Use decay equations, using nuclide notation, to show the emission of α-particles, β-particles and γ-radiation

5.2.3 Radioactive Decay

Learning Objective

Use decay equations, written in nuclide notation, to show the emission of α-particles, β⁻-particles and γ-radiation.

Key Concepts

• Radioactive nuclei are unstable and transform spontaneously to a more stable configuration.
• The moment at which a particular nucleus decays is completely random; the probability of decay is constant for a given isotope, giving rise to exponential decay.
• Three types of nuclear radiation are examined in the IGCSE 0625 syllabus:

  • α‑decay – emission of a helium‑4 nucleus.
  • β⁻‑decay – emission of an electron (beta‑minus particle).
  • γ‑radiation – emission of a photon when an excited nucleus drops to a lower energy state (electromagnetic radiation, not a particle).

• Nuclide notation: \$_{Z}^{A}\text{X}\$ where Z = atomic number (protons), A = mass number (protons + neutrons), and X = chemical symbol.

Quick‑Reference Table – Changes Produced by Each Radiation Type

General Form of a Decay Equation

All decay equations must conserve both mass number (A) and atomic number (Z):

\$\;{Z}^{A}\text{X}\;\rightarrow\;{Z'}^{A'}\text{Y}\;+\;\text{radiation}\$

α‑Decay

• Mass number decreases by 4, atomic number decreases by 2.
• The daughter nuclide is a different element because Z changes.

Equation:

\$\;{Z}^{A}\text{X}\;\rightarrow\;{Z-2}^{A-4}\text{Y}\;+\;_{2}^{4}\text{He}\$

Example – Uranium‑238 → Thorium‑234

\$\;{92}^{238}\text{U}\;\rightarrow\;{90}^{234}\text{Th}\;+\;_{2}^{4}\text{He}\$

β⁻‑Decay

• A neutron is converted into a proton; an electron (β⁻) is emitted.
• Mass number remains unchanged, atomic number increases by 1.
• The daughter nuclide is a different element because Z changes.

Equation:

\$\;{Z}^{A}\text{X}\;\rightarrow\;{Z+1}^{A}\text{Y}\;+\;_{0}^{0}\beta^{-}\$

Example – Carbon‑14 → Nitrogen‑14

\$\;{6}^{14}\text{C}\;\rightarrow\;{7}^{14}\text{N}\;+\;_{0}^{0}\beta^{-}\$

γ‑Radiation

• Occurs when an excited nucleus releases excess energy as a photon.
• No change in mass number or atomic number; the element remains the same.
• γ‑radiation is electromagnetic radiation, not a particle, so it carries no mass or charge.

Equation:

\$\;{Z}^{A}\text{X}^{*}\;\rightarrow\;{Z}^{A}\text{X}\;+\;\gamma\$

Example – De‑excitation of cobalt‑60 after a β⁻‑decay

\$\;{27}^{60}\text{Co}^{*}\;\rightarrow\;{27}^{60}\text{Co}\;+\;\gamma\$

Step‑by‑Step Procedure for Writing a Decay Equation

1. Identify the type of radiation (α, β⁻ or γ).
2. Write the parent nuclide in \$_{Z}^{A}\text{X}\$ form.
3. Apply the appropriate changes to A and Z (see the quick‑reference table).
4. Write the daughter nuclide \$_{Z'}^{A'}\text{Y}\$ on the right‑hand side.
5. Attach the emitted particle using its standard notation (α‑particle, β⁻, or γ).
6. Check that both sides of the equation are balanced for mass number and atomic number.

Common Mistakes to Avoid

• Confusing the charge of the emitted particle with the change in atomic number.
• For β⁻‑decay, forgetting that the atomic number increases (the daughter is a different element).
• Omitting the antineutrino (or neutrino) – not required in IGCSE equations but important conceptually.
• Treating γ‑radiation as a particle with mass; it carries no mass or charge.

Link to Half‑Life (Section 5.2.4)

The exponential decay curve you will sketch later is the graphical representation of the equation:

\$N = N{0}\,e^{-t/\tau}\;=\;N{0}\left(\frac{1}{2}\right)^{t/t_{1/2}}\$

where N is the number of undecayed nuclei at time t, N₀ is the initial number, τ is the mean lifetime and t½ is the half‑life. The time at which the curve falls to half of its initial value (N = N₀/2) is defined as the half‑life. This relationship will be explored in detail in the next subsection.

Brief Derivation & Numerical Example

Because the probability of decay for each nucleus is constant, the rate of decay is proportional to the number of nuclei present:

\$\frac{dN}{dt} = -\lambda N\$

Integrating gives the exponential law shown above. λ is the decay constant (λ = 1/τ = \ln 2 / t_{1/2}).

Example: Start with 1 000 atoms of a radionuclide whose half‑life is 2 h. After 4 h the number remaining is

\$N = 1000\left(\frac{1}{2}\right)^{4/2}=1000\left(\frac{1}{2}\right)^{2}=250\;\text{atoms}.\$

Practice Questions & Model Answers

1. Write the decay equation for the α‑decay of \$^{226}_{88}\text{Ra}\$.
   

   Answer: \$\;{88}^{226}\text{Ra}\;\rightarrow\;{86}^{222}\text{Rn}\;+\;_{2}^{4}\text{He}\$

2. \$_{11}^{23}\text{Na}\$ undergoes β⁻‑decay. Identify the daughter nuclide.
   

   Answer: \$^{23}_{12}\text{Mg}\$ (mass number unchanged, atomic number +1).

3. A nucleus emits a β⁻‑particle and becomes \$^{55}_{26}\text{Fe}\$. What was the original nuclide?
   

   Answer: \$^{55}_{25}\text{Mn}\$ (add 1 to Z to obtain the parent).

4. Explain why the mass number does not change in γ‑radiation.
   

   Answer: γ‑radiation is the emission of a photon, which has no rest mass and carries no nucleons; therefore A (the total number of protons + neutrons) remains the same.

5. (Challenge) Sketch a simple decay curve and label the half‑life, showing that decay events are random and independent.
   

   Answer: (Provide a hand‑drawn or computer‑generated sketch with the curve starting at \$N0\$, falling exponentially, and a vertical line at \$t = t{1/2}\$ where the curve reaches \$N_0/2\$.)

Suggested Diagram (for the teacher)

A single schematic showing a parent nucleus at the centre with three arrows radiating outward:

• Arrow labelled “α” pointing to a small helium‑4 nucleus; beside it write “ΔA = −4, ΔZ = −2”.
• Arrow labelled “β⁻” pointing to an electron; beside it write “ΔA = 0, ΔZ = +1”.
• Arrow labelled “γ” pointing to a wavy line (photon); beside it write “ΔA = 0, ΔZ = 0”.

Below the three arrows, include a small exponential decay curve with the half‑life point marked, reinforcing the link to Section 5.2.4.

Additional Note (Optional for A‑Level)

Positron (β⁺) emission and electron capture are not part of the IGCSE 0625 specification, but they follow the same bookkeeping rules: the atomic number decreases by 1 while the mass number remains unchanged.