Specification: AQA A-level Physics (7408) 3.8.1.2 | Paper: 2 | Data and formulae booklet: given in the exam
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Inverse-square law for gamma radiation: for a point source of gamma radiation, the intensity is inversely proportional to the square of the distance from the source, I = k ÷ x². Doubling the distance reduces the intensity to one quarter.
Why it works for gamma only: gamma photons spread out over the surface of a sphere of area 4πx² and are hardly absorbed by air. Alpha and beta particles are absorbed by air within a short range, so they do not follow the law.
Corrected count rate: the measured count rate minus the background count rate. Only the corrected count rate is proportional to the intensity from the source and follows the inverse-square law.
The equation
intensity = constant ÷ distance² I = k ÷ x²
| Quantity | Symbol | Unit |
|---|---|---|
| intensity of gamma radiation | I | watts per square metre (W m⁻²) |
| constant for the source | k | watts (W), or counts min⁻¹ m² if a count rate is used |
| distance from the source | x | metres (m) |
| corrected count rate | C | counts per minute or counts per second (s⁻¹) |
Rearranged: k = Ix² | x = √(k ÷ I) | ratio form I2 ÷ I1 = (x1 ÷ x2)², and the same for corrected count rates | Subtract the background count rate before you use the inverse-square law, and add it back on if you are asked for a measured reading.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) The corrected count rate 0.10 m from a gamma source is 1200 counts per minute. Calculate the corrected count rate 0.20 m from the source.
Show solution
F: C = k ÷ x², so C2 ÷ C1 = (x1 ÷ x2)²
I: C2 = 1200 × (0.10 ÷ 0.20)²
F: C2 = 1200 × 0.25
A: 300 counts per minute
Q2 (Core) For a gamma source, k = 3.2 × 10−3 W. Calculate the intensity 0.40 m from the source.
Show solution
F: I = k ÷ x²
I: I = 3.2 × 10−3 ÷ 0.40²
F: I is already the subject
A: I = 3.2 × 10−3 ÷ 0.16 = 0.020 W m⁻²
Q3 (Core) The corrected count rate 15 cm from a gamma source is 800 counts per minute. Calculate the distance at which the corrected count rate is 50 counts per minute.
Show solution
Convert first: 15 cm = 0.15 m
F: C = k ÷ x², so x2 = x1 × √(C1 ÷ C2)
I: x2 = 0.15 × √(800 ÷ 50) = 0.15 × √16
F: x2 = 0.15 × 4
A: 0.60 m
Watch out: the count rate falls by a factor of 16, but the distance only increases by a factor of √16 = 4.
Q4 (Core) The intensity of gamma radiation 15 cm from a point source is 4.5 × 10−4 W m⁻². Calculate (a) the constant k for the source and (b) the intensity 50 cm from the source, both to 2 significant figures.
Show solution
Convert first: 15 cm = 0.15 m; 50 cm = 0.50 m
Step 1 – (a) constant
F: I = k ÷ x²
I: 4.5 × 10−4 = k ÷ 0.15²
F: k = 4.5 × 10−4 × 0.0225 = 1.0125 × 10−5 W
A: 1.0 × 10−5 W (2 s.f.)
Step 2 – (b) intensity at 0.50 m
I = 1.0125 × 10−5 ÷ 0.50² = 1.0125 × 10−5 ÷ 0.25 = 4.05 × 10−5 W m⁻²
A: 4.1 × 10−5 W m⁻² (2 s.f.)
Watch out: leaving x in cm gives a k that is 104 times too big. Convert to metres first.
Q5 (Exam) A Geiger–Müller tube is placed 0.20 m from a gamma source and records 750 counts per minute. The background count rate is 30 counts per minute. Calculate the reading the tube would record 0.50 m from the source, to 3 significant figures.
Show solution
Step 1 – correct for background
corrected count rate at 0.20 m = 750 − 30 = 720 counts per minute
Step 2 – inverse-square law
F: C2 = C1 × (x1 ÷ x2)²
I: C2 = 720 × (0.20 ÷ 0.50)² = 720 × 0.16
F: C2 = 115.2 counts per minute (corrected)
Step 3 – add background back
measured reading = 115.2 + 30 = 145.2 counts per minute
A: 145 counts per minute (3 s.f.)
Watch out: scaling the raw reading 750 by 0.16 gives 120 counts per minute. Background does not depend on distance from the source, so it must be removed first and added back at the end.
Q6 (Exam) A technician holds a gamma source with her fingers about 2.0 cm from it. Using tongs, the source is held 30 cm from her fingers. Calculate the factor by which the intensity at her fingers is reduced by using tongs, and state two other ways of reducing her exposure.
Show solution
F: I1 ÷ I2 = (x2 ÷ x1)²
I: I1 ÷ I2 = (30 ÷ 2.0)² = 15²
F: the ratio is already the subject
A: the intensity is reduced by a factor of 225
other ways: keep the exposure time as short as possible, and use lead shielding (store the source in a lead-lined box when not in use); never point the source at anyone
Watch out: the units cancel in a ratio, so there is no need to convert cm to m here.
Q7 (Exam) The dose rate 0.50 m from a gamma source is 120 μSv h⁻¹. A worker must not be exposed to more than 2.5 μSv h⁻¹. Assuming the dose rate is proportional to the intensity, calculate the minimum distance at which the worker can stand, to 2 significant figures.
Show solution
F: I = k ÷ x², so x2 = x1 × √(I1 ÷ I2)
I: x2 = 0.50 × √(120 ÷ 2.5) = 0.50 × √48
F: x2 = 0.50 × 6.928…
A: 3.464… m = 3.5 m (2 s.f.)
Watch out: this is a minimum distance, so rounding down to 3.4 m would give a dose rate above the limit. Round a safety distance up.
Q8 (Exam) A student records corrected count rates of 3600, 900 and 400 counts per minute at distances of 0.10 m, 0.20 m and 0.30 m from a gamma source. (a) Show that the data obey the inverse-square law. (b) Predict the corrected count rate at 0.60 m.
Show solution
Step 1 – (a) test Cx²
if C = k ÷ x², then Cx² = k should be constant
3600 × 0.10² = 36; 900 × 0.20² = 36; 400 × 0.30² = 36
A: Cx² = 36 counts min⁻¹ m² each time, so the data obey the inverse-square law
Step 2 – (b) prediction
F: C = k ÷ x²
I: C = 36 ÷ 0.60² = 36 ÷ 0.36
A: 100 counts per minute
Watch out: showing that C halves when x doubles is not enough; for the inverse-square law it must fall to a quarter. Test Cx² for every pair of values.
Q9 (Stretch) The radioactive material in a gamma source sits an unknown distance d behind the front face of its holder. A student measures distances x from the front face. The background count rate is 25 counts per minute. At x = 0.080 m the measured count rate is 925 counts per minute, and at x = 0.280 m it is 125 counts per minute. (a) Calculate d. (b) Calculate the measured count rate expected at x = 0.480 m.
Show solution
Step 1 – correct for background
corrected count rates: 925 − 25 = 900 and 125 − 25 = 100 counts per minute
Step 2 – (a) find d
F: C = k ÷ (x + d)², so (x2 + d) ÷ (x1 + d) = √(C1 ÷ C2)
I: (0.280 + d) ÷ (0.080 + d) = √(900 ÷ 100) = 3
F: 0.280 + d = 0.240 + 3d, so 2d = 0.040
A: d = 0.020 m (2.0 cm)
Step 3 – (b) constant and prediction
k = C(x + d)² = 900 × (0.080 + 0.020)² = 900 × 0.010 = 9.0 counts min⁻¹ m²
at x = 0.480 m: corrected C = 9.0 ÷ (0.480 + 0.020)² = 9.0 ÷ 0.25 = 36 counts per minute
A: measured count rate = 36 + 25 = 61 counts per minute
Watch out: the inverse-square law uses the distance from the source itself, not from the front of the holder. Ignoring d here would predict 900 × (0.080 ÷ 0.480)² + 25 = 50 counts per minute.
Q10 (Stretch) To find the offset d between a gamma source and the front of its holder, a student plots a graph of 1/√C against x, where C is the corrected count rate in counts per minute and x is the distance measured from the front of the holder. The graph is a straight line passing through (0.085 m, 0.025) and (0.385 m, 0.100), where 1/√C is in (counts per minute)−½. (a) Explain why the graph is a straight line. (b) Use the graph to find d and the constant k. (c) Calculate the corrected count rate at x = 0.185 m.
Show solution
Step 1 – (a) why a straight line
C = k ÷ (x + d)², so 1/√C = (x + d) ÷ √k = (1 ÷ √k)x + d ÷ √k, which has the form y = mx + c
Step 2 – (b) gradient and k
gradient = (0.100 − 0.025) ÷ (0.385 − 0.085) = 0.075 ÷ 0.300 = 0.25
gradient = 1 ÷ √k, so √k = 4.0 and k = 16 counts min⁻¹ m²
Step 3 – (b) intercept and d
y-intercept = 0.025 − 0.25 × 0.085 = 0.00375
d = intercept ÷ gradient = 0.00375 ÷ 0.25 = 0.015 m
A: d = 0.015 m (1.5 cm); the line crosses the x-axis at x = −d = −0.015 m
Step 4 – (c) prediction
C = k ÷ (x + d)² = 16 ÷ (0.185 + 0.015)² = 16 ÷ 0.040
A: 400 counts per minute
Watch out: plotting C against 1/x² gives a curve when d is not zero. Plotting 1/√C against x gives a straight line whether or not d is known, which is why it is used.
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