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-life5 dλ = 0.1386 per d
Mean life τ7.213 d1/λ = 1.44 T½
Left after 15 d12.5%N = 125 of 1,000
Activity now2.01 × 10⁻⁴ Bq5.42 × 10⁻¹⁵ Ci
Time to 25% left10 d2 half-lives
After 15 d, 12.5% is left: 125 of 1000 nuclei.
Show the working
- λ = 0.693 / T½ = 0.693 / 5 = 0.1386 per d
- Mean life τ = 1 / λ = 1.44 T½ = 7.213 d
- N = N₀ e-λt = N₀ (½)n with n = t / T½ = 15 / 5 = 3
- N / N₀ = (½)n = 0.125, so N = 125 of 1,000
- Time to 25% left: t = ln(N₀ / N) / λ = ln(4) / 0.1386 = 10 d