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What are the properties of the three main types of ionising radiation, and how do we detect and shield against them?

The nature, penetration, ionising power and range of alpha, beta and gamma radiation, the inverse square law for gamma, background radiation and the uses and hazards of radiation.

A focused answer to AQA A-Level Physics 3.8.1.2 and 3.8.1.3, covering the nature, penetration, range and ionising power of alpha, beta and gamma radiation, background radiation, the inverse square law for gamma rays and the safe uses of radiation.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The three types of radiation
  3. Ionising power and penetration
  4. Background radiation
  5. The inverse square law for gamma
  6. Uses and hazards
  7. Try this

What this dot point is asking

AQA specification points 3.8.1.2 and 3.8.1.3 want you to compare alpha, beta and gamma radiation by nature, ionising power, penetration and range, explain background radiation, apply the inverse square law to gamma intensity, and discuss the uses and hazards of radiation.

The three types of radiation

In a magnetic field, alpha and beta deflect in opposite directions (opposite charges), with beta deflecting much more for the same field because of its far smaller mass; gamma is not deflected at all.

Ionising power and penetration

The more strongly a radiation ionises, the more energy it loses per unit length and so the less it penetrates. Alpha, with its large charge and low speed, ionises strongly and is quickly absorbed; gamma, being uncharged, interacts only weakly and penetrates far. So ionising power and penetrating power are inversely related: alpha is the most ionising and least penetrating, gamma the least ionising and most penetrating, with beta in between.

Background radiation

The largest contributor in most places is radon gas seeping from the ground. The corrected count rate is the measured count rate minus the background rate.

The inverse square law for gamma

For a point gamma source the intensity falls with the square of the distance, because the radiation spreads out over the surface of an expanding sphere:

This means moving twice as far away reduces the intensity to a quarter, which is why increasing distance is one of the simplest safety measures.

Uses and hazards

Try this

Q1. State which material is used to stop beta radiation. [1 mark]

  • Cue. A few millimetres of aluminium.

Q2. The intensity of gamma radiation at 1.0 m1.0 \text{ m} is 36 units36 \text{ units}. Find the intensity at 3.0 m3.0 \text{ m}. [2 marks]

  • Cue. Distance triples, so intensity falls by 32=93^2 = 9: 369=4.0 units\dfrac{36}{9} = 4.0 \text{ units}.

Q3. State which radiation is most strongly ionising. [1 mark]

  • Cue. Alpha.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20195 marksDescribe how you would use absorption experiments to identify whether an unknown radioactive source emits alpha, beta or gamma radiation, explaining how each type is distinguished.
Show worked answer →

First measure and subtract the background count rate. Then place absorbers of increasing thickness between the source and detector.

If the count rate falls to background when a sheet of paper (or a few centimetres of air) is placed in front, the source emits alpha. If the radiation passes through paper but is stopped by a few millimetres of aluminium, beta is present. If radiation still gets through several millimetres of aluminium and is only reduced by thick lead or concrete, gamma is present.

A source may emit more than one type, so the count rate is recorded at each stage to see which absorbers cause a drop.

Markers reward subtracting background, the sequence of absorbers (paper, aluminium, lead), and the correct identification rule for each radiation.

AQA 20213 marksA gamma detector records an intensity of 90 units90 \text{ units} at a distance of 0.50 m0.50 \text{ m} from a small source. Assuming the inverse square law applies, calculate the intensity at a distance of 1.5 m1.5 \text{ m}.
Show worked answer →

The inverse square law gives I=kx2I = \dfrac{k}{x^2}, so I1x12=I2x22I_1 x_1^2 = I_2 x_2^2.

The distance increases by a factor of 1.50.50=3\dfrac{1.5}{0.50} = 3, so the intensity falls by 32=93^2 = 9.

I2=909=10 unitsI_2 = \dfrac{90}{9} = 10 \text{ units}.

Markers reward using the inverse square relationship and the factor-of-nine reduction for a tripling of distance.

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