Half-life

Half-life of a first order reaction

For a first order reaction t½ = 0.693 / k, whatever the starting amount. With k = 1.15 × 10⁻³ per s, t½ = 603 s, and 5 g falls to 3 g in t = ln(5/3) / k = 444 s.

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.

502 of 1000 left

Half-life602.7 sλ = 1.15 × 10⁻³ per s

Mean life τ869.6 s1/λ = 1.44 T½

Left after 600 s50.2%N = 2.51 of 5

Rate now, kN2.88 × 10⁻³per second

Time to 60% left444.2 s0.737 half-lives

After 600 s, 50.2% is left: 502 of 1000 nuclei.

Show the working

t½ = 0.693 / k = 0.693 / 1.15 × 10⁻³ = 602.7 s
Mean life τ = 1 / k = 1.44 T½ = 869.6 s
N = N₀ e^-λt = N₀ (½)ⁿ with n = t / T½ = 600 / 602.7 = 0.9955
N / N₀ = (½)ⁿ = 0.5016, so N = 2.51 of 5
Time to 60% left: t = ln(N₀ / N) / k = ln(1.667) / 1.15 × 10⁻³ = 444.2 s

Working

Step by step

t½ = 0.693 / k = 0.693 / 1.15 × 10⁻³ = 602.7 s
Mean life τ = 1 / k = 1.44 T½ = 869.6 s
N = N₀ e^-λt = N₀ (½)ⁿ with n = t / T½ = 600 / 602.7 = 0.9955
N / N₀ = (½)ⁿ = 0.5016, so N = 2.51 of 5
Time to 60% left: t = ln(N₀ / N) / k = ln(1.667) / 1.15 × 10⁻³ = 444.2 s

Derivation

Why it works

A first order reaction has rate = k[R], the same form as radioactive decay.
Integrating gives [R] = [R]₀ e^-kt, or kt = ln([R]₀ / [R]).
At the half-life [R] = [R]₀ / 2, so t½ = ln 2 / k = 0.693 / k.
t½ does not depend on the starting amount: that is the mark of a first order reaction.

Questions