Modern Physics: Real-World Scenarios (6)

Question

A radioactive sample has a half-life of 20 days. Initially it has $4 \times 10^{20}$ atoms. How many atoms remain after 60 days, and what is the activity at that moment?

Solution — Step by Step

Final answer: $N = 5 \times 10^{19}$ atoms, $A \approx 2.0 \times 10^{13}$ Bq

Why This Works

Radioactive decay is a first-order process: the rate at which atoms decay is proportional to how many are left. This gives the exponential law $N(t) = N_0 e^{-\lambda t}$, equivalent to halving every $T_{1/2} = \ln 2 / \lambda$.

The activity (decays per second) is just $\lambda N$, so it shrinks by the same factor of 8 over 3 half-lives.

Alternative Method

For non-integer numbers of half-lives, use $N = N_0 e^{-\lambda t}$ directly. For integer multiples, the powers-of-2 trick is much faster.

Common Mistake