k m_0 = -(\frac{\ln 2}{h}) m_0 = -\frac{m_0}{h} \ln 2$.

$A$ radioactive substance has a half-life of $h \ days$. Its initial decay rate is given by (Note that at $t = 0, M = m_0$):

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