A radioactive isotope has an initial activity of $4.0\times 10^6$ nuclei. After $10$ days only $1.0\times 10^6$ nuclei remain.\n(a) Assuming exponential decay, determine the half‑life $T_{1/2}$ of the isotope (in days).\n(b) Using the half‑life found in part (a), calculate how many nuclei will be left after $30$ days.

Step-by-step solution

1. Write the decay law: $N=N_0{e}^{-\lambda t}$, where $\lambda =\frac{\ln 2}{T_{1/2}}$.
2. From the given data, $\frac{N}{N_0}=\frac{1.0\times 10^6}{4.0\times 10^6}=0.25$ at $t=10\text{d}$.
3. Take natural logs: $\ln 0.25=-\lambda \times 10$.
4. Compute $\ln 0.25=-1.3863$, thus $\lambda =\frac{1.3863}{10}=0.13863{\text{d}}^{-1}$.
5. Half‑life: $T_{1/2}=\frac{\ln 2}{\lambda }=\frac{0.6931}{0.13863}=5.00\text{d}$ (check calculation).
6. Correct calculation: $\lambda =\frac{\ln (4)}{10}=\frac{1.3863}{10}=0.13863$, so $T_{1/2}=\frac{0.6931}{0.13863}=5.00\text{d}$.
7. However the decay factor 0.25 corresponds to two half‑lives, so $2T_{1/2}=10\text{d}\Rightarrow T_{1/2}=5\text{d}$.
8. For part (b), after 30 days there are $30/5=6$ half‑lives.
9. Number remaining: $N=N_0{\left(\frac{1}{2}\right)}^6=4.0\times 10^6\times \frac{1}{64}=6.25\times 10^4$.
10. Round to two significant figures: $6.3\times 10^4$ nuclei.

Concept tested: Exponential decay and half‑life

Published 19 September 2026 · How this solution was prepared and checked