Physics · Units 3 & 4

Atomic Spectra and Energy Levels

Understand atomic spectra and energy levels the easy way, with plain English intuition, an interactive simulation, energy level diagrams, emission and absorption spectra, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.

Learn

Hold a flame under some table salt and the light turns a sharp orange. Heat a different element and you get a different colour, every single time. That happens because the electrons inside an atom can only sit at certain fixed energies, like rungs on a ladder, and the light an atom gives off is set entirely by the size of the jumps between those rungs. Learn the rungs and you can read which element you are looking at from its light alone.

Electrons live on a ladder

Picture the electron in an atom standing on a ladder. It can stand on rung 1, or rung 2, or rung 3, but never floating halfway between two rungs. These allowed energies are called energy levels, and the fact that only certain values are allowed is what we mean by quantised.

  • To climb up a rung, the electron has to take in a chunk of energy. It does this by absorbing a photon.
  • To drop down a rung, the electron has to get rid of energy. It does this by emitting a photon.

The photon that is absorbed or released always carries an energy exactly equal to the gap between the two rungs.

energyE₄E₃E₂E₁dropphotona bright spectral line

The blue dot is the electron. When it drops from a higher rung to a lower one, the red arrow, it sheds energy as a single photon, the blue wavy arrow. That photon shows up as one bright line of a single colour in the atom’s spectrum.

The gap is everything

This is the rule that ties the whole topic together. The energy of the photon equals the size of the jump, nothing more and nothing less. Write the higher level energy minus the lower level energy, and that is your photon energy.

ΔE=hf=Ehigh−Elow\Delta E = hf = E_{high} - E_{low}

Because the photon energy is fixed by the gap, and the gap is fixed by the atom, each element can only ever emit or absorb its own special set of photon energies. That fixed set of lines is the atom’s fingerprint.

Emission and absorption: two sides of the same coin

The same set of energy gaps shows up in two opposite ways, depending on whether the atom is giving out light or taking it in.

  • An emission line spectrum is a set of bright lines on a black background. A hot gas glows, its electrons drop down, and each drop sends out a photon of one fixed colour.
  • An absorption spectrum is a set of dark lines on a continuous rainbow. White light passes through a cooler gas, the gas absorbs the photons that match its own gaps, and those exact colours go missing.

The clever part is that both patterns sit at the same wavelengths for a given element, because both are controlled by the same energy gaps. A line you see glowing in emission is the same line you see missing in absorption.

See it for yourself

Switch on the light source, fire photons at a single hydrogen atom, and watch the electron jump up when it absorbs a photon and drop back down when it emits one. Try the different models and notice that only certain photon energies cause a jump, exactly the ones that match a gap between levels.

Interactive simulation, Models of the Hydrogen Atom Source: PhET Interactive Simulations, University of Colorado Boulder (CC BY 4.0)

How to actually solve one

Most exam questions are a short, repeatable recipe.

  1. Read off the two energy levels involved in the jump, keeping their signs.
  2. Find the photon energy from the gap, ΔE=Ehigh−Elow\Delta E = E_{high} - E_{low}.
  3. If you need joules, multiply the answer in electronvolts by 1.6×10−191.6 \times 10^{-19}.
  4. If you need the frequency, use ΔE=hf\Delta E = hf rearranged to f=ΔEhf = \dfrac{\Delta E}{h}, with ΔE\Delta E in joules.

Watch the units. Energy levels are usually quoted in electronvolts, but ΔE=hf\Delta E = hf only works when the energy is in joules, so convert before you reach for Planck’s constant.

Try one: an electron drops from a level at −1.5-1.5 eV to a level at −6.5-6.5 eV. Find the energy of the emitted photon in electronvolts, then in joules. Take 1 eV=1.6×10−191 \text{ eV} = 1.6 \times 10^{-19} J.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

What does it mean that electron energies are quantised?
When does an atom emit a photon, and when does it absorb one?
Write the rule linking photon energy to the energy levels.
Why does each element have its own set of spectral lines?
What is an emission spectrum?
What is an absorption spectrum, and where do its dark lines sit?
Recall · The Photoelectric Effect
Write the energy of a photon in terms of its frequency.
Recall · Matter Waves
Why can an electron only sit on certain orbits (de Broglie picture)?

See the recipe in action in the Worked Examples tab, then test yourself in Try It.

Worked examples

Worked Example 1Energy of an emitted photon

An electron in an atom drops from an energy level of −2.0-2.0 eV to −5.0-5.0 eV. Find the energy of the emitted photon, in electronvolts and in joules. Take 1 eV=1.6×10−191 \text{ eV} = 1.6 \times 10^{-19} J.

  1. 1

    The photon carries away exactly the gap between the two levels, so subtract the lower energy from the higher one.

    ΔE=Ehigh−Elow=(−2.0)−(−5.0)=3.0 eV\Delta E = E_{high} - E_{low} = (-2.0) - (-5.0) = 3.0 \text{ eV}
  2. 2

    Convert electronvolts to joules by multiplying by the size of one electronvolt.

    ΔE=3.0×1.6×10−19=4.8×10−19 J\Delta E = 3.0 \times 1.6 \times 10^{-19} = 4.8 \times 10^{-19} \text{ J}
Answer
ΔE=3.0 eV=4.8×10−19 J\Delta E = 3.0 \text{ eV} = 4.8 \times 10^{-19} \text{ J}
Worked Example 2Frequency of that photon

Find the frequency of the photon emitted in the previous example. Take h=6.63×10−34h = 6.63 \times 10^{-34} J s and use the photon energy ΔE=4.8×10−19\Delta E = 4.8 \times 10^{-19} J.

  1. 1

    Photon energy and frequency are linked by ΔE=hf\Delta E = hf, so rearrange to make frequency the subject.

    f=ΔEhf = \frac{\Delta E}{h}
  2. 2

    Substitute the photon energy in joules and Planck's constant.

    f=4.8×10−196.63×10−34=7.2×1014 Hzf = \frac{4.8 \times 10^{-19}}{6.63 \times 10^{-34}} = 7.2 \times 10^{14} \text{ Hz}
Answer
f=7.2×1014 Hzf = 7.2 \times 10^{14} \text{ Hz}

Practice questions

Practice test

Try it yourself

6 questions, 9 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.An electron moves from a higher energy level to a lower energy level inside an atom. During this jump the atom:

1mark
Need a hint?
Dropping down means the atom is shedding energy. Where does that energy go?

Q2.The energy of the photon released or taken in during a transition is equal to:

1mark
Need a hint?
The photon bridges the gap. Think about what fills the space between the two levels.

Q3.An electron drops from a level at −1.0-1.0 eV to a level at −4.0-4.0 eV. The energy of the emitted photon is:

1mark
Need a hint?
Subtract the lower energy from the higher one, keeping the negative signs.

Q4.A cool gas is placed in front of a hot white light source. The spectrum seen through the gas shows:

1mark
Need a hint?
The cool gas takes out only the photons whose energy matches its own gaps.

Q5.An electron drops from a level at −3.0-3.0 eV to a level at −9.0-9.0 eV. Find the energy of the emitted photon in electronvolts, then convert it to joules. Take 1 eV=1.6×10−191 \text{ eV} = 1.6 \times 10^{-19} J. Show your working.

3marks

Work this on paper. The worked solution appears once you submit.

Q6.A hydrogen atom emits a photon of energy 2.92.9 eV. Calculate the wavelength of the emitted photon, in nanometres (h=6.63×10−34h = 6.63 \times 10^{-34} J s, c=3.0×108c = 3.0 \times 10^{8} m/s, 11 eV =1.6×10−19= 1.6 \times 10^{-19} J).

2marks

Work this on paper. The worked solution appears once you submit.

VCAA 2025 Physics Exam, Section B Q18

Frequently asked questions

Why does each element have its own set of spectral lines?
Every element has its own fixed ladder of energy levels, so the gaps between its levels are unique. Those gaps set the exact photon energies it can emit or absorb, which is why the line pattern works like a fingerprint.
What is the difference between an emission and an absorption spectrum?
An emission spectrum is bright coloured lines on a black background, made when a hot gas releases photons as electrons drop down. An absorption spectrum is dark lines on a continuous rainbow, made when a cool gas removes those same photons from white light passing through it.
Why can an electron only sit at certain energies?
Inside an atom the electron energies are quantised, meaning only certain discrete values are allowed and nothing in between. The electron can never have an energy that falls in a gap, so it must jump from one allowed level straight to another.