An initially prepared sample contains 5.0$\times 10^5$ atoms of a radioactive isotope. After 24 days only 6.25$\times 10^4$ atoms remain. Determine the half‑life of the isotope.

Step-by-step solution

1. Write the decay law: N = $N_0$\,(1/2)${}^{t/T_{1/2}}$.
2. Form the ratio N/$N_0$=6.25$\times 10^4$/5.0$\times 10^5$=0.125.
3. Recognise that 0.125 = (1/2)^{3}.
4. Set the exponent equal to t/$T_{1/2}$: (1/2)${}^{t/T_{1/2}}$=(1/2)^{3} $\Rightarrow$ t/$T_{1/2}$=3.
5. Insert the given time t=24 days: 24/$T_{1/2}$=3 $\Rightarrow$ $T_{1/2}$=24/3=8$\text{ days}$.

Concept tested: Half‑life from decay data

Published 8 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked