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Atomic models and energy levels: WACE Year 12 Physics Unit 4

Syllabus dot point

“Describe the development of atomic models and explain quantised electron energy levels”

8 min read

Quick answer

Rutherford's experiments showed the atom has a tiny dense positive nucleus with electrons around it, but classical physics said orbiting electrons should radiate energy and spiral in. Bohr fixed this by proposing that electrons occupy only certain allowed energy levels and radiate only when jumping between them. An electron drops a level by emitting a photon of energy E=hf=Ehigh−ElowE=hf=E_{\text{high}}-E_{\text{low}}, and absorbs one to jump up.

What this dot point is asking

WACE wants you to trace how atomic models developed and explain quantised energy levels in terms of photon emission and absorption. This is the bridge between the photon idea and the line spectra of atoms.

From Rutherford to a problem

Rutherford's gold-foil scattering experiment showed that most of an atom is empty space, with nearly all the mass and the positive charge concentrated in a tiny central nucleus. Electrons orbit this nucleus. The trouble is that, by classical electromagnetism, an accelerating (orbiting) charge should continuously radiate energy, lose speed and spiral into the nucleus in a fraction of a second. Real atoms are stable, so the classical picture was incomplete.

Bohr's quantised levels

Bohr postulated that electrons can occupy only certain discrete, stable energy levels (orbits) without radiating, and that radiation occurs only when an electron jumps between levels. Each level has a fixed energy, with the lowest (ground state) most tightly bound and higher levels closer together near the top. The energies are negative, measured relative to a free electron at zero, because the electron is bound to the atom.

Emission and absorption

When an electron falls from a higher level EhighE_{\text{high}} to a lower level ElowE_{\text{low}}, it emits a single photon carrying exactly the energy difference:

hf=Ehigh−Elow.hf=E_{\text{high}}-E_{\text{low}}.

To jump up, the atom must absorb a photon of exactly that energy. Because only specific differences exist, atoms emit and absorb only specific frequencies, which is why spectra are discrete lines rather than a continuous band.

Why the levels are quantised

De Broglie's matter waves give a physical reason for Bohr's rule: an electron orbit is stable only when a whole number of electron wavelengths fits around the circumference, forming a standing wave. Orbits that do not fit a whole number of wavelengths interfere destructively and cannot persist, so only certain radii (and hence energies) are allowed.

Photon emitted in an energy-level transition

Wavelength of an emission line

An electron drops from a level at −1.5 eV-1.5\ \text{eV} to a level at −3.4 eV-3.4\ \text{eV}. Find the wavelength of the emitted photon (h=6.63×10−34 J sh=6.63\times10^{-34}\ \text{J s}, 1 eV=1.6×10−19 J1\ \text{eV}=1.6\times10^{-19}\ \text{J}).

Energy of the photon equals the level difference:

E=(−1.5)−(−3.4)=1.9 eV=1.9×1.6×10−19=3.04×10−19 J.E=(-1.5)-(-3.4)=1.9\ \text{eV}=1.9\times1.6\times10^{-19}=3.04\times10^{-19}\ \text{J}.

Wavelength from E=hc/λE=hc/\lambda:

λ=hcE=(6.63×10−34)(3.0×108)3.04×10−19=6.5×10−7 m=650 nm.\lambda=\frac{hc}{E}=\frac{(6.63\times10^{-34})(3.0\times10^8)}{3.04\times10^{-19}}=6.5\times10^{-7}\ \text{m}=650\ \text{nm}.

This red line is part of the visible hydrogen spectrum.

The development of the models in order

WACE expects you to be able to outline how the picture of the atom evolved, not just state the final model. Thomson's plum-pudding model treated the atom as positive charge spread through a sphere with electrons embedded in it. Rutherford's gold-foil scattering, in which a few alpha particles bounced almost straight back, showed this was wrong and that the positive charge and nearly all the mass sit in a tiny dense nucleus. Bohr then added quantised energy levels to explain why orbiting electrons do not radiate and why atoms produce discrete line spectra. De Broglie supplied a physical reason for the quantisation through standing matter waves, and the later quantum-mechanical model replaced fixed orbits with probability distributions (orbitals). A short chronological sketch, naming the experiment or idea that forced each change, is a common short-answer question.

Getting the energy difference right

Subtract the lower (more negative) energy from the higher one and take the magnitude; the photon energy is always positive. Convert electronvolts to joules before using hh unless you are working in eV throughout. Emission is a downward jump, absorption an upward jump.

Forgetting that bound energy levels are negative

Energy levels are quoted as negative values because the electron is bound, with zero set at the point where it just escapes the atom. The ground state is the most negative. A transition energy is the difference between two levels, for example (−1.5)−(−3.4)=1.9 eV(-1.5)-(-3.4)=1.9\ \text{eV}, not −1.5−3.4-1.5-3.4. Mishandling the signs gives the wrong photon energy.

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 2022-style6 marks
A hydrogen atom has energy levels including E1=−13.6 eVE_1=-13.6\ \text{eV}, E2=−3.40 eVE_2=-3.40\ \text{eV} and E3=−1.51 eVE_3=-1.51\ \text{eV}. (a) Calculate the wavelength of the photon emitted when an electron falls from E3E_3 to E2E_2. (b) Determine the minimum energy, in joules, needed to ionise the atom from its ground state.
Show worked answer →

A 6 mark calculation rewards the transition wavelength and the ionisation energy.

(a) Transition wavelength. Photon energy E=∣E3−E2∣=∣−1.51−(−3.40)∣=1.89 eV=3.02×10−19 JE=|E_3-E_2|=|-1.51-(-3.40)|=1.89\ \text{eV}=3.02\times10^{-19}\ \text{J}. Then

λ=hcE=(6.63×10−34)(3.0×108)3.02×10−19=6.6×10−7 m.\lambda=\frac{hc}{E}=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{3.02\times10^{-19}}=6.6\times10^{-7}\ \text{m}.

(b) Ionisation energy. Ionising from the ground state means raising the electron from −13.6 eV-13.6\ \text{eV} to 00, so E=13.6 eV=13.6×1.6×10−19=2.18×10−18 JE=13.6\ \text{eV}=13.6\times1.6\times10^{-19}=2.18\times10^{-18}\ \text{J}.

Markers reward the level difference of 1.89 eV1.89\ \text{eV}, λ=hc/E\lambda=hc/E near 6.6×10−7 m6.6\times10^{-7}\ \text{m} and the ionisation energy of 2.18×10−18 J2.18\times10^{-18}\ \text{J}.

WACE 2021-style5 marks
Explain the problem with the Rutherford model of the atom according to classical physics, and explain how Bohr's postulates resolved it.
Show worked answer →

A 5 mark explanation needs the classical instability and Bohr's two key postulates.

The classical problem. In the Rutherford model electrons orbit the nucleus. Classical electromagnetism says an accelerating charge (an orbiting electron is accelerating because its direction changes) must radiate energy continuously. The electron would therefore lose energy, spiral inward and collapse into the nucleus in a tiny fraction of a second, so the atom could not be stable, contradicting observation.

Bohr's resolution. Bohr postulated that electrons occupy only certain discrete, stable energy levels in which they do not radiate, and that radiation occurs only when an electron jumps between levels, emitting or absorbing a photon of energy hf=Ehigh−Elowhf=E_{\text{high}}-E_{\text{low}}. This both kept atoms stable and explained their discrete line spectra.

Markers reward the radiating-accelerating-charge collapse problem and Bohr's stable-levels plus quantised-jump postulates.

Practise this

Sources & how we know this

Based on the official SCSA syllabus. Not yet individually reviewed by a person; always check SCSA and your school's materials for current rules.

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