A sample contains $5.0\times 10^5$ nuclei of a radioactive isotope. After $12$ days only $6.0\times 10^4$ nuclei remain. Assuming exponential decay, find the half‑life of the isotope (in days).

Step-by-step solution

1. Write the exponential decay law $N=N_0{e}^{-kt}$ and identify $N_0=5.0\times 10^5$, $N=6.0\times 10^4$, $t=12$ days.
2. Solve for the decay constant $k$ using $k=\frac{ln(N_0/N)}{t}$.
3. Compute $k$ numerically: $ln(5.0\times 10^5/6.0\times 10^4)/12$.
4. Use the relation $T_{1/2}=\frac{ln2}{k}$ to find the half‑life.
5. Insert the value of $k$ and evaluate $T_{1/2}$.
6. Round to one decimal place.

Concept tested: Half‑life calculation

Published 21 September 2026 · How this solution was prepared and checked