Half-life

Activity of a radioactive sample

The activity is the number of decays per second, R = λN. For 1.0 × 10¹⁸ nuclei of iodine-131 (T½ = 8 days), λ = 1.0 × 10⁻⁶ per second, so R = 1.0 × 10¹² Bq = 27.1 Ci.

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.

1000 of 1000 left

Half-life8 dλ = 0.08664 per d

Mean life τ11.54 d1/λ = 1.44 T½

Left after 0 d100%N = 1.0 × 10¹⁸ of 1.0 × 10¹⁸

Activity now1.0 × 10¹² Bq27.1 Ci

Time to 50% left8 d1 half-lives

After 0 d, 100% is left: 1000 of 1000 nuclei.

Show the working

λ = 0.693 / T½ = 0.693 / 8 = 0.08664 per d
Mean life τ = 1 / λ = 1.44 T½ = 11.54 d
N = N₀ e^-λt = N₀ (½)ⁿ with n = t / T½ = 0 / 8 = 0
N / N₀ = (½)ⁿ = 1, so N = 1.0 × 10¹⁸ of 1.0 × 10¹⁸
Activity R = λN = 1.003 × 10⁻⁶ per s × 1.0 × 10¹⁸ = 1.0 × 10¹² Bq (27.1 Ci)
Time to 50% left: t = ln(N₀ / N) / λ = ln(2) / 0.08664 = 8 d

Working

Step by step

λ = 0.693 / T½ = 0.693 / 8 = 0.08664 per d
Mean life τ = 1 / λ = 1.44 T½ = 11.54 d
N = N₀ e^-λt = N₀ (½)ⁿ with n = t / T½ = 0 / 8 = 0
N / N₀ = (½)ⁿ = 1, so N = 1.0 × 10¹⁸ of 1.0 × 10¹⁸
Activity R = λN = 1.003 × 10⁻⁶ per s × 1.0 × 10¹⁸ = 1.0 × 10¹² Bq (27.1 Ci)
Time to 50% left: t = ln(N₀ / N) / λ = ln(2) / 0.08664 = 8 d

Derivation

Why it works

Activity is the number of decays per second: R = -dN/dt.
From dN/dt = -λN, R = λN, with λ per second.
So R falls with the same half-life as N: R = R₀ e^-λt.
The SI unit is the becquerel, 1 Bq = 1 decay per second. The older curie is 1 Ci = 3.7 × 10¹⁰ Bq.

Questions