All subjects · Physics · WPH15 (Unit 5)

Exponential decay equations and log graphs

Because each nucleus has the same chance of decaying, the number left falls exponentially. Logarithms turn the curve into a straight line.

The decay equations

Solving dN/dt = -λN gives

N = N_0e^-λt and A = A_0e^-λt

where N_0 and A_0 are the values at t = 0. The same shape describes capacitor discharge (Unit 4) and gamma absorption.

Logarithmic form

Taking natural logs:

ln N = ln N_0 - λt and ln A = ln A_0 - λt

A graph of ln A against t is a straight line with gradient -λ and y-intercept ln A_0. This is the best way to find λ (and so t_1/2 = ln 2 / λ) from experimental data, because a best-fit line uses all the readings.

Solving for time

To find how long it takes to fall to a fraction f: e^-λt = f, so t = ln(1/f) / λ. In radiocarbon dating, the ratio of carbon-14 activity in a dead sample to that in living material gives f, and the half-life of carbon-14 (5730 years) gives λ.

Worked example

A source has an initial activity of 2000 Bq and a half-life of 6.0 h. Find its activity after 4.0 h.

λ = 0.693 / 6.0 = 0.1155 h^-1
A = 2000e^(-0.1155 x 4.0) = 2000 x 0.630

Answer: A = 1260 Bq (about 1300 Bq)

Key idea: N = N_0e^-λt, A = A_0e^-λt; ln A against t has gradient -λ; t = ln(1/f)/λ.

Check yourself

A sample has N_0 = 1.0 x 10¹² nuclei and λ = 0.10 day^-1. Find N after 10 days, in units of 10¹¹, to 3 s.f.
Show answer
3.68 x 10¹¹. N = 1.0 x 10¹² x e^-1.0 = 3.68 x 10¹¹.
A source has initial activity 2000 Bq and half-life 6.0 h. Find its activity after 4.0 h, in Bq, to 3 s.f.
Show answer
1260 Bq. λ = 0.693 / 6.0 = 0.1155 h^-1; A = 2000e^-0.462 = 2000 x 0.630 = 1260 Bq.
Wood from an old boat has 30% of the carbon-14 activity of living wood. The half-life of carbon-14 is 5730 years. Find its age in years, to 3 s.f.
Show answer
9950 years. λ = 0.693 / 5730 = 1.21 x 10^-4 per year; t = ln(1 / 0.30) / λ = 1.204 / 1.21 x 10^-4 = 9950 years.
A graph of ln A against t (in minutes) has a gradient of -0.0231 min^-1. Find the half-life in minutes, to 3 s.f.
Show answer
30.0 min. λ = 0.0231 min^-1, so t_1/2 = 0.693 / 0.0231 = 30.0 min.
The activity of a source falls from 800 Bq to 500 Bq in 20 minutes. Find λ in min^-1, to 3 s.f.
Show answer
0.0235 min^-1. 500 = 800e^-20λ, so λ = ln(1.6) / 20 = 0.470 / 20 = 0.0235 min^-1.
What does a graph of ln A against t look like for a radioactive source?
- A curve that falls steeply then levels off
- A straight line through the origin with gradient λ
- A straight line with gradient -λ
- A horizontal line
Show answer
A straight line with gradient -λ. ln A = ln A_0 - λt is in the form y = c + mx.
What does radiocarbon dating compare?
- The carbon-14 activity of a dead sample with that of living material
- The mass of carbon in a sample with its volume
- The number of carbon-12 atoms with the number of oxygen atoms
- The temperature of a sample with its activity
Show answer
The carbon-14 activity of a dead sample with that of living material. Living things keep a steady C-14 level; after death it decays.

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Last updated 7 November 2026. Spotted a mistake? Tell us.