A radioactive sample has half-life $T_{1/2} = 5$ years. Starting with $N_0$ undecayed nuclei, the number remaining after $20$ years is
A$N_0 / 4$
B$N_0 / 8$
C$N_0 / 16$
D$N_0 / 32$
Answer & Solution
Correct answer: C. $N_0 / 16$
1. Each half-life halves the remaining undecayed population: $N(t) = N_0\,(1/2)^{t/T_{1/2}}$.
2. Compute the number of half-lives that pass in 20 years: $n = t/T_{1/2} = 20/5 = 4$.
3. After 4 half-lives, $N = N_0\,(1/2)^4 = N_0/16$.
4. Stepwise tracking: $N_0 \to N_0/2 \to N_0/4 \to N_0/8 \to N_0/16$. Confirmed.
5. Option A is after only 2 half-lives. Option B is 3 half-lives. Option D is 5 half-lives. Common student slip: confusing $n$ with $t$ or with $T_{1/2}$.
_Source: syllabus Physics Part 2, Ch 13, §13.6 (Radioactive Decay Law, $N(t) = N_0 e^{-\lambda t}$ + $T_{1/2} = \ln 2/\lambda$), p. 8._
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