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Atomic spectra and the Bohr model (SACE Stage 2 Physics)

Syllabus dot point

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

9 min read

Quick answer

In the Bohr model, electrons occupy fixed, discrete energy levels in an atom. An electron emits a photon when it drops to a lower level and absorbs one when it jumps up. The photon's energy exactly equals the difference between the two levels: Ephoton=hf=E2−E1E_{\text{photon}} = hf = E_2 - E_1. Because the levels are discrete, only specific photon energies (and so wavelengths) appear, producing line spectra.

Jump to a section
  1. What this dot point is asking
  2. The Bohr model
  3. Why spectra are lines, not continuous
  4. Reading energy-level diagrams
  5. How SACE assesses this
  6. Exam-style questions

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 eV-3.4\ \text{eV} to the ground state at −13.6 eV-13.6\ \text{eV}. This example finds the wavelength of the emitted photon, using h=6.63×10−34 J sh = 6.63\times10^{-34}\ \text{J s}, c=3.00×108 m s−1c = 3.00\times10^8\ \text{m s}^{-1}, and 1 eV=1.60×10−19 J1\ \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):

ΔE=(−3.4)−(−13.6)=10.2 eV.\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:

ΔE=10.2×1.60×10−19=1.63×10−18 J.\Delta E = 10.2\times1.60\times10^{-19} = 1.63\times10^{-18}\ \text{J}.

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

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

λ=hcΔE=(6.63×10−34)(3.00×108)1.63×10−18=1.99×10−251.63×10−18=1.22×10−7 m=122 nm.\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: λ=122 nm\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.

Photon emission and absorption

When an electron moves between energy levels, the photon involved carries exactly the energy difference:

Ephoton=hf=hcλ=Ehigher−ElowerE_{\text{photon}} = hf = \frac{hc}{\lambda} = E_{\text{higher}} - E_{\text{lower}}

  • Emission: an electron drops from a higher to a lower level, releasing a photon of that energy.
  • Absorption: an electron jumps up by absorbing a photon of exactly the right energy.

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.

Emission and absorption spectra
  • An emission spectrum is a set of bright coloured lines on a dark background, produced when excited atoms drop to lower levels and emit photons.
  • An absorption spectrum is a set of dark lines on a continuous bright background, produced when atoms absorb specific wavelengths from white light passing through them.

The dark lines in an absorption spectrum occur at exactly the same wavelengths as the bright lines in the emission spectrum of the same element.

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 00 eV. The energy is negative because the electron is bound; raising it to 00 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=Ehigher−ElowerE = E_\text{higher} - E_\text{lower}, convert from eV to joules using 1 eV=1.60×10−19 J1\ \text{eV} = 1.60\times10^{-19}\ \text{J}, then apply E=hfE = hf or E=hcλE = \dfrac{hc}{\lambda}. Sign handling is where most marks are lost, because the levels are negative, so a drop from −0.85-0.85 to −13.6-13.6 eV releases 12.7512.75 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.

Common errors
Using level energies instead of their difference
The photon energy is the difference between the two levels, Ehigher−ElowerE_{\text{higher}} - E_{\text{lower}}, not a single level's value.
Sign confusion
Energy levels are negative. Subtract carefully: a drop from −3.4-3.4 to −13.6-13.6 eV releases 10.210.2 eV.
Mixing up emission and absorption
Emission = electron falls and a photon is released (bright lines). Absorption = a photon is taken in and the electron rises (dark lines).
Forgetting the eV-to-joule conversion
Convert energy-level differences to joules before finding a wavelength with λ=hc/ΔE\lambda = hc/\Delta E.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2024-style4 marks
A hydrogen atom has electron energy levels n=1n=1 at −13.6 eV-13.6\ \text{eV}, n=2n=2 at −3.40 eV-3.40\ \text{eV}, n=3n=3 at −1.51 eV-1.51\ \text{eV} and n=4n=4 at −0.85 eV-0.85\ \text{eV}. An electron transitions from n=4n=4 to n=1n=1. Calculate the frequency of the emitted photon. Use h=6.63×10−34 J sh = 6.63\times10^{-34}\ \text{J s} and 1 eV=1.60×10−19 J1\ \text{eV} = 1.60\times10^{-19}\ \text{J}.
Show worked answer →

The photon energy equals the difference between the levels:

E=E4−E1=(−0.85)−(−13.6)=12.75 eV.E = E_4 - E_1 = (-0.85) - (-13.6) = 12.75\ \text{eV}.

Convert to joules: E=12.75×1.60×10−19=2.04×10−18 JE = 12.75\times1.60\times10^{-19} = 2.04\times10^{-18}\ \text{J}.
Use E=hfE = hf:
f=Eh=2.04×10−186.63×10−34=3.08×1015 Hz.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=hfE = hf to reach about 3.08×1015 Hz3.08\times10^{15}\ \text{Hz} (ultraviolet).

SACE 2023-style4 marks
An electron transitions from n=3n=3 (at −8.0 eV-8.0\ \text{eV}) to n=2n=2, emitting a photon of frequency 8.45×1014 Hz8.45\times10^{14}\ \text{Hz}. Determine the energy of the n=2n=2 level in eV.
Show worked answer →

Photon energy: E=hf=(6.63×10−34)(8.45×1014)=5.60×10−19 JE = hf = (6.63\times10^{-34})(8.45\times10^{14}) = 5.60\times10^{-19}\ \text{J}.

Convert to eV: E=5.60×10−191.60×10−19=3.50 eVE = \dfrac{5.60\times10^{-19}}{1.60\times10^{-19}} = 3.50\ \text{eV}.

This equals the gap, E=E3−E2E = E_3 - E_2, so E2=E3−E=(−8.0)−3.50=−11.5 eVE_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 E2=E3−EE_2 = E_3 - E, 1 mark for the answer of −11.5 eV-11.5\ \text{eV}. The n=2n=2 level is lower (more negative) than n=3n=3, as expected.

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Sources & how we know this

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

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