OCR A Level Physics A: Radioactive decay, activity, half-life and dating (H556) – Revision Notes

These notes condense section 6.4.3 (Radioactivity), statements (a) to (h), of the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017. H556 is the full A level. For explanations and longer worked examples, read the study guide, then use the practice set. The course hub is OCR A Level Physics, the printable list is the checklist, and the free 10-minute diagnostics find gaps.

Equations at a glance

Statement	Equation	Notes
(d)	A = λN	A in Bq, λ in s⁻¹, N a number of nuclei
(d)	ΔN/Δt = −λN	decay rate; minus sign as N falls
(e)	λt½ = ln 2	t½ and λ in matching time units
(f)	N = N₀e^(−λt)	N₀ is the number at t = 0
(f)	A = A₀e^(−λt)	same for corrected count rate
(g)	ln N = ln N₀ − λt	gradient of ln N against t is −λ
(h)	t = (1/λ) ln(A₀/A)	age from activities
(g)	N(new) = N(old) − λN(old)Δt	the spreadsheet step

6.4.3(a) Spontaneous and random

• Spontaneous: no trigger, unaffected by temperature, pressure or chemical state.
• Random: cannot predict which nucleus decays or when.
• Every undecayed nucleus has the same fixed chance of decaying each second; that chance is λ.
• Large samples give smooth curves; small samples scatter.

6.4.3(b) The three radiations

	α	β⁻	β⁺	γ
Nature	helium nucleus, +2e	electron, −e	positron, +e	photon
Ionising	strongest	moderate	moderate	weakest
Range in air	a few cm	metres	short	no fixed range
Absorbed by	paper, skin	few mm aluminium	few mm aluminium	reduced by thick lead

Absorption investigation, in steps:

1. Measure the background count rate with no source.
2. Fix the source-to-tube distance.
3. Add absorber, count for a fixed time, record thickness.
4. Subtract background from each count rate.
5. Plot corrected count rate against thickness; repeat and average.

Expect a sharp cut-off for α and β and an exponential fall for γ. Safety: tongs, source stored away, short times, large distances.

6.4.3(c) Decay equations

Decay	A	Z	Particle emitted
α	−4	−2	⁴₂α
β⁻	0	+1	⁰₋₁e with antineutrino
β⁺	0	−1	⁰₊₁e with neutrino

Check that the nucleon numbers and the proton numbers both add up on each side.

6.4.3(d)-(e) Activity, decay constant and half-life

• Activity: decays per second, in becquerel.
• Decay constant λ: the chance per second that a given nucleus decays.
• Half-life: time for the number of undecayed nuclei, or the activity, to halve.
• Number of half-lives n = t / t½; fraction left = (½)ⁿ.

Measuring a half-life:

• Short (protactinium generator): shake the bottle, wait for the layers to separate, count every 10 s over the upper layer, subtract background, then plot and read repeated halving times.
• Long: find N from mass and molar mass, measure A, then use λ = A/N and t½ = ln 2 / λ.

6.4.3(f) Exponential decay and dice

• Dice: each die has a probability p of “decaying” per throw. The expected survivors after n throws are N₀(1 − p)ⁿ, so λ = −ln(1 − p) per throw.
• For p = 1/6, λ = 0.182 and t½ = 3.8 throws.
• λ is close to p only when p is small.

6.4.3(g) Graphs and spreadsheet

• Tangent gradient on an N–t graph = −λN = −A.
• A straight ln N against t line confirms exponential decay; gradient = −λ.
• Spreadsheet: put λ and Δt in fixed cells; each row is old N minus λ × old N × Δt.
• Smaller Δt gives a model closer to N₀e^(−λt); the model with a large step always falls too fast.

6.4.3(h) Dating

• Living organisms take in carbon-14, so activity per gram of carbon stays at a constant level.
• After death, carbon-14 decays (half-life 5730 years) and the activity per gram falls.
• Age t = (1/λ) ln(A₀/A).
• Assumptions: unchanged past ratio, no contamination. Limit: activity too low against background after many half-lives.

Method in steps

Finding an activity from a half-life:

1. Convert t½ to seconds if the answer must be in becquerel.
2. Calculate λ = ln 2 / t½ and keep three or four figures.
3. Calculate A = λN.
4. Give the answer with its unit and sensible significant figures.

Finding a time from two activities:

1. Write the ratio A₀/A.
2. Take natural logs: λt = ln(A₀/A).
3. Divide by λ, with λ in the time unit you want.
4. Check against whole half-lives as a quick sense test.

Reading a ln graph:

1. Draw a line of best fit through the plotted ln values.
2. Choose two well-separated points on the line itself, not on the data.
3. Gradient = change in ln A / change in t; this is −λ.
4. Use the intercept at t = 0 to get ln A₀, then A₀ = e^(intercept).

Worked reminders

Activity from N. An isotope has half-life 2.0 hours and a sample has 9.0 × 10¹² nuclei. Then λ = ln 2 / (2.0 × 3600) = 9.63 × 10⁻⁵ s⁻¹ and A = λN = 8.7 × 10⁸ Bq.

Time to reach a fraction. A source has half-life 9.0 days. To fall to one twentieth of its activity, λ = 0.0770 day⁻¹ and t = ln 20 / 0.0770 = 38.9 days, which is 4.32 half-lives. This is sensible, since 4 half-lives give 1/16 and 5 give 1/32.

One spreadsheet row. With λ = 0.040 s⁻¹, Δt = 3.0 s and N = 2000 at the start of a row, the next value is 2000 − 0.040 × 2000 × 3.0 = 1760. The exact value from 2000e^(−0.12) is 1774, so the model is a little low.

Reading an intercept. If a ln N graph cuts the vertical axis at 9.21, then N₀ = e^(9.21) = 1.0 × 10⁴ nuclei.

Fraction remaining. For a half-life of 6.0 hours, 40% of the sample remains after t = ln(1/0.4) / (ln 2 / 6.0) = 7.9 hours.

Must-know distinctions

• Activity (decays per second) versus count rate (what the detector registers, which is lower).
• λ (per unit time) versus t½ (a time).
• Decay constant versus probability per throw in a dice model.
• Spontaneous (no trigger) versus random (no prediction).

Quick self-test

1. A source has half-life 6.0 hours. What percentage of its nuclei remain after 15 hours?
2. Find λ in days⁻¹ for an isotope of half-life 8.0 days.
3. A sample of that isotope has activity 500 Bq. What is its activity after 20 days?
4. Find λ in s⁻¹ for an isotope of half-life 3.0 minutes.
5. A corrected count rate falls from 800 s⁻¹ to 200 s⁻¹ in 40 s. Find λ and the half-life.
6. State the change in nucleon number and proton number in β⁺ decay.
7. A wooden object has one eighth of the carbon-14 activity per gram of living wood. Using a half-life of 5730 years, state its age.
8. Why is the count from a source always measured with background subtracted?
9. A graph of ln A against t has gradient −0.012 s⁻¹. State λ and find t½.
10. Why does a spreadsheet model with a large time step give values below the exact curve?

Answers

1. 15/6.0 = 2.5 half-lives, so (½)^2.5 = 0.177, which is 17.7%.
2. λ = ln 2 / 8.0 = 0.0866 day⁻¹.
3. A = 500 e^(−0.0866 × 20) = 88 Bq.
4. 3.0 minutes = 180 s, so λ = ln 2 / 180 = 3.85 × 10⁻³ s⁻¹.
5. A factor of four is two half-lives in 40 s, so t½ = 20 s and λ = ln 2 / 20 = 0.0347 s⁻¹.
6. Nucleon number unchanged, proton number down by one.
7. Three half-lives, so 17 190 years.
8. Background radiation is always present and adds to the count, so it must be removed to get the count from the source alone.
9. λ = 0.012 s⁻¹, so t½ = 0.693 / 0.012 = 58 s.
10. Each step uses the N from the start of the step, which is larger than the average N during the step, so the model removes too many nuclei.

Where marks are usually lost

• Not subtracting background before calculating a ratio or plotting a graph.
• Mixing time units, such as t½ in years with t in days.
• Omitting the minus sign in the exponent or in ΔN/Δt = −λN.
• Leaving out the antineutrino or neutrino from a β equation.
• Stating that half-life is the time for all nuclei to decay.
• Quoting a range for γ-rays as if it were a fixed value.
• Giving only “wear gloves” for safety, instead of tongs, distance and short exposure.
• Writing “random” without saying that every undecayed nucleus has the same fixed chance of decaying per second.

Official syllabus

OCR Advanced GCE in Physics A (H556), specification Version 3.0 (March 2026), first assessment 2017, published by OCR: section 6.4.3, Radioactivity.