A sample contains $3.0\times 10^5$ atoms of a radioactive isotope. After $12$ days only $6.0\times 10^4$ atoms remain. Find the half‑life of the isotope.

Step-by-step solution

1. Write the exponential decay law $N=N_0,2^{-t/T_{1/2}}$ and substitute $N_0=3.0\times 10^5$, $N=6.0\times 10^4$, $t=12$ days.
2. Compute the ratio $\frac{N}{N_0}=\frac{6.0\times 10^4}{3.0\times 10^5}=0.20$.
3. Rearrange: $0.20=2^{-12/T_{1/2}}$.
4. Take natural logs: $ln0.20=(-12/T_{1/2})ln2$.
5. Solve for $T_{1/2}=-,\frac{12ln2}{ln0.20}$.
6. Evaluate numerically: $T_{1/2}approx5.17$ days.

Concept tested: exponential decay, half‑life

Published 15 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked