What is sign confusion?

Energy levels are negative. Subtract carefully: a drop from -3.4 to -13.6 eV releases 10.2 eV.

SACE Board SACE 2024-style practice: A hydrogen atom has electron energy levels n=1 at -13.6\ eV, n=2 at -3.40\ eV, n=3 at -1.51\ eV and n=4 at -0.85\ eV. An electron transitions from n=4 to n=1. Calculate the frequency of the emitted photon. Use h = 6.6310^-34\ J s and 1\ eV = 1.6010^-19\ J.

The photon energy equals the difference between the levels: $E = E_4 - E_1 = (-0.85) - (-13.6) = 12.75\ eV.$ Convert to joules: E = 12.751.6010^-19 = 2.0410^-18\ J. Use E = hf: $f = Eh = 2.0410^-186.6310^-34 = 3.0810^15\ Hz.$ 1 mark for the energy difference in eV, 1 mark for the sign handling, 1 mark for converting to joules, 1 mark for using E = hf to reach about 3.0810^15\ Hz (ultraviolet).

SACE Board SACE 2023-style practice: An electron transitions from n=3 (at -8.0\ eV) to n=2, emitting a photon of frequency 8.4510^14\ Hz. Determine the energy of the n=2 level in eV.

Photon energy: E = hf = (6.6310^-34)(8.4510^14) = 5.6010^-19\ J. Convert to eV: E = 5.6010^-191.6010^-19 = 3.50\ eV. This equals the gap, E = E_3 - E_2, so E_2 = E_3 - E = (-8.0) - 3.50 = -11.5\ eV. 1 mark for the photon energy in joules, 1 mark for converting to eV, 1 mark for E_2 = E_3 - E, 1 mark for the answer of -11.5\ eV. The n=2 level is lower (more negative) than n=3, as expected.

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SAPhysicsSyllabus dot point

Why do atoms emit and absorb light only at specific wavelengths, and how does the Bohr model explain this?

Explain atomic emission and absorption spectra using the Bohr model and the relationship between photon energy and energy-level differences.

How discrete atomic energy levels in the Bohr model produce line emission and absorption spectra, and the link between photon energy and the difference between energy levels.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The Bohr model
Why spectra are lines, not continuous
Reading energy-level diagrams
How SACE assesses this

What this dot point is asking

You need to explain emission and absorption line spectra using the Bohr model and the relationship between photon energy and the difference between atomic energy levels.

Wavelength of an emitted photon

An electron in hydrogen drops from the level at $-3.4\ \text{eV}$ to the ground state at $-13.6\ \text{eV}$ . This example finds the wavelength of the emitted photon, using $h = 6.63\times10^{-34}\ \text{J s}$ , $c = 3.00\times10^8\ \text{m s}^{-1}$ , and $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ .

Step 1: find the energy difference between the two levels

The photon carries away exactly the energy the electron loses in the transition. Taking the difference between the higher and lower level energies (remembering both are negative):

\Delta E = (-3.4) - (-13.6) = 10.2\ \text{eV}.

Step 2: convert the energy from eV to joules

The wavelength formula requires energy in SI units. Multiplying by the conversion factor:

\Delta E = 10.2\times1.60\times10^{-19} = 1.63\times10^{-18}\ \text{J}.

Step 3: compute the wavelength using $\lambda = hc/\Delta E$

A photon's energy and wavelength are related by $E = hc/\lambda$ , which rearranges to give wavelength directly:

\lambda = \frac{hc}{\Delta E} = \frac{(6.63\times10^{-34})(3.00\times10^8)}{1.63\times10^{-18}} = \frac{1.99\times10^{-25}}{1.63\times10^{-18}} = 1.22\times10^{-7}\ \text{m} = 122\ \text{nm}.

Final answer: $\lambda = 122\ \text{nm}$ , an ultraviolet line in the hydrogen Lyman series (transitions ending at the ground state).

The Bohr model

Bohr proposed that electrons can only exist in certain allowed orbits, each with a fixed energy - the atom's energy levels. The lowest level is the ground state; higher levels are excited states. An electron cannot have an energy between the allowed levels.

A photon whose energy does not match a level difference is simply not absorbed.

Why spectra are lines, not continuous

Because the energy levels are discrete, only certain transitions - and therefore only certain photon energies and wavelengths - are possible. Each element has its own unique set of levels, so it produces a unique pattern of spectral lines, like a fingerprint. If the levels were continuous, every transition energy would be possible and the spectrum would be an unbroken band; the existence of sharp lines is direct evidence for quantised energy levels.

This is how astronomers identify the elements in stars: dark absorption lines in starlight reveal which elements are present in the star's atmosphere.

Reading energy-level diagrams

Energy levels are usually drawn as horizontal lines, with the ground state at the bottom (most negative energy) and the ionisation level at $0$ eV. The energy is negative because the electron is bound; raising it to $0$ eV frees it (ionisation). Larger downward jumps emit higher-energy, shorter-wavelength photons.

How SACE assesses this

SACE Stage 2 questions give a set of energy levels (in eV) and ask for the frequency or wavelength of the photon emitted in a stated transition, or work backwards from a measured photon to find an unknown level. The dependable method is: take the difference between the two levels, $E = E_\text{higher} - E_\text{lower}$ , convert from eV to joules using $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ , then apply $E = hf$ or $E = \dfrac{hc}{\lambda}$ . Sign handling is where most marks are lost, because the levels are negative, so a drop from $-0.85$ to $-13.6$ eV releases $12.75$ eV. A common explanation part asks why discrete spectral lines are evidence for quantised levels; the answer is that each line corresponds to one fixed energy gap, which is only possible if the electron can occupy only certain discrete energies.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20244 marksA hydrogen atom has electron energy levels

n=1

-13.6\ \text{eV}

n=2

-3.40\ \text{eV}

n=3

-1.51\ \text{eV}

and

n=4

-0.85\ \text{eV}

. An electron transitions from

n=4

n=1

. Calculate the frequency of the emitted photon. Use

h = 6.63\times10^{-34}\ \text{J s}

and

1\ \text{eV} = 1.60\times10^{-19}\ \text{J}

Show worked answer →

The photon energy equals the difference between the levels:

E = E_4 - E_1 = (-0.85) - (-13.6) = 12.75\ \text{eV}.

Convert to joules:

E = 12.75\times1.60\times10^{-19} = 2.04\times10^{-18}\ \text{J}

.
Use

E = hf

f = \dfrac{E}{h} = \dfrac{2.04\times10^{-18}}{6.63\times10^{-34}} = 3.08\times10^{15}\ \text{Hz}.

1 mark for the energy difference in eV, 1 mark for the sign handling, 1 mark for converting to joules, 1 mark for using

E = hf

to reach about

3.08\times10^{15}\ \text{Hz}

(ultraviolet).

SACE 20234 marksAn electron transitions from

n=3

(at

-8.0\ \text{eV}

) to

n=2

, emitting a photon of frequency

8.45\times10^{14}\ \text{Hz}

. Determine the energy of the

n=2

level in eV.

Show worked answer →

Photon energy: $E = hf = (6.63\times10^{-34})(8.45\times10^{14}) = 5.60\times10^{-19}\ \text{J}$ .

Convert to eV: $E = \dfrac{5.60\times10^{-19}}{1.60\times10^{-19}} = 3.50\ \text{eV}$ .

This equals the gap, $E = E_3 - E_2$ , so $E_2 = E_3 - E = (-8.0) - 3.50 = -11.5\ \text{eV}$ .
1 mark for the photon energy in joules, 1 mark for converting to eV, 1 mark for $E_2 = E_3 - E$ , 1 mark for the answer of $-11.5\ \text{eV}$ . The $n=2$ level is lower (more negative) than $n=3$ , as expected.