Calculate photon frequency from atomic energy levels
Use hf = E₁ − E₂ to calculate the photon frequency for an atomic transition and interpret its direction.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 22. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use wave frequency and wavelength, energy, momentum, SI prefixes, electronvolts and scientific notation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What sets the energy of a photon emitted in an atomic transition?
What you will learn
- Use hf = E₁ − E₂ for a photon transition
- Calculate frequency from an energy-level difference
- Distinguish emission and absorption transitions
A level difference fixes the photon energy
For two atomic energy levels E₁ and E₂, the photon energy equals the magnitude of the energy difference. The syllabus relation is hf = E₁ − E₂ when E₁ is the higher level in the transition. Thus f = (E₁ − E₂)/h.
If the atom moves from the higher level to the lower one, the energy leaves as an emitted photon. If it moves upward, the atom absorbs a photon with the same energy difference. State which energy is larger before subtracting so the photon energy is positive.
Convert units before dividing by Planck's constant
When a level gap is given in electronvolts, convert it to joules if using h = 6.63 × 10⁻³⁴ J s. Alternatively use h = 4.14 × 10⁻¹⁵ eV s and keep the level gap in eV. Do not mix joules with eV s.
After finding f, check its scale. An energy gap of a few eV corresponds to a frequency of order 10¹⁴ Hz. If wavelength is requested as a follow-on, use λ = c/f and convert metres to nanometres where useful.
Pause and explain
An atom falls from E₁ to E₂, with E₁ > E₂. Which expression gives the photon frequency?
Worked example
An atom makes a downward transition with energy gap 2.0 eV. Find the emitted photon frequency. Use h = 4.14 × 10⁻¹⁵ eV s.
Show the worked solution
- Use hf = ΔE.
- Divide ΔE by h, keeping eV units consistent.
- Identify the transition as emission because the atom moves down.
Answer f = 2.0/(4.14 × 10⁻¹⁵) ≈ 4.83 × 10¹⁴ Hz; the photon is emitted.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
An atom emits a photon when it drops through a 3.0 eV energy gap. Calculate the frequency using h = 4.14 × 10⁻¹⁵ eV s.
Give me a hint
Use f = ΔE/h with both values expressed using eV.
Compare my reasoning
- f = 3.0/(4.14 × 10⁻¹⁵).
- f ≈ 7.25 × 10¹⁴ s⁻¹.
- State hertz and emission direction.
f ≈ 7.2 × 10¹⁴ Hz; the atom emits the photon as it moves to a lower level.
Look for these in your work
- Energy gap used
- Planck constant divided correctly
- Frequency and process stated
A photon of frequency 6.0 × 10¹⁴ Hz is absorbed by an atom. Find the energy-level gap in eV. Use h = 4.14 × 10⁻¹⁵ eV s.
Give me a hint
Use ΔE = hf and keep h in eV s.
Compare my reasoning
- ΔE = (4.14 × 10⁻¹⁵)(6.0 × 10¹⁴) eV.
- ΔE = 2.484 eV.
- Absorption moves the electron to a level higher by this energy.
The level gap is about 2.48 eV, and the atom moves upward by absorbing the photon.
Look for these in your work
- Energy relation applied
- eV unit retained
- Absorption direction interpreted
An atom emits a photon of frequency 5.0 × 10¹⁴ Hz. Estimate its wavelength and state how this compares with a 600 nm line.
Give me a hint
Use λ = c/f after finding the photon frequency; convert metres to nanometres.
Compare my reasoning
- λ = 3.00 × 10⁸/(5.0 × 10¹⁴) m.
- λ = 6.0 × 10⁻⁷ m = 600 nm.
- The line matches a wavelength of 600 nm.
λ = 600 nm, the same as the comparison line.
Look for these in your work
- Wave relation selected
- Unit conversion correct
- Comparison made
Common mistakes
- Subtracting the lower energy from the higher in the wrong order
- Treating a downward transition as absorption
- Mixing ΔE in eV with h in J s
- Reporting the energy difference as a frequency without dividing by h
What can you explain now?
What sets the energy of a photon emitted in an atomic transition?
Compare with the explanation
The difference between the two energy levels
The photon carries the energy lost by the atom.
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Make it stick
How do an atomic energy gap and Planck's constant determine the transition-photon frequency?
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