Free Pearson Edexcel International A Level Physics lesson

Nuclear radiation and measurement

Identify emissions and separate the source from background.

What you’ll learn

  • Link particle nature to its interactions, and state the limits of typical range comparisons.
  • Conserve A and charge and complete beta equations with the correct neutrino.
  • Control geometry, subtract background and compare the signal with its uncertainty.

Nature explains ionisation and penetration

Alpha particles are helium nuclei with charge +2e and relatively large mass. They cause dense ionisation along short paths and are typically stopped by thin paper or the outer dead layer of skin. Limited external penetration does not mean that an alpha-emitting substance is harmless if it enters living tissue.

Beta-minus particles are electrons emitted during nuclear transformation; beta-plus particles are positrons. Typical beta radiation has less dense ionisation and greater range than alpha radiation. Gamma radiation consists of high-energy photons from nuclei, with no charge or rest mass. Thick shielding reduces its intensity probabilistically rather than imposing one fixed stopping distance.

These comparisons depend on particle energy and absorber material. A detector registers only some source emissions, so count rate is not automatically activity. Educational sources belong in authorised supervised facilities with suitable handling, distance and shielding procedures. The supplied comparison tables support analysis; they are not instructions to acquire or use a radioactive source.

alpha: ⁴₂He beta−: ⁰₋₁e beta+: ⁰₊₁e gamma: photon

Symbols and units

  • In a nuclear symbol, the upper number A is nucleon number and the lower number is charge in units of the positive elementary charge e; these numbers are dimensionless.
  • ⁴₂He denotes an alpha particle: two protons and two neutrons, with charge +2e. Here e is the elementary-charge magnitude, approximately 1.60 × 10⁻¹⁹ C.
  • ⁰₋₁e and ⁰₊₁e denote a beta-minus electron and beta-plus positron. The upper zero means zero nucleon number, not zero particle rest mass.
  • Gamma denotes a photon with no charge and zero rest mass. Detector count rate is registered events per second (s⁻¹); activity is source transformations per second (Bq).

How the equation explains the idea

  1. Read the symbols as particle identities and conservation labels. They do not provide a numerical stopping distance; range also depends on energy and absorber material.
  2. In the example, a component removed by paper is consistent with short-range alpha radiation. A second component transmitted by paper but reduced by aluminium is consistent with beta radiation under the stated measurement conditions.
  3. A penetrating remainder is not proof that every remaining count is gamma from the selected source. Measure and subtract background rates, while keeping geometry and detector response comparable.
  4. Link the observations to interactions: charged particles cause ionisation along their paths, while gamma transmission decreases probabilistically. Penetration alone does not determine the effect of an exposure.

Worked example

Interpret successive absorbers

  1. A component disappears after paper; another survives paper but is greatly reduced by thin aluminium.
  2. These observations are consistent with alpha and beta components under appropriate energies and detector response.
  3. A penetrating remainder may contain gamma and background; measure the background separately.

Use energy, geometry and detector response as well as absorber identity.

Keep this idea. Link particle nature to its interactions, and state the limits of typical range comparisons.

Balance nuclear transformations

A nuclear symbol gives nucleon number A above proton number Z. Nuclear equations conserve total nucleon number and total charge. Alpha emission reduces the daughter A by four and Z by two. Gamma emission lowers nuclear excitation energy without changing either number, so it does not change the element or isotope.

In beta-minus decay, a neutron becomes a proton, emitting an electron and an electron antineutrino. Daughter A stays fixed and Z rises by one. Beta-plus changes a proton to a neutron, emitting a positron and electron neutrino; daughter Z falls by one. A beta electron is created in the nuclear process rather than removed from the atom’s outer shell.

Neutrinos and antineutrinos carry energy, momentum and lepton number even though their charge and nucleon number are zero. Include them in complete beta equations, especially in Unit 4 synoptic conservation questions. Balancing A and Z identifies a missing symbol, but it does not alone establish that the proposed reaction is energetically possible.

beta−: n → p + e⁻ + electron antineutrino beta+: p → n + e⁺ + electron neutrino

Symbols and units

  • A is nuclear nucleon number and Z is nuclear proton number, both dimensionless. Particle charge labels are expressed in units of e, the positive elementary-charge magnitude.
  • n and p denote a neutron and proton here, not particle number or gas pressure; e⁻ and e⁺ denote an electron and positron.
  • The electron antineutrino in beta-minus and electron neutrino in beta-plus each carry zero charge and zero nucleon number, but carry energy, momentum and lepton number.
  • An alpha particle has A = 4 and charge +2e; a gamma photon has A = 0 and charge zero.

How the equation explains the idea

  1. Balance nucleon number and charge separately. For the alpha example, 226 = Adaughter + 4 and 88 = Zdaughter + 2, giving Adaughter = 222 and Zdaughter = 86.
  2. For beta-minus, n → p + e⁻ + antineutrino preserves one nucleon and charge 0 = (+1) + (−1) + 0. Thus daughter A is unchanged and Z increases by one.
  3. For beta-plus within an allowed nuclear transformation, p → n + e⁺ + neutrino gives daughter A unchanged and Z reduced by one. These nuclear shorthand relations do not assert that an isolated free proton can spontaneously decay this way.
  4. Gamma emission changes excitation energy without changing A or Z. A balanced symbol equation is necessary bookkeeping, but energy and other conservation conditions must also permit the transformation.

Worked example

Find an alpha daughter

  1. Parent A = 226, Z = 88.
  2. Subtract alpha A = 4 and Z = 2.
  3. Daughter A = 222, Z = 86, preserving both total columns.

Balance before identifying the chemical element.

Keep this idea. Conserve A and charge and complete beta equations with the correct neutrino.

Core practical 15: measure a decreasing signal

Compare gamma count rates through several measured lead thicknesses with source, absorber and detector positions fixed. First measure background without the selected source. Use sufficiently long intervals to manage statistical scatter. Convert every observation to a rate before subtracting the background; equal raw counts over unequal times are not equal signals.

Plot corrected rate against thickness. An approximate monoenergetic narrow-beam model is R = R₀ exp(−μx), giving a straight ln R against x plot with slope −μ. Scattering, mixed photon energies and changing geometry can invalidate the simple model. A half-value thickness halves corrected transmission and is different from a half-life, which concerns radioactive change over time.

For independent counts N, √N is an approximate standard counting uncertainty. Combine the variances of separately measured gross and background rates. A corrected signal close to zero has a large relative uncertainty and may have no meaningful logarithm. Do not replace negative corrected fluctuations with invented positive values; collect longer readings or report a justified detection limit.

Rsource = Ngross/tgross − Nbackground/tbackground ln R = ln R₀ − μx

Symbols and units

  • Ngross and Nbackground are observed counts; tgross and tbackground are their measurement durations in s. Counts are dimensionless event totals.
  • Rsource is the background-corrected source rate in s⁻¹; R or R₀ denotes corrected rate at absorber thickness x or at zero thickness.
  • x is absorber thickness in m and μ is the attenuation coefficient in m⁻¹. Their product μx is dimensionless.
  • ln denotes the natural logarithm. Read the shorthand ln R − ln R₀ as ln(R/R₀), or use a common reference rate so each logarithm has a dimensionless argument.
  • The approximate standard uncertainty in corrected rate is √(Ngross/tgross² + Nbackground/tbackground²), in s⁻¹, for independent Poisson counts.

How the equation explains the idea

  1. Convert each count to its own rate before subtracting: Rsource = 150/50 − 80/200 = 3.00 − 0.400 = 2.60 s⁻¹. Unequal durations prevent direct subtraction of the raw counts.
  2. Combine independent rate variances: uncertainty = √(150/50² + 80/200²) = √0.062 = 0.249 s⁻¹. Background correction reduces the signal but does not remove its statistical uncertainty.
  3. For approximately monoenergetic narrow-beam transmission with fixed geometry, divide R = R₀ exp(−μx) by R₀ and take natural logs: ln(R/R₀) = −μx. A fitted slope against x is −μ.
  4. The corrected example rate can form one measurement, but several known thicknesses are needed to estimate μ. Do not take a logarithm of zero or negative corrected fluctuations or invent a replacement signal.
Background can hide exponential attenuation

Conditions. Computed narrow-beam monoenergetic model: Rsource = 40 exp(−μx) s⁻¹, μ = ln(2)/0.005 m⁻¹, and constant background = 5 s⁻¹. Source–detector geometry and efficiency stay fixed. Curves are expected rates without simulated counting scatter.

Absorber thickness / m; Expected count rate / s⁻¹.

Every 0.005 m halves the corrected source contribution. Gross rate approaches the 5 s⁻¹ background rather than zero. The horizontal spacing measures a half-value thickness, not a radioactive half-life; scattering or mixed energies can change the model.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Unequal timing intervals

  1. Gross count is 150 in 50 s; background is 80 in 200 s.
  2. Corrected rate = 150/50 − 80/200 = 2.60 s⁻¹.
  3. Approximate standard uncertainty = √(150/50² + 80/200²) = 0.249 s⁻¹.

Subtract rates and combine independent variances.

Keep this idea. Control geometry, subtract background and compare the signal with its uncertainty.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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