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 to $\alpha$% of the original mass of radium decay.

Answer 1

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

Answer 2

The decay rate is given by $\frac{dm}{dt} = -\lambda m$. The solution is $m(t) = m_0 e^{-\lambda t}$. At $t=t_0$, $\alpha\%$ has decayed, so the remaining mass is $m(t_0) = m_0(1 - \frac{\alpha}{100})$. Substituting this into the decay law: $m_0(1 - \frac{\alpha}{100}) = m_0 e^{-\lambda t_0}$. Thus, $e^{-\lambda t_0} = 1 - \frac{\alpha}{100}$. We can write $e^{-\lambda t} = (e^{-\lambda t_0})^{t/t_0} = \left(1 - \frac{\alpha}{100}\right)^{t/t_0}$. Therefore, the law of variation is $m(t) = m_0 \left(1 - \frac{\alpha}{100}\right)^{t/t_0}$.