A radioactive sample has half-life $T_{1/2} = 5$ years. Starting with $N_0$ undecayed nuclei, the number remaining after $20$ years is

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: NCERT Class 12 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._