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Decay Equations (Oxford AQA IGCSE Combined Science Double Award)

Revision Note

Caroline Carroll

Written by: Caroline Carroll

Reviewed by: Lucy Kirkham

Oxford AQA Combined Science Double Award: PhysicsOxford AQA Physics

Decay Equations

Extension Tier only

  • The process of radioactive decay can be shown as a decay equation

  • A decay equation is similar to a chemical reaction equation

    • The particles present before the decay are shown before the arrow

    • The particles produced in the decay are shown after the arrow

  • During decay equations, the sum of the mass and atomic numbers before the reaction must be equal to the sum of the mass and atomic numbers after the reaction

Decay equations involving alpha particles

  • Remember that an alpha particle is a helium nucleus

    • It is made of 2 protons and 2 neutrons

  • When the alpha particle is emitted from the unstable nucleus, the mass number and atomic number of the nucleus changes

    • The mass number decreases by 4

      • The mass of the nucleus decreases

    • The atomic number decreases by 2

      • The charge of the nucleus decreases by 2

      • This is because protons have a charge of +1 each

  • A new element is formed

  • This can be shown by the general decay equation:

    straight X presubscript straight Z presuperscript straight A space rightwards arrow space straight Y presubscript straight Z minus 2 end presubscript presuperscript straight A minus 4 end presuperscript space plus space He presubscript 2 presuperscript 4

  • For example, the following equation shows Polonium-212 undergoing alpha decay

    • It emits an alpha particle causing its mass and charge to decrease

    • It forms lead-208 and an alpha particle

    • An alpha particle can also be written asstraight alpha presubscript 2 presuperscript 4

Po presubscript 84 presuperscript 212 space rightwards arrow space Pb presubscript 82 presuperscript 208 space plus space He presubscript 2 presuperscript 4

Worked Example

A nucleus with 84 protons and 126 neutrons undergoes alpha decay. It forms lead, which has the element symbol Pb.

Which of the isotopes of lead below is the correct one formed during the decay?

A

B

C

D

Pb presubscript 82 presuperscript 206

Pb presubscript 82 presuperscript 208

Pb presubscript 84 presuperscript 210

Pb presubscript 86 presuperscript 214

Answer:  A

Step 1: Calculate the mass number of the original nucleus

  • The mass number is equal to the number of protons plus the number of neutrons

  • The original nucleus has 84 protons and 126 neutrons

84 + 126 = 210

  • The mass number of the original nucleus is 210

Step 2: Calculate the new atomic number

  • The alpha particle emitted is made of two protons and two neutrons

  • Protons have an atomic number of 1, and neutrons have an atomic number of 0

  • Removing two protons and two neutrons will reduce the atomic number by 2

84 – 2 = 82

  • The new nucleus has an atomic number of 82

Step 3: Calculate the new mass number

  • Protons and neutrons both have a mass number of 1

  • Removing two protons and two neutrons will reduce the mass number by 4

210 – 4 = 206

  • The new nucleus has a mass number of 206

Examiner Tips and Tricks

It is easy to forget that an alpha particle is a helium nucleus. The two are interchangeable, so don’t be surprised to see either used in the exam.

Decay equations involving beta particles

  • Remember that a beta particle is a high-speed electron

  • It has a mass number of 0

  • Therefore, the mass number of the decaying nuclei remains the same

    • Therefore the mass of the nuclei remains the same

  • Electrons have an atomic number of -1

    • This means that the new nuclei will increase its atomic number by 1 in order to maintain the overall atomic number before and after the decay

    • Therefore the charge of the nuclei increases

  • This can be shown by the general decay equation:

    straight X presubscript straight Z presuperscript straight A space rightwards arrow space straight Y presubscript straight Z plus 1 end presubscript presuperscript straight A space plus space straight e presubscript negative 1 end presubscript presuperscript 0

  • The following equation shows carbon-14 undergoing beta decay

    • The carbon nucleus emits a beta particle causing the charge of the nucleus to increase but the mass remains the same

    • It forms nitrogen-14 and a beta particle

    • Beta particles can also be written as straight beta presubscript negative 1 end presubscript presuperscript 0

    straight C presubscript 6 presuperscript 14 space rightwards arrow space straight N presubscript 7 presuperscript 14 space plus space straight e presubscript negative 1 end presubscript presuperscript 0

Worked Example

A nucleus with 11 protons and 13 neutrons undergoes beta decay. It forms magnesium, which has the element symbol Mg.

A

B

C

D

Mg presubscript 9 presuperscript 20

Mg presubscript 10 presuperscript 24

Mg presubscript 11 presuperscript 23

Mg presubscript 12 presuperscript 24

Which is the correct isotope of magnesium formed during the decay?

Answer:  D

Step 1: Calculate the mass number of the original nucleus

  • The mass number is equal to the number of protons plus the number of neutrons

  • The original nucleus has 11 protons and 13 neutrons

11 + 13 = 24

  • The mass number of the original nucleus is 24

Step 2: Calculate the new atomic number

  • During beta decay a neutron changes into a proton and an electron

  • The electron is emitted as a beta particle

  • The neutron has an atomic number of 0 and the proton has an atomic number of 1

  • So the atomic number increases by 1

11 + 1 = 12

  • The new nucleus has an atomic number of 12

Step 3: Calculate the new mass number

  • Protons and neutrons both have a mass number of 1

  • Changing a neutron to a proton will not affect the mass number

  • The new nucleus has a mass number of 24 (the same as before)

Examiner Tips and Tricks

You are not expected to know the names of the elements produced during radioactive decays, but you do need to be able to calculate the mass and atomic numbers by making sure they are balanced on either side of the reaction.

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    Author
    Reviewer
    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.

    Lucy Kirkham

    Author: Lucy Kirkham

    Expertise: Head of STEM

    Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels.Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all.