Why do atoms and molecules absorb selected photon energies?
You will be able to: Relate photon energy and direction of energy transfer to allowed transitions.
Why do atoms and molecules absorb selected photon energies?
A simplified species has two allowed states separated by 4.00 × 10⁻¹⁹ J. A photon matching that gap can be absorbed in an allowed transition; an arbitrary smaller-energy photon cannot supply the missing energy in this one-photon model.
A useful starting point: How do wavelength and photon energy connect? →
Words and symbols before equations
- Energy level
- An allowed energy state of a species.
- Energy gap ΔE
- Difference between two levels; Δ means “change in.”
- Excitation
- Promotion to a higher energy state.
- Emission
- Release of a photon as a species moves to a lower state.
What this picture assumes
Two-level, single-photon, allowed-transition teaching model. Ground level is assigned zero as a reference. Real species have additional levels, selection rules and line/band broadening.
Read the picture in three steps
- Read the species and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- Absorption: species gains 4e-19 J. Photon energy is positive 4e-19 J; wavelength 497 nm.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
During absorption, the species gains energy equal to the photon’s energy. For a matching transition, Ephoton = Eupper − Elower. The positive gap corresponds to an upward change in the species’ energy.
During emission, the species loses that amount of energy while the outgoing photon carries positive energy. A downward arrow is not a negative-energy photon.
In the simplified isolated-level model, matching energy is necessary and the transition must also be allowed. Real spectra may contain many transitions and broadened bands; the model does not claim infinitely sharp two-line behavior for every material.
Keep photon energy separate from the sign of the species’ energy change. This prevents an algebraic sign from becoming a false physical statement.
A worked example, step by step
For an allowed gap of 4.00 × 10⁻¹⁹ J, find the corresponding photon wavelength using h = 6.626 × 10⁻³⁴ J·s and c = 3.00 × 10⁸ m/s.
- Set the positive photon energy equal to the gap.
- Rearrange E = hc/λ to λ = hc/E.
- λ = (6.626 × 10⁻³⁴)(3.00 × 10⁸)/(4.00 × 10⁻¹⁹) = 4.97 × 10⁻⁷ m.
- Convert to about 497 nm. Absorption raises the species’ energy; emission across that gap lowers it and produces the same photon energy.
A downward transition gives a negative change for the species, but the emitted photon’s energy is positive.
Can a photon with half the required gap excite that transition in this one-photon model?
Compare with an explanation
No. It does not supply the required energy. Multiphoton processes are outside this model.
Predict. Change one thing. Explain.
Change the gap and switch absorption/emission. Compare the arrow direction with the positive photon energy and wavelength readouts.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Absorption: species gains 4e-19 J. Photon energy is positive 4e-19 J; wavelength 497 nm.
Two-level, single-photon, allowed-transition teaching model. Ground level is assigned zero as a reference. Real species have additional levels, selection rules and line/band broadening.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using particle interactions, concentration, gas behavior or energy transfer. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA species drops by 3.00 × 10⁻¹⁹ J in an allowed transition. State the signs of the species’ energy change and photon energy, then calculate wavelength.
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Compare with the answer and four-point rubric
- 1 point: ΔEspecies = −3.00 × 10⁻¹⁹ J.
- 1 point: Ephoton = +3.00 × 10⁻¹⁹ J.
- 1 point: λ = hc/E = (6.626 × 10⁻³⁴)(3.00 × 10⁸)/(3.00 × 10⁻¹⁹).
- 1 point: λ = 6.63 × 10⁻⁷ m, about 663 nm; the photon carries away the lost energy.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What does Δ mean?
A change or difference between specified values.
RECALL 2What is conserved in a one-photon transition?
Energy transferred between the species and radiation.
RECALL 3Is every energetically matching transition guaranteed to occur?
No; the transition must also be allowed.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why do atoms and molecules absorb selected photon energies?
- Ephoton =
- ΔEspecies
- = hν.
- Absorption: species gains energy. Emission: species loses energy.
Remember: A downward transition gives a negative change for the species, but the emitted photon’s energy is positive.
Conditions: Two-level, single-photon, allowed-transition teaching model. Ground level is assigned zero as a reference. Real species have additional levels, selection rules and line/band broadening.
Refresh Kid · AP Chemistry Unit 3 · Objectives 3.12.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.12, objective 3.12.A. CED effective Fall 2024 and June 2026 clarifications checked September 16, 2026. Unit 3: Properties of Substances and Mixtures, Topics 3.1–3.13. Focused lesson names, examples, models and assessments are original Refresh Kid teaching materials, not additional official topics or official AP questions. Official corrections.
The model states its assumptions beside the diagram. Colligative-property calculations and solution molality/mass-percent/volume-percent calculations are not required here. The optional speed-density model illustrates distributions; it does not require memorizing its mathematical derivation.
Teaching resources: The Organic Chemistry Tutor video titles/descriptions and topic coverage were checked for optional links; no claim is made to have watched every video. No creator scripts, examples, worksheets or artwork were copied. GitHub’s 3D website collection and its Three.js camera-control example informed the idea of controllable spatial inspection. Scientific diagrams, geometry and interactions here are original. The self-hosted Three.js runtime retains its MIT license. Camera rotation changes the view, not the chemistry.
Independent teacher review and observation of students remain pending. Implementation checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Chemistry questions and scoring guides. This archive contains questions across units; it is not an assignment of every question to this lesson.
The teaching sequence is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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