Chemical Kinetics: Real-World Scenarios (2)

Question

The half-life of a radioactive isotope is $20$ years. How long will it take for $87.5\%$ of a sample to decay?

This applies first-order kinetics to nuclear decay — relevant for both chemistry and physics syllabi.

Solution — Step by Step

Final answer: $60$ years.

Why This Works

For first-order processes, time is divided into equal half-life intervals. After $n$ half-lives, the fraction remaining is $(1/2)^n$. So for any “fraction remaining = $1/2^n$” question, the answer is $n \times t_{1/2}$.

This trick avoids computing the rate constant. Use it whenever the fraction is a power of $1/2$ — $50\%, 25\%, 12.5\%, 6.25\%$ correspond to $1, 2, 3, 4$ half-lives.

Alternative Method

Direct rate-equation solution. $\ln(N_0/N) = kt$, so $t = \ln 8 / k$. Compute $k = \ln 2/20 = 0.0347$. Then $t = 2.079/0.0347 = 60$ years. Same answer, longer.

Common Mistake

Treating decay as zero-order and saying “half decays in 20 years, so all decays in 40 years”. Decay is exponential, not linear. The sample never fully disappears — it keeps halving forever.