Interactive practical
Gamma absorption by lead
Measure lead sheets with Vernier callipers, collect repeated counts and determine attenuation and half-value thickness.
Open interactive simulationA-Level Physics
Gamma absorption by lead
OCR H556 — matched to your course.
These revision notes are free. Premium unlocks the full question and retrieval-card bank for this topic.
Sign in to save this revision for later.
Revision hub
At a glance
Practical focus
- Measure corrected gamma count rate through different lead thicknesses and investigate exponential attenuation.
Practical method
Aim
Investigate the absorption of gamma radiation by lead and determine the linear attenuation coefficient and half-value thickness.
Variables
Independent variable: thickness of lead x
Dependent variable: corrected count rate C
Control variables: source–detector distance; counting time; the same source.
Apparatus
- Sealed cobalt-60 source
- GM tube and counter
- Lead sheets of known thickness
- Micrometer
- Fixed holder for source and tube
- Tongs
Method
- Measure the background count rate.
- Fix the source and GM tube a set distance apart.
- Measure the count rate with no lead, then with lead thicknesses up to 20 mm (measured with a micrometer).
- Subtract background. I = I₀\(e^{-\mu x}\), so plot ln C against x: gradient −μ.
- Half-value thickness x½ = \(\dfrac{\ln 2}{\mu }\).
Sample results
Background subtracted. Fixed source–tube distance.
| x / mm | C / counts s⁻¹ | ln(\(\dfrac{C}{\text{counts}}\) s⁻¹) |
|---|---|---|
| 0 | 120.0 | 4.787 |
| 5 | 88.9 | 4.487 |
| 10 | 65.9 | 4.188 |
| 15 | 48.8 | 3.888 |
| 20 | 36.1 | 3.586 |
Analysis
ln C falls linearly with x, gradient −0.060 mm⁻¹, so μ = 60 m⁻¹ and the half-value thickness is \(\dfrac{\ln 2}{60}\) = 0.0116 m = 11.6 mm. Every 11.6 mm of lead halves the count rate.
Uncertainty
Counting errors (√N) dominate; at the thickest lead, longer counts are needed. Lead thickness by micrometer is precise (±0.01 mm).
Errors and improvements
| Problem | Effect | Improvement |
|---|---|---|
| Random decay | Scatter in ln C | Longer counts; repeat |
| Scattered radiation reaching the tube | Count rate too high | Keep geometry fixed; collimate |
| Background | Distorts thick-lead readings | Subtract background |
Risk assessment
| Hazard | Risk | Control measure |
|---|---|---|
| Ionising radiation | Cell damage | Tongs, distance, short exposure, lead box |
| Lead is toxic | Poisoning | Wear gloves or wash hands after handling lead |
Common mistakes
- Plotting C against x and reading a 'gradient'.
- Forgetting background.
- Changing the source–tube distance between readings.
Worked analysis
Gradient = \(\dfrac{3.586 - 4.787}{20}\) = −0.060 mm⁻¹, so μ = 0.060 mm⁻¹ = 60 m⁻¹. x½ = \(\dfrac{\ln 2}{0.060}\) = 11.6 mm.
Common misconception
No thickness of lead stops all gamma radiation; it reduces the intensity exponentially, halving for each half-value thickness.
Assessment coverage
Practice this practical
2 questions available, worth 4 marks in total.
Retrieval cards
Gamma absorption equation?
C = C₀\(e^{-\mu x}\).
This card is available with A-Level Physics premium access.
This card is available with A-Level Physics premium access.
This card is available with A-Level Physics premium access.
This card is available with A-Level Physics premium access.
This card is available with A-Level Physics premium access.
How this practical fits your course
The label below uses the wording from your course specification.
- OCR H556 — Suggested PAG activity: PAG 7 – Investigating ionising radiation / example: absorption of alpha, beta or gamma. OCR requires safe use of ionising radiation and detectors in PAG 7; absorption of alpha, beta or gamma is the specification example, and this shared resource uses supervised gamma absorption by lead. Checked 18 September 2026. View official course source