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.

Answer 1

The law of variation of the mass of radium as a function of time is given by: $m(t) = m_0 \left(1 - \frac{\alpha}{100}\right)^{\frac{t}{t_0}}$

Answer 2

The decay rate is given by $\frac{dm}{dt} = -\lambda m$. The solution is $m(t) = m_0 e^{-\lambda t}$. Given that $\alpha$% decays in time $t_0$, the remaining mass is $m(t_0) = m_0(1 - \frac{\alpha}{100})$. So, $m_0(1 - \frac{\alpha}{100}) = m_0 e^{-\lambda t_0}$, which implies $1 - \frac{\alpha}{100} = e^{-\lambda t_0}$. Thus, $e^{-\lambda} = (1 - \frac{\alpha}{100})^{\frac{1}{t_0}}$. Substituting this into the decay law: $m(t) = m_0 (e^{-\lambda})^t = m_0 \left(\left(1 - \frac{\alpha}{100}\right)^{\frac{1}{t_0}}\right)^t = m_0 \left(1 - \frac{\alpha}{100}\right)^{\frac{t}{t_0}}$.