All subjects · Physics · WPH15 (Unit 5)

Core practical 15: absorption of gamma radiation by lead

Core practical 15 investigates how the intensity of gamma radiation falls as it passes through increasing thicknesses of lead.

Method

Measure the background count with a Geiger-Müller (GM) tube and counter over several minutes, with no source nearby.
Clamp a sealed gamma source (such as cobalt-60) at a fixed distance from the GM tube.
Place lead sheets of increasing thickness between them. Measure each total thickness with a micrometer.
For each thickness, count for a fixed, fairly long time and repeat. Subtract background to get the corrected count rate C.

Analysis

The corrected count rate falls exponentially with thickness x: C = C_0e^-μx, where μ is the absorption coefficient. A graph of ln C against x is a straight line with gradient -μ. The half-thickness (the thickness that halves the count rate) is ln 2 / μ.

Safety and precision

Handle the source with long tongs, point it away from people, keep your distance and minimise exposure time. Return it to its lead-lined box straight after use.
Keep the source-to-detector distance fixed, because gamma intensity also falls with distance (inverse square).
Decay is random, so count for long times and repeat to reduce the percentage uncertainty.

Worked example

850 counts are recorded in 5.0 minutes behind some lead. The background is 0.50 counts per second. Find the corrected count rate.

Measured rate = 850 / 300 = 2.83 counts/s
Corrected = 2.83 - 0.50

Answer: 2.3 counts/s

Key idea: Subtract background; fixed distance; ln C against x has gradient -μ; half-thickness = ln 2 / μ; tongs, distance, short exposure.

Check yourself

850 counts are recorded in 5.0 minutes. The background count rate is 0.50 counts per second. Find the corrected count rate in counts/s, to 3 s.f.
Show answer
2.33 counts/s. 850 / 300 = 2.833 counts/s; corrected = 2.833 - 0.50 = 2.33 counts/s.
Lead has an absorption coefficient μ = 0.050 mm^-1 for some gamma rays. Find the half-thickness in mm, to 3 s.f.
Show answer
13.9 mm. x_1/2 = ln 2 / μ = 0.693 / 0.050 = 13.9 mm.
With no lead the corrected count rate is 400 counts/min. Find the corrected count rate behind 20 mm of lead with μ = 0.060 mm^-1, in counts/min, to 3 s.f.
Show answer
120 counts/min. C = 400e^{-0.060 x 20} = 400e^-1.2 = 400 x 0.301 = 120 counts/min.
Why is the background count measured first?
- To check the source is working
- To warm up the GM tube
- To find the half-life
- So it can be subtracted to find the count rate due to the source alone
Show answer
So it can be subtracted to find the count rate due to the source alone. Background adds to every reading.
Why must the distance between the source and the GM tube be kept the same?
- The GM tube only works at one distance
- Gamma intensity also falls with distance, so changing it would affect the results
- To keep the lead cool
- To reduce background radiation
Show answer
Gamma intensity also falls with distance, so changing it would affect the results. Distance is a control variable because of the inverse square law.
Which is a correct safety precaution when using the gamma source?
- Hold it in your hand briefly
- Leave it out of its box between readings
- Stand close to it to read the counter
- Handle it with long tongs and keep it pointed away from people
Show answer
Handle it with long tongs and keep it pointed away from people. Distance and short exposure time reduce the dose.
Why are counts taken over a long time and repeated?
- The source needs time to warm up
- Radioactive decay is random, so longer counts reduce the percentage uncertainty
- The lead absorbs more over time
- To increase the activity
Show answer
Radioactive decay is random, so longer counts reduce the percentage uncertainty. Larger counts have a smaller relative random fluctuation.

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