3.12 Photons

Properties of Photons

Fire a photon at a 3-level atom — it absorbs only when hν = ΔE; mismatches pass through. Switch to emit and drop the electron to send the matching photon out.

E = hνc = λνAbsorb / Emit3 Energy Levels
Topic 3.12

Properties of Photons

Explain the properties of an absorbed or emitted photon in relationship to an electronic transition in an atom or molecule.

When a photon is absorbed by an atom or molecule, the energy of that species increases by exactly the energy of the photon; when a photon is emitted, the energy decreases by exactly that amount. Energy is conserved, and because the transition energies are quantized, only photons of matching energy interact.

Two equations connect the wave and particle descriptions, and both are on the equation sheet:

c = λν and E = hν

Combining them gives the form you will use most: E = hc/λ. Note what this says — energy is inversely proportional to wavelength. Short wavelength (UV) means high energy; long wavelength (infrared, microwave) means low energy.

Units are the whole battle here. Wavelengths are quoted in nanometres but c is in m·s⁻¹, so convert: 1 nm = 10⁻⁹ m. And E = hν gives the energy of one photon in joules; multiply by Avogadro's number to get joules per mole, then divide by 1000 for kJ/mol.

Constants on the sheet: h = 6.626 × 10⁻³⁴ J·s and c = 2.998 × 10⁸ m·s⁻¹.

Key points

  • E = hν = hc/λ. Energy is inversely proportional to wavelength.
  • The absorbed or emitted photon energy equals the transition energy exactly — nothing partial.
  • Convert nm to m before using c = λν.
  • Per-photon energies are ~10⁻¹⁹ J; per-mole energies are ~10² kJ. If your answer is neither, check units.

Equations

  • on the exam sheet
    • 2.998 × 10⁸ m s⁻¹
    • wavelength (m)
    • frequency (Hz = s⁻¹)
  • on the exam sheet
    • photon energy (J)
    • 6.626 × 10⁻³⁴ J s
  • not on the sheetNot printed separately, but it is just the two sheet equations combined.

Common mistakes

  • Wavelength in nanometres. Convert to metres or every answer is off by 10⁹.
  • Per photon vs. per mole. Read the units in the question before deciding whether to multiply by NA.
  • Higher frequency means higher energy, but shorter wavelength. Do not mix the two directions.
  • Intensity is not energy per photon. Brighter light means more photons, not more energetic ones.

Worked example

A hydrogen atom emits light at 486 nm. Calculate (a) the frequency, (b) the energy of one photon, and (c) the energy per mole of photons.

(a) Frequency. First convert: λ = 486 nm = 4.86 × 10⁻⁷ m.
ν = c/λ = (2.998 × 10⁸ m/s) ÷ (4.86 × 10⁻⁷ m) = 6.17 × 10¹⁴ s⁻¹

(b) Energy of one photon.
E = hν = (6.626 × 10⁻³⁴ J·s)(6.17 × 10¹⁴ s⁻¹) = 4.09 × 10⁻¹⁹ J

(c) Per mole.
E = (4.09 × 10⁻¹⁹ J)(6.022 × 10²³ mol⁻¹) = 2.46 × 10⁵ J/mol = 246 kJ/mol

Both magnitudes are exactly where visible-light photons should land, which is a good check.

Full notes for topic 3.12 →