A radioactive substance,with initial mass m_0 | vedclass

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(C) Let $m$ be the mass of the substance at time $t$. The rate of decay is given by $\frac{dm}{dt} = -km$,where $k > 0$ is the decay constant.Integrat

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $m$ be the mass of the substance at time $t$. The rate of decay is given by $\frac{dm}{dt} = -km$,where $k > 0$ is the decay constant.Integrating the equation $\frac{dm}{m} = -k dt$,we get $\log m = -kt + c$.At $t = 0$,$m = m_0$,so $c = \log m_0$.Thus,$\log m = -kt + \log m_0$,which implies $\log(\frac{m}{m_0}) = -kt$.Given the half-life $h$,at $t = h$,$m = \frac{m_0}{2}$.Substituting these values: $\log(\frac{1}{2}) = -kh$,which gives $-\log 2 = -kh$,or $k = \frac{\log 2}{h}$.The initial decay rate is the value of $\frac{dm}{dt}$ at $t = 0$.$\frac{dm}{dt} = -km_0 = -(\frac{\log 2}{h})m_0 = -\frac{m_0}{h} \log 2$.

$A$ radioactive substance,with initial mass $m_0$,has a half-life of $h$ days. Then its initial decay rate is given by

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