E.3.22 (HL)—Half-life and decay constant

Syllabus
First assessment 2025
Objective
Level
HL

Relate Half-Life to Lambda

HL only

Use the half-life relation

Half-life and decay constant are related by T1/2=ln2λT_{1/2}=\frac{\ln2}{\lambda}. A larger decay constant means a shorter half-life.

T_{1/2}=\frac{\ln 2}{\lambda}\qquad\text{or}\qquad\lambda=\frac{\ln2}{T_{1/2}}

Worked example — convert time first

For T1/2=6.0h=2.16×104sT_{1/2}=6.0\,\mathrm{h}=2.16\times10^4\,\mathrm{s}, λ=0.693/(2.16×104)=3.21×105s1\lambda=0.693/(2.16\times10^4)=3.21\times10^{-5}\,\mathrm{s^{-1}}. The inverse-second unit matches a probability rate.

Convert units first

If half-life is given in days, hours or years but lambda is required in s1\mathrm{s^{-1}}, convert the time to seconds before dividing ln2\ln2 by it.

Check the scale

The product λT1/2\lambda T_{1/2} should equal approximately 0.693. Use this as a quick unit and order-of-magnitude check.

Common trap

Do not use 1/λ1/\lambda as the half-life; it is the characteristic time and differs by the factor ln2\ln2.

E.3.22 (HL) Exam Analysis

HL only

Assessment in practice

1–2 marks
How it is assessed

Questions calculate lambda from a half-life or identify the expression for the time at which a sample has halved.

Command terms

Calculate / Identify

What earns marks

Use T_half=ln2/lambda, convert the half-life to the requested time unit, and retain the correct inverse relationship.

Watch for

Using lambda/ln2 or omitting unit conversion.