Question

It is known that the decay rate of radium is directly proportional to its quantity at each given instant. Find the law of variation of a mass of radium as a function of time if at t = 0 , the mass of the radius was $m_0$ and during time $t_0\alpha$ % of the original mass of radium decay.

• A. $m=m_0{e}^{-kt}$ where $k=\frac{1}{t_0}ln(1-\frac{\alpha }{100})$
• B. $m=m_0{e}^{-k(t-1)}$ where $k=\frac{1}{t_0}ln(1+\frac{\alpha }{100})$
• C. $m=m_0{e}^{-k(t-1)}$ where $k=\frac{1}{t_0}ln(1-\frac{\alpha }{100})$
• D. $m=m_0{e}^{-kt}$ where $k=\frac{1}{t_0}ln(1+\frac{\alpha }{100})$

Answer 1

The correct option is A $m=m_0{e}^{-kt}$ where $k=\frac{1}{t_0}ln(1-\frac{\alpha }{100})$

$\frac{dm}{dt}=-\lambda m$
$\frac{1}{m}dm=-\lambda dt$
$lnm=-\lambda t+c$
$m=k{e}^{-\lambda t}$ $(att=0,m=m0)$
$\Rightarrow k=m_0$
$m=m_0{e}^{-\lambda t}$ $(att=t_0,m=m_0-\frac{\alpha m_o}{100})$
$\Rightarrow \lambda =\frac{-1}{t_0}ln(1-\frac{\alpha }{100})$