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Alpha, Beta & Gamma Particles (CIE A Level Physics)

Revision Note

Leander

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Leander

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Conservation of nucleon number & charge

  • Nuclear processes such as fission and fusion are represented using nuclear equations (similar to chemical reactions in chemistry)
  • The number of protons and neutrons in atom is known as its constituents
  • For example:

straight U presubscript 92 presuperscript 235 space plus space straight n presubscript 0 presuperscript 1 space rightwards arrow space Sr presubscript 38 presuperscript 90 space plus space Xe presubscript 54 presuperscript 144 space plus space 2 straight n presubscript 0 presuperscript 1

  • The above equation represents a fission reaction in which a uranium nucleus is hit with a neutron and splits into two smaller nuclei – a Strontium nucleus and Xenon nucleus, releasing two neutrons in the process
  • In nuclear equations, the nucleon number and charge are always conserved
  • This means that:
    • the sum of the nucleons on the left hand side must equal the sum of the nucleons on the right hand side
    • the sum of the charge on the left hand side must equal the sum of the charge on the right hand side
  • In the above equation, the sum of the nucleon (top) numbers on both sides are equal

235 space plus space 1 space equals space 90 space plus space 144 space plus space open parentheses 2 space cross times space 1 close parentheses

236 space equals space 236

  • The same is true for the proton (bottom) numbers

92 space plus space 0 space equals space 38 space plus space 54 space plus space open parentheses 2 space cross times space 0 close parentheses

92 space equals space 92

  • By balancing equations in this way, you can determine the nucleon number, proton number or the number of missing elements

  • Let's consider another example:

straight U presubscript 92 presuperscript 235 space plus space straight n presubscript 0 presuperscript 1 space rightwards arrow space Rb presubscript 37 presuperscript 96 space plus space Cs presubscript 55 presuperscript 137 space plus space straight N straight n presubscript 0 presuperscript 1

  • Determine the total nucleon number
    • This is determined from the side of the equation where all the values are known
    • In this example, from the reactants

straight U presubscript 92 presuperscript 235 space plus space straight n presubscript 0 presuperscript 1

    • The total nucleon number = 235 + 1 = 236
  • Equate the total nucleon number to the total nucleon number of the products including the unknown N

Rb presubscript 37 presuperscript 96 space plus space Cs presubscript 55 presuperscript 137 space plus space straight N straight n presubscript 0 presuperscript 1

    • Total nucleon number of reactants = 96 + 137 + (N ×1) = 236
  • Rearrange to solve for N

straight N space equals space fraction numerator 236 space minus space 96 space minus space 137 over denominator 1 end fraction

straight N space equals space 3

  • Balancing the equation shows that 3 neutrons must be released in the reaction

Worked example

When a californium atom reacts with an unknown element X, the following reaction occurs.

Cf presubscript 98 presuperscript 252 space plus space straight X presubscript straight Y presuperscript 10 space rightwards arrow space 3 straight n presubscript 0 presuperscript 1 space plus space Lr presubscript 103 presuperscript straight Z

Determine the missing values of Y and Z.

Answer: 

Step 1: Identify what the value of Y represents

  • Y is the proton number of element X

Step 2: Determine the value of Y

  • Determine the number of protons on both sides of the equation

98 space plus space straight Y space equals space open parentheses 3 space cross times space 0 close parentheses space plus space 103

straight Y space equals space 0 space plus space 103 space minus space 98

straight Y space equals space 5

Step 3: Identify what the value of Z represents

  • Z is the nucleon number of the element Lr

Step 4: Determine the value of Z

  • Determine the total nucleon numbers on both sides of the equation

252 space plus space 10 space equals space open parentheses 3 cross times 1 close parentheses space plus space straight Z

straight Z space equals space 252 space plus space 10 space minus space 3

straight Z space equals space 259

Alpha, beta & gamma particles

  • Some elements have nuclei that are unstable
    • This tends to be when the number of nucleons does not balance
  • In order to become more stable, they emit particles and/or electromagnetic radiation
    • These nuclei are said to be radioactive
  • There are three main types of radioactive emission: alpha, beta and gamma

Alpha particles

  • Alpha (α) particles are high energy particles made up of 2 protons and 2 neutrons (the same as a helium nucleus)
  • They are usually emitted from nuclei that are too large

Alpha decay diagram, downloadable AS & A Level Physics revision notes

During alpha decay, a parent nucleus becomes a daughter nucleus by emitting an alpha particle (helium nucleus)

  • The nuclide notion for an alpha particle is:

straight alpha presubscript 2 presuperscript 4 or He presubscript 2 presuperscript 4

  • Alpha is a highly ionising form of radiation
    • It has a large charge of +2e
  • Alpha is a weakly penetrating form of radiation
    • This is because it is so ionising, it readily interacts with any object instead of passing through it
    • Alpha particles have a range of a few cm in air
  • Alpha particles can be blocked by skin or a piece of paper

Beta particles

  • Beta (β) particles are high energy electrons emitted from the nucleus
    • β particles are emitted by nuclei that have too many neutrons

Beta minus decay diagram, downloadable AS & A Level Physics revision notes

During beta-minus decay, a neutron in a parent nucleus becomes a proton in a daughter nucleus by emitting a beta-minus particle (an electron) and an anti-electron neutrino

  • The nuclide notion for a beta minus particle is:

straight beta presubscript negative 1 end presubscript presuperscript 0

  • Beta (β+) particles are high energy positrons emitted from the nucleus
    • β+ particles are emitted by nuclei that have too many protons

Beta plus decay diagram, downloadable AS & A Level Physics revision notes

During beta-plus decay, a proton in a parent nucleus becomes a neutron in a daughter nucleus by emitting a beta-plus particle (a positron) and an electron neutrino

  • The nuclide notion for a beta plus particle is:

straight beta presubscript plus 1 end presubscript presuperscript 0

  • Beta is a moderately ionising type of radiation
    • This is due to it having a charge of ±1e
    • This means it is able to do some slight damage to cells (less than alpha but more than gamma)

  • Beta is a moderately penetrating type of radiation
    • Beta particles have a range of around 20 cm - 3 m in air, depending on their energy

  • Beta particles can be stopped by a few millimetres of aluminium foil

 

Gamma rays

  • Gamma (γ) rays are high energy electromagnetic waves
  • They are emitted by nuclei that need to lose some energy

  • The nuclide notation for a gamma particle is:

straight gamma presubscript 0 presuperscript 0

 

  • If these particles hit other atoms, they can knock out electrons, ionising the atom
  • This can cause chemical changes in materials and can damage or kill living cells

Ionisation by radiation

Ionisation, downloadable AS & A Level Physics revision notes

When radiation passes close to atoms, it can knock out electrons, ionising the atom

  • The properties of the different types of radiation are summarised in the table below

Properties of types of radiation

Particle Composition Mass / u Charge / e Speed / c
Alpha (α) 2 protons + 2 neutrons 4 +2 0.05
Beta minus (β) Electron (e) 0.0005 −1 >0.99
Beta plus (β+) Positron (e+) 0.0005 +1 >0.99
Gamma (γ) Electromagnetic wave 0 0 1

  • u is the atomic mass unit (see “Atomic Mass Unit (u)”)
  • e is the charge of the electron: 1.60 × 10-19 C
  • c is the speed of light: 3 × 108 m s-1

Worked example

Three successive radioactive decays are shown in the diagram below. Each one results in a particle being emitted.

The first decay results in the emission of a β-particle.

The second decay results in the emission of an α-particle.

The third decay results in the emission of another β-particle.

11-1-4-we-alpha-beta-gamma-particles---cie-new

Nuclides W and Z are compared. 

Which nuclide of Z is formed at the end of this decay?

A.  straight Z presubscript 90 presuperscript 237      B:  straight Z presubscript 92 presuperscript 233      C:  straight Z presubscript 89 presuperscript 237      D:  straight Z presubscript 90 presuperscript 233

Answer: D

Step 1: Write the equation for the β− decay

  • A β− particle is an electron
  • The nucleon number stays the same
  • The proton number increases by 1

straight W presubscript 92 presuperscript 237 space rightwards arrow space straight X presubscript 93 presuperscript 237 space plus space straight beta presubscript 1 presuperscript 0

Step 2: Write the equation for the α decay

  • An α particle is a helium nucleus
  • The nucleon number reduces by 4
  • The proton number reduces by 2

straight X presubscript 93 presuperscript 237 space rightwards arrow space straight Y presubscript 91 presuperscript 233 space plus space straight alpha presubscript 2 presuperscript 4

Step 3: Write the equation for the β+ decay

  • A β+ particle is a positron
  • The nucleon number stays the same
  • The proton number reduces by 1

straight Y presubscript 91 presuperscript 233 space rightwards arrow space straight Z presubscript 90 presuperscript 233 space plus space straight beta presubscript plus 1 end presubscript presuperscript 0

Step 4: Determine the final nucleon Z

  • The final nucleon, Z will be:

straight Z presubscript 90 presuperscript 233

Examiner Tip

It is important to be familiar the properties of each type of radiation and their symbols.

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Leander

Author: Leander

Expertise: Physics

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.