Students will be able to:
⁽ᴬ⁾₍ᶻ₎X
Example: ¹⁴₆C (6 p, 8 n) vs ¹²₆C (6 p, 6 n).
²³⁸₉₂U).| Nuclide | Mass number A | Atomic number Z | Element |
|---|---|---|---|
¹⁴₆C | 14 | 6 | Carbon |
¹²₆C | 12 | 6 | Carbon |
⁴₂He | 4 | 2 | Helium (α‑particle) |
Key inference for the syllabus: the atom consists of a central nucleus (protons + neutrons) surrounded by orbiting electrons.
Emission of an α‑particle (⁴₂α ≡ ⁴₂He).
General equation
\$\$
\,^{A}{Z}\text{X} \;\rightarrow\; \,^{A-4}{Z-2}\text{Y} \;+\; \,^{4}_{2}\alpha
\$\$
Example
\$\$
\,^{238}{92}\text{U} \;\rightarrow\; \,^{234}{90}\text{Th} \;+\; \,^{4}_{2}\alpha
\$\$
A neutron transforms into a proton, emitting an electron and an antineutrino (often omitted in syllabus work).
General equation
\$\$
\,^{A}{Z}\text{X} \;\rightarrow\; \,^{A}{Z+1}\text{Y} \;+\; \,^{0}{-1}\beta^{-} \;+\; \bar{\nu}e
\$\$
Example
\$\$
\,^{14}{6}\text{C} \;\rightarrow\; \,^{14}{7}\text{N} \;+\; \,^{0}{-1}\beta^{-} \;+\; \bar{\nu}e
\$\$
A proton converts into a neutron, emitting a positron and a neutrino.
General equation
\$\$
\,^{A}{Z}\text{X} \;\rightarrow\; \,^{A}{Z-1}\text{Y} \;+\; \,^{0}{+1}\beta^{+} \;+\; \nue
\$\$
Example
\$\$
\,^{22}{11}\text{Na} \;\rightarrow\; \,^{22}{10}\text{Ne} \;+\; \,^{0}{+1}\beta^{+} \;+\; \nue
\$\$
An inner‑shell electron is captured by the nucleus, turning a proton into a neutron and emitting a neutrino.
General equation
\$\$
\,^{A}{Z}\text{X} \;+\; e^{-} \;\rightarrow\; \,^{A}{Z-1}\text{Y} \;+\; \nu_e
\$\$
Example
\$\$
\,^{7}{4}\text{Be} \;+\; e^{-} \;\rightarrow\; \,^{7}{3}\text{Li} \;+\; \nu_e
\$\$
An excited nucleus releases excess energy as a high‑energy photon.
General equation
\$\$
\,^{A}{Z}\text{X}^{*} \;\rightarrow\; \,^{A}{Z}\text{X} \;+\; \gamma
\$\$
Example
\$\$
\,^{60}{27}\text{Co}^{*} \;\rightarrow\; \,^{60}{27}\text{Co} \;+\; \gamma
\$\$
\$N = N_{0}\,e^{-\lambda t}\$
\$t_{½} = \frac{\ln 2}{\lambda}\$
\$A = \lambda N\$
\$\Delta m = Z\,m{p} + (A-Z)\,m{n} - M_{\text{nuc}}\$
\$E_{b} = \Delta m\,c^{2}\;\;\;( \approx 931.5\;\text{MeV per atomic mass unit})\$
⁵⁶₍₂₆₎Fe).| Decay mode | Particle(s) emitted | Change in A | Change in Z | Typical example |
|---|---|---|---|---|
| α‑decay | ⁴₂α (He nucleus) | –4 | –2 | ⁸⁸₂U → ⁸⁴₂Th + α |
| β⁻‑decay | e⁻ (β⁻) + \$\bar{\nu}_e\$ | 0 | +1 | ¹⁴₆C → ¹⁴₇N + β⁻ |
| β⁺‑decay | e⁺ (β⁺) + νe | 0 | –1 | ²²₁₁Na → ²²₁₀Ne + β⁺ |
| Electron capture (EC) | νe (no charged particle) | 0 | –1 | ⁷₄Be + e⁻ → ⁷₃Li + νe |
| γ‑decay | γ photon | 0 | 0 | ⁶⁰₂₇Co* → ⁶⁰₂₇Co + γ |
| Syllabus Requirement | Coverage in these notes | Suggested further activity / emphasis |
|---|---|---|
| α‑particle scattering experiment & inference of a small nucleus | Section 2.1 gives a concise description and the key inference. | Include a labelled diagram of the gold‑foil experiment for visual learners. |
| Simple nuclear model (protons, neutrons, orbital electrons) | Section 2.2 outlines the ball model and distinguishes nuclear from electronic structure. | Add a schematic “nucleus + electron cloud” illustration. |
| Distinguish nucleon number (A) from proton number (Z) | Table 1.1 (side‑by‑side example) reinforces the distinction. | Use a short classroom quiz: “Identify A and Z for ⁶⁴₈Gd”. |
| Isotopes & notation A Z X | Covered in Section 1 with examples. | None needed. |
| Conservation of nucleon number & charge in reactions | Explicitly stated in Section 1 and applied in each decay equation. | Add a worked example showing how to balance a mixed‑decay reaction. |
| Write decay equations for α, β⁻, β⁺, EC and γ | Sections 3.1–3.5 provide general equations and worked examples. | Practice worksheet: complete the missing products for given parent nuclides. |
| Radioactive decay law, half‑life, activity | Section 4 presents the formulas with definitions. | Include a sample calculation of the activity of a 1 g sample of ⁶⁰Co. |
| Mass‑defect and binding energy (introductory) | Section 5 introduces the concepts and key equations. | Demonstrate a simple calculation for the binding energy of ⁴He. |
⁽ᴬ⁾₍ᶻ₎X → ⁽ᴬ⁻⁴⁾₍ᶻ⁻²₎Y + ⁴₂α⁽ᴬ⁾₍ᶻ₎X → ⁽ᴬ⁾₍ᶻ⁺¹₎Y + ⁰₋₁β⁻ (+ ν̅ₑ)⁽ᴬ⁾₍ᶻ₎X → ⁽ᴬ⁾₍ᶻ⁻₁₎Y + ⁰₊₁β⁺ (+ νₑ)⁽ᴬ⁾₍ᶻ₎X + e⁻ → ⁽ᴬ⁾₍ᶻ⁻₁₎Y + νₑ⁽ᴬ⁾₍ᶻ₎X* → ⁽ᴬ⁾₍ᶻ₎X + γN = N₀ e⁻ˡᵃᵐᵇᵈᵃ t, t½ = ln 2 / λ, A = λNEb = Δm c², with Δm = Z mp + (A‑Z) mn – MnucYour generous donation helps us continue providing free Cambridge IGCSE & A-Level resources, past papers, syllabus notes, revision questions, and high-quality online tutoring to students across Kenya.