Radioactive decay
N = N₀ e-λt, T½ = 0.693 / λ, mean life τ = 1 / λ and activity R = λN in becquerel. The same law fits a first order reaction, with k in place of λ. The dots are 1000 nuclei.
Half-life5730 yλ = 1.21 × 10⁻⁴ per y
Mean life τ8267 y1/λ = 1.44 T½
Left after 11460 y25%N = 250 of 1,000
Activity now9.58 × 10⁻¹⁰ Bq2.59 × 10⁻²⁰ Ci
Time to 25% left11460 y2 half-lives
After 11460 y, 25% is left: 250 of 1000 nuclei.
Show the working
- λ = 0.693 / T½ = 0.693 / 5730 = 1.21 × 10⁻⁴ per y
- Mean life τ = 1 / λ = 1.44 T½ = 8267 y
- N = N₀ e-λt = N₀ (½)n with n = t / T½ = 11460 / 5730 = 2
- N / N₀ = (½)n = 0.25, so N = 250 of 1,000
- Time to 25% left: t = ln(N₀ / N) / λ = ln(4) / 1.21 × 10⁻⁴ = 11460 y