Cambridge Notes, Past Papers, Revision Questions

Atoms, Nuclei and Radiation

1. Nuclide notation

The standard way of writing a nuclide is

^{A}_{Z}\!X

where

\(Z\) – proton number (also called the atomic number)

\(A\) – nucleon number (mass number)

\(X\) – chemical symbol of the element

\(N = A - Z\) – neutron number

Example: carbon‑12 is written \(^{12}{6}\!C\). The same element with a different number of neutrons, carbon‑13, is written \(^{13}{6}\!C\).

2. Proton number (\(Z\)) and nucleon number (\(A\))

Proton number (\(Z\)) – the total count of protons in the nucleus. It uniquely determines the chemical element.

Nucleon number (\(A\)) – the total count of nucleons (protons + neutrons). It distinguishes between isotopes of the same element.

The three numbers are related by

A = Z + N

3. Quick comparison

Aspect	Proton number (\(Z\))	Nucleon number (\(A\))
What it counts	Protons only	Protons + neutrons
Symbol	\(Z\)	\(A\)
Determines	Chemical element (e.g. carbon, oxygen)	Specific isotope (e.g. \(^{12}\)C vs \(^{13}\)C)
Typical natural range	1 – 118 (known elements)	1 – ≈ 300 (heaviest naturally occurring nuclides)
Conservation in reactions	Conserved unless a nuclear transmutation changes the element	Conserved in every nuclear reaction

4. Conservation laws

Conservation of nucleon number (\(A\)): the total mass number on the left‑hand side of a nuclear equation equals that on the right‑hand side.

Conservation of proton number (\(Z\)): the total atomic number is also conserved, except when the reaction involves a change of element (e.g. β‑decay).

Both laws must be satisfied when writing or balancing any nuclear equation.

5. Types of nuclear radiation

Radiation	Particle emitted	Charge	Mass (u)	Typical energy	Effect on \(A\) and \(Z\)	Penetration / shielding
α (alpha)	\(^{4}_{2}\!He\) (helium nucleus)	+2 e	4 u	≈ 5 MeV	\(A\) − 4, \(Z\) − 2	Stopped by a sheet of paper or a few cm of air
β⁻ (beta‑minus)	Electron \(e^{-}\)	−1 e	≈ 0 u	≈ 0.1–10 MeV	\(A\) unchanged, \(Z\) + 1	Requires a few mm of aluminium or plastic
β⁺ (beta‑plus, positron emission)	Positron \(e^{+}\)	+1 e	≈ 0 u	≈ 0.1–10 MeV	\(A\) unchanged, \(Z\) − 1	Similar shielding to β⁻ (aluminium, plastic)
γ (gamma)	High‑energy photon	0 e	≈ 0 u	≈ 0.1–10 MeV (often higher)	\(A\) and \(Z\) unchanged	Requires dense material – lead or several cm of concrete

6. Antiparticles and neutrinos in β‑decay

In β⁻ decay a neutron converts to a proton, emitting an electron and an antineutrino \(\bar\nu_{e}\).

In β⁺ decay a proton converts to a neutron, emitting a positron and a neutrino \(\nu_{e}\).

Neutrinos (\(\nu{e}\)) and antineutrinos (\(\bar\nu{e}\)) carry away the missing energy and momentum; they have essentially zero mass and do not affect \(A\) or \(Z\).

7. Sample nuclear equations

All of the following obey the conservation of \(A\) and \(Z\).

α‑decay (e.g. uranium‑238)
\[
^{238}{92}\!U \;\longrightarrow\; ^{234}{90}\!Th \;+\; ^{4}_{2}\!\alpha
\]
\(A: 238 \to 234+4\) \(Z: 92 \to 90+2\)

β⁻ decay (e.g. carbon‑14)
\[
^{14}{6}\!C \;\longrightarrow\; ^{14}{7}\!N \;+\; e^{-} \;+\; \bar\nu_{e}
\]
\(A\) unchanged (14) \(Z\) increases from 6 to 7.

β⁺ decay (e.g. fluorine‑18)
\[
^{18}{9}\!F \;\longrightarrow\; ^{18}{8}\!O \;+\; e^{+} \;+\; \nu_{e}
\]
\(A\) unchanged (18) \(Z\) decreases from 9 to 8.

γ emission (often follows α or β decay)
\[
^{60}{27}\!Co^{*} \;\longrightarrow\; ^{60}{27}\!Co \;+\; \gamma
\]
Both \(A\) and \(Z\) remain 60 and 27; only excess nuclear energy is released as a photon.

8. Mass defect and nuclear binding energy

The mass of a nucleus is slightly less than the sum of the individual masses of its protons and neutrons. This “mass defect” \(\Delta m\) is released as binding energy:

E_{\text{binding}} = \Delta m \, c^{2}

Binding energy explains why nuclear reactions (e.g. fission, fusion) can release large amounts of energy compared with chemical reactions.

9. Real‑world applications

α particles – used in smoke detectors (‑\(^{241}\)Am source).
β⁻ emitters – radiotracers in medical imaging (e.g., \(^{14}\)C‑labelled compounds).
β⁺ emitters – positron emission tomography (PET) scanners use \(^{18}\)F.
γ rays – radiotherapy for cancer, sterilisation of medical equipment, and industrial radiography.

10. Practice question

Balance the following nuclear equation and state how \(A\) and \(Z\) change:

^{226}{88}\!Ra \;\longrightarrow\; \; ? \;+\; ^{4}{2}\!\alpha

Solution: The daughter nucleus is \(^{222}_{86}\!Rn\). \(A\) decreases by 4 and \(Z\) decreases by 2, consistent with α‑decay.

Suggested diagram: a schematic nucleus showing red circles for protons, blue circles for neutrons, with labels for \(Z\), \(N\) and \(A\). Place the notation \(^{A}_{Z}\!X\) beside the drawing.

distinguish between nucleon number and proton number