Atomic and nuclear physics: WACE Year 12 Physics Unit 4

Syllabus dot point

“Explain atomic energy levels and spectra, the Standard Model, mass-energy equivalence and nuclear reactions”

9 min readUpdated 2 Jun 2026

Quick answer

Electrons occupy discrete energy levels, so atoms emit and absorb photons only at energies equal to the gaps between levels, producing line spectra. Matter is built from quarks and leptons (the Standard Model), and nuclear reactions convert a tiny mass defect into large energy through $E=mc^2$ , which is the source of fission and fusion energy.

What this dot point is asking

WACE wants you to link three ideas: that quantised atomic energy levels explain spectral lines, that the Standard Model classifies fundamental particles, and that mass-energy equivalence accounts for the energy released in nuclear reactions through mass defect and binding energy. The recurring tool is energy conservation, now allowing mass to count as energy.

Atomic energy levels and spectra

Electrons in an atom can only occupy discrete energy levels. An electron drops from a higher level $E_i$ to a lower level $E_f$ by emitting a single photon whose energy equals the gap,

hf=E_i-E_f.

Because only certain gaps exist, only certain photon energies (and so certain wavelengths) appear: an emission spectrum is a set of bright lines on a dark background. Absorption works in reverse, with photons of exactly the right energy lifting electrons to higher levels, leaving dark lines in a continuous spectrum. The line pattern is unique to each element, which is how the composition of stars is identified.

The Standard Model

The Standard Model groups fundamental particles into quarks and leptons. Protons and neutrons are not fundamental: each is made of three quarks (a proton is two up and one down, a neutron is one up and two down). Leptons include the electron and the neutrino. Forces are carried by exchange particles (for example the photon for the electromagnetic force). For WACE you should recognise that everyday matter is built from up quarks, down quarks and electrons, and that antiparticles exist with opposite charge.

Mass-energy equivalence

Mass and energy are interchangeable,

E=mc^2,

with $c=3.0\times10^{8}\ \text{m s}^{-1}$ . Because $c^2$ is enormous, a very small mass corresponds to a very large energy. In nuclear reactions the total mass of the products is slightly less than that of the reactants, and this missing mass appears as released energy.

Mass defect and binding energy

The mass of a nucleus is less than the sum of the masses of its separate protons and neutrons. This difference is the mass defect $\Delta m$ , and the energy equivalent

E_b=\Delta m\,c^2

is the binding energy, the energy needed to pull the nucleus apart. Dividing by the number of nucleons gives the binding energy per nucleon, which peaks near iron. Nuclei lighter than iron release energy by joining together (fusion), while heavier nuclei release energy by splitting (fission), because both move toward more tightly bound products.

Nuclear reactions

In every nuclear reaction, nucleon number and charge are conserved. Radioactive decay includes alpha emission (a helium nucleus), beta emission (an electron from a neutron changing to a proton) and gamma emission (a high-energy photon as the nucleus de-excites). Fission splits a heavy nucleus into lighter fragments plus neutrons; fusion combines light nuclei. In each case the energy released equals the mass defect times $c^2$ .

Energy released from a mass defect

Energy from a 0.0186 u mass loss

A nuclear reaction loses $\Delta m=0.0186\ \text{u}$ , where $1\ \text{u}=1.66\times10^{-27}\ \text{kg}$ .

Step 1: Convert the mass defect to kilograms.

The equation $E = \Delta m\,c^2$ requires $\Delta m$ in kilograms and $c$ in metres per second to give an answer in joules. Atomic mass units must always be converted before substituting:

\Delta m=0.0186\times1.66\times10^{-27}=3.09\times10^{-29}\ \text{kg}.

Step 2: Apply $E = \Delta m\,c^2$ to find the energy released.

The speed of light is $c=3.0\times10^{8}\ \text{m s}^{-1}$ , which must be squared. The large value of $c^2$ is why even a tiny mass defect produces an enormous amount of energy:

E=\Delta m\,c^2=(3.09\times10^{-29})(3.0\times10^{8})^2.

Since $(3.0\times10^{8})^2=9.0\times10^{16}$ ,

E=3.09\times10^{-29}\times9.0\times10^{16}=2.8\times10^{-12}\ \text{J},

which is about $17\ \text{MeV}$ , a typical fusion energy release.

Choosing the right energy relationship

Decide whether a question concerns the atom (use $hf=E_i-E_f$ with energy levels) or the nucleus (use $E=\Delta m\,c^2$ with masses). Convert atomic mass units to kilograms and electronvolts to joules before substituting, and check that nucleon number and charge balance in any reaction equation.

Forgetting to square c, or not converting mass units

Two errors dominate. First, students multiply by $c$ instead of $c^2$ in $E=mc^2$ , which underestimates the energy by a factor of $3\times10^{8}$ . Second, they leave the mass defect in atomic mass units instead of converting to kilograms, so the joule answer is meaningless. Always square the speed of light and convert $\Delta m$ to kilograms (multiply u by $1.66\times10^{-27}$ ) before calculating the energy released.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2023-style7 marks

A nucleus of uranium-235 undergoes fission. The total mass of the products is

0.186\ \text{u}

less than the mass of the original nucleus plus the absorbed neutron, where

1\ \text{u}=1.66\times10^{-27}\ \text{kg}

. (a) Calculate the energy released in this single fission event, in joules and in MeV. (b) Explain, in terms of binding energy, why energy is released.

Show worked answer →

A 7 mark calculation rewards the mass-energy conversion in two units and a binding-energy explanation.

(a) Energy released. Mass defect $\Delta m=0.186\times1.66\times10^{-27}=3.09\times10^{-28}\ \text{kg}$ . Then

E=\Delta m c^2=(3.09\times10^{-28})(3.0\times10^{8})^2=2.78\times10^{-11}\ \text{J}.

Converting,

E=\dfrac{2.78\times10^{-11}}{1.6\times10^{-13}}=174\ \text{MeV}

(b) Binding energy. The fission products have a higher binding energy per nucleon than uranium-235, so the nucleons are more tightly bound after fission. The increase in total binding energy is released as kinetic energy of the fragments and neutrons, which is the mass that disappears via $E=\Delta mc^2$ .

Markers reward $\Delta m$ in kg, $E=\Delta mc^2$ , the conversion to about $174\ \text{MeV}$ and the higher-binding-energy-per-nucleon explanation.

WACE 2020-style5 marks

An electron in a hydrogen atom drops from an energy level of

-1.51\ \text{eV}

to a level of

-3.40\ \text{eV}

. (a) Calculate the energy of the emitted photon. (b) Calculate the wavelength of the emitted light and state the region of the spectrum it lies in.

Show worked answer →

A 5 mark calculation rewards the energy difference and the photon wavelength.

(a) Photon energy. The photon energy equals the magnitude of the level difference: $E=|-1.51-(-3.40)|=1.89\ \text{eV}=1.89\times1.6\times10^{-19}=3.02\times10^{-19}\ \text{J}$ .

(b) Wavelength. From $E=\dfrac{hc}{\lambda}$ , $\lambda=\dfrac{hc}{E}=\dfrac{(6.63\times10^{-34})(3.0\times10^{8})}{3.02\times10^{-19}}=6.6\times10^{-7}\ \text{m}$ . This is about $660\ \text{nm}$ , in the visible (red) region.

Markers reward the level difference of $1.89\ \text{eV}$ , conversion to joules, $\lambda=hc/E$ near $6.6\times10^{-7}\ \text{m}$ and identifying visible red light.

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Physics ATAR course: syllabus and support materials — SCSA
Physics: past ATAR course examinations — SCSA