Skip to content
CHEM 121 Studioby Learn4Less · UBC CHEM 121

5.1 · Light & matter

Light as photons

E = hν = hc/λ, per photon and per mole.

By the end you should be able to:

  • Convert between wavelength, frequency and photon energy, and place light in the EM spectrum

Formula

The energy of a photon

E=hν=hcλ,hc=1.986×10−25 J mE = h\nu = \frac{hc}{\lambda}, \qquad hc = 1.986\times10^{-25}\ \mathrm{J\,m}

Emol=E×NA,NA=6.022×1023 mol−1E_{\text{mol}} = E \times N_A, \qquad N_A = 6.022\times10^{23}\ \mathrm{mol^{-1}}

Light is absorbed and emitted in packets (photons). Higher frequency, or shorter wavelength, means a more energetic photon. EE is per photon, in J; EmolE_{\text{mol}} is per mole of photons, usually quoted in kJ/mol.

Key idea

The electromagnetic spectrum

RegionWavelengthWhat it typically excites
gamma raysbelow 0.01 nmnuclear transitions
X-rays0.01 to 10 nmcore electrons
ultraviolet10 to 400 nmvalence electrons; can break bonds
visible400 to 750 nmvalence electrons; colour
infrared750 nm to 1 mmmolecular vibrations
microwave1 mm to 10 cmmolecular rotations
radioabove 10 cmnuclear spins (NMR)

Photon energy increases up the table. Across the visible range, from long to short wavelength: red (about 700 nm), orange, yellow, green (about 530 nm), blue (about 470 nm), violet (about 400 nm).

Method

Converting between λ, ν and E

  1. Convert the wavelength to metres (nm × 10−910^{-9}).
  2. Frequency: ν=c/λ\nu = c/\lambda.
  3. Energy per photon: E=hν=hc/λE = h\nu = hc/\lambda.
  4. Per mole: multiply by NAN_A, then divide by 1000 for kJ/mol.
  5. Working backwards from kJ/mol: multiply by 1000 and divide by NAN_A to get J per photon, then λ=hc/E\lambda = hc/E.

Common mistake

Photon-energy traps

Wrong: leaving λ\lambda in nm in E=hc/λE = hc/\lambda. Right: convert to metres first: 500 nm = 5.00×10−75.00\times10^{-7} m.

Wrong: comparing the energy of one photon (J) directly with a bond energy (kJ/mol). Right: convert one of them with NAN_A first.

Wrong: "brighter light has more energetic photons." Right: brightness is the number of photons; the energy of each photon depends only on its frequency.

Worked example

Worked example: a 500 nm photon

ν=2.998×1085.00×10−7=6.00×1014 Hz\nu = \frac{2.998\times10^{8}}{5.00\times10^{-7}} = 6.00\times10^{14}\ \mathrm{Hz}

E=(6.626×10−34)(2.998×108)5.00×10−7=3.97×10−19 JE = \frac{(6.626\times10^{-34})(2.998\times10^{8})}{5.00\times10^{-7}} = 3.97\times10^{-19}\ \mathrm{J}

Per mole: (3.97×10−19)(6.022×1023)=2.39×105(3.97\times10^{-19})(6.022\times10^{23}) = 2.39\times10^{5} J/mol.

Answer: 3.97×10−193.97\times10^{-19} J per photon, or 239 kJ/mol.

Worked example

Worked example: which light can break Cl₂?

The Cl–Cl bond energy is 242 kJ/mol. Per molecule:

E=242×103 J mol−16.022×1023 mol−1=4.02×10−19 JE = \frac{242\times10^{3}\ \mathrm{J\,mol^{-1}}}{6.022\times10^{23}\ \mathrm{mol^{-1}}} = 4.02\times10^{-19}\ \mathrm{J}

λmax⁡=hcE=1.986×10−254.02×10−19=4.94×10−7 m\lambda_{\max} = \frac{hc}{E} = \frac{1.986\times10^{-25}}{4.02\times10^{-19}} = 4.94\times10^{-7}\ \mathrm{m}

Answer: photons with λ≤494\lambda \le 494 nm (blue, violet and ultraviolet) can break the bond; longer wavelengths cannot, however intense the light.

Worked example

Worked example: counting photons

How many photons are in a 1.00 mJ pulse of 532 nm laser light?

Ephoton=1.986×10−255.32×10−7=3.73×10−19 JE_{\text{photon}} = \frac{1.986\times10^{-25}}{5.32\times10^{-7}} = 3.73\times10^{-19}\ \mathrm{J}

N=1.00×10−3 J3.73×10−19 J=2.68×1015N = \frac{1.00\times10^{-3}\ \mathrm{J}}{3.73\times10^{-19}\ \mathrm{J}} = 2.68\times10^{15}

Answer: 2.68×10152.68\times10^{15} photons.

Check yourself

Fresh questions every time you visit. Answers count toward your progress.

Keep practising this topic