A LevelPhysicsCambridge (CIE)Revision Notes11. Particle PhysicsAtoms, Nuclei & RadiationDecay Equations

Decay Equations (Cambridge (CIE) A Level Physics): Revision Note

Exam code: 9702

Written by: Leander Oates

Reviewed by: Caroline Carroll

Updated on 7 October 2026

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Neutrino emission

An electron neutrino is a type of subatomic particle with no charge and negligible mass which is emitted from the nucleus
The electron anti-neutrino is the antiparticle of an electron neutrino
- Electron anti-neutrinos are produced during β– decay
- Electron neutrinos are produced during β+ decay

Electron neutrino and anti-electron neutrino

Symbols for the electron neutrino, ν with subscript e, and the anti-electron neutrino, the same symbol with a bar above it

The anti neutrino is denoted by a line above the symbol

Examiner Tips and Tricks

One way to remember which particle decays into which depends on the type of beta emission, think of beta ‘plus’ as the ‘proton’ that turns into the neutron (plus an electron neutrino)

α & β decay equations

Alpha decay

Alpha decay is common in large, unstable nuclei with too many protons
The decay involves a nucleus emitting an alpha particle and decaying into a different nucleus
An alpha particle consists of 2 protons and 2 neutrons (the nucleus of a Helium atom)

Alpha decay

A parent nucleus of protons and neutrons emits an alpha particle (two protons and two neutrons, helium-4), leaving a smaller daughter nucleus

Alpha decay produces a daughter nucleus and an alpha particle (helium nucleus)

When an unstable nucleus (the parent nucleus) emits radiation, the constitution of its nucleus changes
As a result, the isotope will change into a different element (the daughter nucleus)
Alpha decay can be represented by the following radioactive decay equation:

${}_{Z}^{A}X \to {}_{Z - 2}^{A - 4}Y + {}_{2}^{4}α$

An example would be:

${}_{84}^{212}{Po} \to {}_{82}^{208}{Pb} + {}_{2}^{4}α$

When an alpha particle is emitted from a nucleus:
- The nucleus loses 2 protons: proton number decreases by 2
- The nucleus loses 4 nucleons: nucleon number decreases by 4

Worked Example

The radioactive nucleus ${}_{86}^{222}{Rn}$ undergoes alpha decay into a daughter nucleus Po.

Grid of proton number (83 to 87) against neutron number (132 to 136) with four labelled points: A at 132 neutrons and 86 protons, B at 132 and 84, C at 134 and 84, and D at 136 and 84

(a) Which letter in the diagram represents the daughter product?

(b) Which is the nucleon and proton number of the final nucleus Po?

Answer:

(a)

Step 1: Determine the number of neutrons

This is alpha decay
The nucleon number of the parent nucleus decreases by 4
The proton number of the parent nucleus decreases by 2
Therefore, the number of neutrons is:

$222 - 86 = 136$

Step 2: Determine the change in the proton and neutrons

A helium nucleus has 2 protons and 2 neutrons
The proton and neutron numbers will decrease by 2
This reaches the point of the diagram marked C

(b)

Step 1: Write the equation for the alpha decay

The same grid with radon-222 marked at 136 neutrons and 86 protons and an arrow labelled α going down to point C at 134 neutrons and 84 protons, labelled Po

${}_{Z}^{A}X \to {}_{2}^{4}α + {}_{Z - 2}^{A - 4}Y$

${}_{86}^{222}{Rn} \to {}_{2}^{4}α + {}_{84}^{218}{Po}$

Step 2: Determine the nucleon and proton numbers of Po

Nucleon number = 218
Proton number = 84

β^- decay

A β^-particle is a high energy electron emitted from the nucleus
β^-decay is when a neutron turns into a proton emitting an electron and an anti-electron neutrino

Beta decay

A neutron changes into a proton, emitting an electron (beta particle) and an anti-electron neutrino: n → p + e⁻ + anti-electron neutrino

In beta-minus decay, a neutron decays into a proton, an electron and an anti-electron neutrino

When a β^- is emitted from a nucleus:
- The number of protons increases by 1: proton number increases by 1
- The total number of nucleons stays the same: nucleon number remains the same
The decay equation for beta-minus decay is:

${}_{Z}^{A}X \to {}_{Z + 1}^{A}Y + {}_{- 1}^{0}β + {}_{0}^{0}{\bar{v}}_{e}$

An example would be:

${}_{6}^{14}C \to {}_{7}^{14}N + {}_{- 1}^{0}β + {}_{0}^{0}{\bar{v_{e}}}$

The new nucleus formed from the decay is called the “daughter” nucleus (nitrogen in the example above)

β⁺ decay

A β⁺particle is a high energy positron emitted from the nucleus
β⁺decay is when a proton turns into a neutron emitting a positron (anti-electron) and an electron neutrino

Beta-plus decay

A proton changes into a neutron, emitting a positron (beta particle) and an electron neutrino: p → n + e⁺ + electron neutrino

In beta plus decay, a proton decays into a neutron, a positron and an electron neutrino

When a β⁺ is emitted from a nucleus:
- The number of protons decreases by 1: proton number decreases by 1
- The total number of nucleons stays the same: nucleon number remains the same
The decay equation for beta-plus decay is:

${}_{Z}^{A}X \to {}_{Z - 1}^{A}Y + {}_{+ 1}^{0}β + {}_{0}^{0}v_{e}$

An example of this would be:

${}_{6}^{11}C \to {}_{5}^{11}B + {}_{+ 1}^{0}β + {}_{0}^{0}v_{e}$

Worked Example

A radioactive substance with a nucleon number of 212 and a proton number of 82 decays by β-plus emission into a daughter product which in turn decays by further β-plus emission into a granddaughter product.

Grid of proton number (80 to 84) against neutron number with four labelled points: D at the top left, C at 130 neutrons and 84 protons, B at 130 neutrons and 80 protons, and A at 132 neutrons and 80 protons

State the letter in the diagram that represents the granddaughter product.

Answer: A

In β+ decay, a proton decays into a neutron, a positron, and an electron neutrino
The number of neutrons is:

$212 - 82 = 130$

The proton number decreases by one and the number of neutrons increases by one
- This happens twice, so it reaches point A on the diagram

The same grid with two arrows labelled β+ starting at 130 neutrons and 82 protons, each going down one proton and across one neutron, ending at point A (132 neutrons, 80 protons)