A radioactive substance decays at a rate proportional to the amount present, that is, \(\displaystyle\frac{dm}{dt} = -km\), where m is the mass (in g) at time t (in years) and k is a positive constant. If half of the initial mass decays in 5 years, then the fraction of the initial mass left after 15 years is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{8}\)
The variables separate:
\( \displaystyle \frac{dm}{m} = -k\,dt \)
Integrating both sides:
\(\log m = -kt + c\)
\(m = m_0e^{-kt}\)
Here \(m_0\) is the initial mass. At t = 5, half is left:
\( \displaystyle m_0e^{-5k} = \frac{m_0}{2},\quad e^{-5k} = \frac{1}{2} \)
At t = 15:
\( \displaystyle e^{-15k} = \left(e^{-5k}\right)^3 = \frac{1}{8} \)
\( \displaystyle \frac{m}{m_0} = \frac{1}{8} \)
The mass halves every 5 years (1, then \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\)); it does not fall by equal amounts, which is why 0 is wrong.