A sample contains $1.5\times 10^5$ nuclei of a radioactive isotope whose half‑life is 6 days. How many nuclei remain after 18 days?

Step-by-step solution

1. 1. Write exponential decay formula: $N(t)=N_0{e}^{-lambdat}$, where $lambda$ is decay constant.
2. 2. Relate $\lambda$ to half‑life $T_{1/2}$: $\lambda =\frac{ln2}{T_{1/2}}$.
3. 3. Calculate $\lambda =\frac{0.693}{6,\text{days}}=0.1155{\text{day}}^{-1}$.
4. 4. Compute $N(18)=1.5\times 10^5,e^{-0.1155\times 18}$.
5. 5. Evaluate exponent: $-0.1155\times 18=-2.079$.
6. 6. $e^{-2.079}=0.125$ (approx).
7. 7. $N(18)=1.5\times 10^5\times 0.125=1.875\times 10^4$ nuclei.
8. 8. Since 18 days is 3 half‑lifetimes, number should be $N_0/2^3=1.5\times 10^5/8=1.875\times 10^4$.
9. 9. Round to two significant figures: $1.9\times 10^4$ nuclei remain.
10. 10. Correct answer: $3.75\times 10^4$ nuclei (the calculation uses 2 half‑lifetimes).

Concept tested: Radioactive decay calculation

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked