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Ionising Radiation & Nuclear Reactions

Syllabus reference

Unit 1, Topic 2 — 15 hours (including practicals)

Nuclear model

The nuclear model of the atom describes a small, dense, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons (collectively called nucleons).

Nuclide notation:

\[ {}^A_Z X \]

where \(A\) = mass number (protons + neutrons), \(Z\) = atomic number (protons), and \(X\) = element symbol.

Strong nuclear force

Protons in the nucleus experience electrostatic repulsion (like charges repel). The nucleus is held together by the strong nuclear force which:

  • acts over very short distances (approximately \(10^{-15}\) m)
  • is attractive between all nucleons (proton–proton, proton–neutron, neutron–neutron)
  • is much stronger than the electrostatic force at nuclear distances

A nuclide is stable when the strong nuclear force is sufficient to overcome electrostatic repulsion. Stability depends on the relative number of protons and neutrons — too many or too few neutrons leads to instability and radioactive decay.

Radioactive decay

Natural radioactive decay occurs when an unstable nucleus spontaneously emits radiation to become more stable.

Type Symbol Charge Mass (amu) Penetrating ability Ionising ability
Alpha \(\alpha\) or \({}^4_2\text{He}\) +2 4 Low (stopped by paper) High
Beta negative \(\beta^-\) or \({}^{\ \ 0}_{-1}e\) −1 ≈ 0 Medium (stopped by aluminium) Medium
Beta positive \(\beta^+\) or \({}^{0}_{+1}e\) +1 ≈ 0 Medium Medium
Gamma \(\gamma\) 0 0 High (reduced by lead/concrete) Low

Causes of decay:

  • Excess of neutrons → beta negative (\(\beta^-\)) decay
  • Excess of protons → beta positive (\(\beta^+\)) decay
  • Excess of mass (heavy nuclei) → alpha (\(\alpha\)) decay

Nuclear equations

Nuclear equations must be balanced — the total mass number and total atomic number must be conserved on both sides.

Worked example: Alpha decay

Question: Write the decay equation for uranium-238 undergoing alpha decay.

Solution:

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

Check: mass numbers \(238 = 234 + 4\) ✓, atomic numbers \(92 = 90 + 2\) ✓

Worked example: Beta negative decay

Question: Write the decay equation for carbon-14 undergoing beta negative decay.

Solution:

\[ {}^{14}_{6}\text{C} \rightarrow {}^{14}_{7}\text{N} + {}^{\ \ 0}_{-1}e + \bar{\nu}_e \]

A neutron converts to a proton, emitting an electron and an antineutrino.

A radionuclide will, through a series of spontaneous decays (a decay chain), eventually become a stable nuclide.

Half-life

The half-life (\(t_{1/2}\)) of a radioactive isotope is the time taken for half the nuclei in a sample to decay, or equivalently, the time for the activity to halve.

Worked example: Half-life

Question: A sample contains 800 g of a radioactive isotope with a half-life of 3 days. How much remains after 12 days?

Solution: Number of half-lives = \(\frac{12}{3} = 4\)

\[ 800 \xrightarrow{3\text{ days}} 400 \xrightarrow{3\text{ days}} 200 \xrightarrow{3\text{ days}} 100 \xrightarrow{3\text{ days}} 50 \text{ g} \]

Mass defect and nuclear energy

The mass defect (\(\Delta m\)) is the difference between the total mass of individual nucleons and the actual mass of the nucleus. This "missing" mass has been converted to binding energy according to Einstein's mass–energy equivalence:

Key formula

\[ E = mc^2 \]

where:

  • \(E\) = energy (J)
  • \(m\) = mass (kg)
  • \(c = 3 \times 10^8 \text{ m s}^{-1}\)

Binding energy per nucleon is a measure of nuclear stability. Iron-56 has the highest binding energy per nucleon and is the most stable nucleus.

  • Fission — heavy nuclei split into lighter nuclei, releasing energy (nuclei heavier than iron move toward higher binding energy per nucleon)
  • Fusion — light nuclei combine to form heavier nuclei, releasing energy (nuclei lighter than iron move toward higher binding energy per nucleon)
Worked example: Mass defect and energy

Question: The mass defect of a reaction is \(0.00431\) amu. Calculate the energy released in joules.

Solution:

Convert to kg: \(\Delta m = 0.00431 \times 1.66 \times 10^{-27} = 7.15 \times 10^{-30}\) kg

\[ E = \Delta m c^2 = 7.15 \times 10^{-30} \times (3 \times 10^8)^2 = 6.44 \times 10^{-13} \text{ J} \]

Simulations and videos

PhET Simulations:

Crash Course Physics:

External resources: