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Activity, Half-life and Decay constant

Key idea: Define activity, decay constant and half-life, and solve problems using A=λN, N=N0 e^{-λt}, and t1/2 = ln2/λ (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Analyse random radioactive decay, activity, decay constant and half-life.

1. Definitions (Must Know)

A. Activity, A

The activity, A, of a radioactive sample is the number of decays per unit time.

Unit: becquerel (Bq), where 1 Bq = 1 decay s⁻¹.

B. Decay constant, λ

The decay constant, λ, is the probability of decay per unit time for a nucleus.

Unit: s⁻¹.

C. Relationship between A and N

If N is the number of undecayed nuclei: A = -dN/dt = λ N

D. Half-life, t_(1/2)

The half-life, t_(1/2), is the time taken for N (or A) to fall to half its initial value.

Useful relation: t_(1/2) = (ln 2)/λ

2. Key Ideas (What Earns Marks)

Random individual decays and predictable population half-lifeThree groups contain sixteen, eight and four undecayed nuclei at zero, one and two half-lives, illustrating statistical decay of a large population.t = 0N/N₀ = 16/16t = t½N/N₀ = 8/16t = 2t½N/N₀ = 4/16which nucleus?unpredictablepopulation trend?predictable
Scroll diagram horizontally to read all labels.
Individual decay times are random, but a large population follows a predictable exponential law: the expected number remaining halves after each half-life.
  • A = λ N and A has unit Bq.
  • Exponential decay laws:
    • N = N₀e^(-λ t)
    • A = A₀e^(-λ t)
  • Half-life relation: t_(1/2) = (ln 2)/λ
  • When using measured count rates, subtract background first (see Random Nature of Radioactive Decay).
Exam pitfall: wrong decay quantity in calculations

Decide whether the data column is N, A, or count rate before using formulas. Activity and undecayed nuclei follow the same exponential form but represent different physical quantities.

3. Detailed Explanations

A. Why the decay is exponential

If each nucleus has the same probability of decaying per unit time, then the rate of decay is proportional to how many nuclei remain: -dN/dt ∝ N ⇒ -dN/dt = λ N

Solving gives: N = N₀e^(-λ t)

B. Half-life from the exponential law

At t = t_(1/2), N = N₀/2: 1/2 = e^(-λ t_(1/2)) ⇒ t_(1/2) = (ln 2)/λ

Graph skill: straight line using ln

From N = N₀e^(-λ t): ln N = ln N₀ - λ t

So plotting ln N (or ln A) against t gives a straight line whose gradient is -λ.

Linearising radioactive decay: ln(A) vs time (example)

A straight-line ln(A) against time plot; the negative gradient equals the decay constant λ.

Scroll across the graph to read all labels.

A straight-line ln(A) against time plot; the negative gradient equals the decay constant λ.A straight-line ln(A) against time plot; the negative gradient equals the decay constant λ.
If the gradient is −0.0040 s⁻¹, then the decay constant is λ = 0.0040 s⁻¹.
Open full-size graph
View figure data
Values for Linearising radioactive decay: ln(A) vs time (example)
Time, t (s)Example data (gradient = −λ)
010
1009.6
2009.2
3008.8
4008.4
5008

4. Common Mistakes

  • Mixing up counts with count rate (divide by time first).
  • Forgetting to subtract background before doing half-life/decay-constant calculations.
  • Using A = λ N with inconsistent units for A (must be decays per second).

5. Exam Tips

  • If the question gives half-life, find λ using t_(1/2) = ln 2/λ before anything else.
  • When plotting data, use ln A vs t for a straight line (gradient = -λ) if asked.
  • State whether you are using N or A (both follow the same exponential form).

6. Worked Examples

Modelled example 1

Find decay constant from half-life

Core

Problem

A nuclide has half-life t_(1/2) = 5.0 days. Find λ in s⁻¹.
Study the worked solution

  1. Convert time to seconds

    Method

    t_(1/2) = 4.32 × 10⁵ s.

    Reason

    The requested decay-constant unit is per second.

    Working

    t_(1/2) = 5.0(24)(3600) = 4.32 × 10⁵ s

  2. Use the half-life relation

    Method

    λ = 1.60 × 10⁻⁶ s⁻¹.

    Reason

    Half-life is inversely related to decay probability per unit time.

    Working

    λ = (ln 2)/(t_(1/2)) = 0.693/(4.32 × 10⁵) = 1.60 × 10⁻⁶ s⁻¹

Guided practice 2

Activity after a time

About 5 min

Problem

A sample has initial activity A₀ = 800 Bq and decay constant λ = 2.0 × 10⁻⁶ s⁻¹. Find its activity after t = 3.0 × 10⁵ s.

Try this before viewing the solution

Unit: Bq

Hints

Hint 1: form the dimensionless exponent
Calculate λ t first, then substitute into A = A₀e^(-λ t).
View solution step by step

  1. Calculate the exponent

    Method

    λ t = 0.60.

    Reason

    The product of s⁻¹ and seconds is dimensionless.

    Working

    (2.0 × 10⁻⁶)(3.0 × 10⁵) = 0.60

  2. Apply exponential decay

    Method

    A = 4.4 × 10² Bq.

    Reason

    The negative exponent makes activity decrease from its initial value.

    Working

    A = 800e^(-0.60) = 4.4 × 10² Bq

Common misconception 3

Half-life from decay constant

Find and correct the mistake

Learner claim

A nuclide has λ = 5.0 × 10⁻⁵ s⁻¹. A learner multiplies by ln 2 and concludes that a larger decay constant gives a longer half-life. Diagnose the relationship and find the half-life in hours.

Try this before viewing the solution

Unit: h

View solution step by step

  1. Repair the relationship

    Method

    t_(1/2) = ln 2/λ.

    Reason

    A larger per-time decay probability makes the population halve sooner, not later.

    Working

    t_(1/2) = 0.693/(5.0 × 10⁻⁵) = 1.39 × 10⁴ s

  2. Convert to hours

    Method

    t_(1/2) = 3.85 h ≈ 3.9 h.

    Reason

    Divide seconds by 3600 s h⁻¹.

    Working

    t_(1/2) = (1.39 × 10⁴)/3600 = 3.85 h

Examiner practice 4

Number of nuclei from activity

2 marks

Examination question

A sample has activity A = 3.0 × 10⁶ Bq and decay constant λ = 2.5 × 10⁻⁴ s⁻¹. Find the number of undecayed nuclei N. [2 marks]

Try this before viewing the solution

View solution step by step

  1. Rearrange activity relation

    1 mark

    Method

    N = A/λ.

    Reason

    Activity is the decay probability per unit time multiplied by the number remaining.

    Working

    A = λ N ⇒ N = A/λ

  2. Evaluate

    1 mark

    Method

    N = 1.2 × 10¹⁰ nuclei.

    Reason

    The number of nuclei is a count and therefore dimensionless.

    Working

    N = (3.0 × 10⁶)/(2.5 × 10⁻⁴) = 1.2 × 10¹⁰

Challenge 5

Finding λ from a straight-line plot

Minimal support

Independent transfer

A graph of ln A against t is a straight line through (0 s,10.0) and (200 s,9.2). Find the decay constant λ.

Try this before viewing the solution

Hints

Hint 1: connect equation and graph
Compare ln A = ln A₀-λ t with y = c + mx.
View solution step by step

  1. Calculate graph gradient

    Method

    m = -0.0040 s⁻¹.

    Reason

    Use the change in ln A divided by the time change.

    Working

    m = (9.2-10.0)/(200-0) = -0.0040 s⁻¹

  2. Interpret the sign

    Method

    λ = 4.0 × 10⁻³ s⁻¹.

    Reason

    The linearised equation has gradient -λ, while the decay constant itself is positive.

    Working

    λ = -m = 4.0 × 10⁻³ s⁻¹

7. Mind Stretchers

Mind stretcher 1: Half-life counting shortcutExtension

Without using exponentials, how would you estimate the activity after 4 half-lives?

Show Answer

Each half-life halves the activity, so after 4 half-lives: A = A₀/2⁴ = A₀/16

Mind stretcher 2: Why does half-life stay constant?Extension

If activity decreases with time, why doesn’t the half-life “get longer” as the sample gets weaker?

Show Answer

Half-life depends on the decay constant λ, which is a property of the nuclide (probability of decay per unit time) and does not change with how many nuclei you have.

As N decreases, activity decreases because A = λ N, but λ (and hence t_(1/2) = ln 2/λ) stays constant.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027