Wavelength, Frequency and Photon Energy Calculations

Three equations connect everything about a photon: its wavelength, its frequency and its energy. With them, you can work out the colour of a flame from an energy level diagram, find whether ultraviolet light can break a chemical bond, or count the photons coming out of a laser pointer. This article collects the equations and works through every common type of problem, with careful attention to units, which is where most mistakes happen.

The equations

1. Wave equation: c = f × λ

• c = speed of light = 3.00 × 10⁸ m s⁻¹
• f = frequency, in hertz (Hz = s⁻¹)
• λ = wavelength, in metres

2. Planck’s equation: E = h × f

• E = energy of one photon, in joules (J)
• h = Planck’s constant = 6.626 × 10⁻³⁴ J s

3. Combined: E = hc ÷ λ

Energy per mole of photons: E_molar = E × N_A, where N_A = 6.022 × 10²³ mol⁻¹

Unit conversions you’ll need

Unit	In SI
1 nm	10⁻⁹ m
1 pm	10⁻¹² m
1 μm	10⁻⁶ m
1 cm	10⁻² m
1 kHz / MHz / GHz	10³ / 10⁶ / 10⁹ Hz
1 eV	1.602 × 10⁻¹⁹ J
1 kJ	10³ J

Useful shortcut: hc = 6.626 × 10⁻³⁴ × 3.00 × 10⁸ = 1.988 × 10⁻²⁵ J m. And for energies in eV: E (eV) ≈ 1240 ÷ λ (nm).

Worked example 1: wavelength → frequency

Sodium emits yellow light at 589 nm. Calculate its frequency.

λ = 589 × 10⁻⁹ m = 5.89 × 10⁻⁷ m

f = c ÷ λ = 3.00 × 10⁸ ÷ 5.89 × 10⁻⁷ = 5.09 × 10¹⁴ Hz

Worked example 2: frequency → wavelength

A radio station broadcasts at 98.5 MHz. Calculate the wavelength.

f = 98.5 × 10⁶ Hz

λ = c ÷ f = 3.00 × 10⁸ ÷ 9.85 × 10⁷ = 3.05 m

Worked example 3: frequency → energy

Calculate the energy of one photon with frequency 7.50 × 10¹⁴ Hz.

E = hf = 6.626 × 10⁻³⁴ × 7.50 × 10¹⁴ = 4.97 × 10⁻¹⁹ J

That’s violet light (λ = 400 nm).

Worked example 4: wavelength → energy per photon and per mole

Calculate the energy of red light at 650 nm, per photon and per mole.

E = hc ÷ λ = 1.988 × 10⁻²⁵ ÷ 6.50 × 10⁻⁷ = 3.06 × 10⁻¹⁹ J per photon

Per mole: 3.06 × 10⁻¹⁹ × 6.022 × 10²³ = 1.84 × 10⁵ J/mol = 184 kJ/mol

In electronvolts: 1240 ÷ 650 = 1.91 eV

Worked example 5: energy → wavelength

An electron in an atom drops through an energy gap of 3.03 × 10⁻¹⁹ J. What wavelength of light is emitted, and what colour?

λ = hc ÷ E = 1.988 × 10⁻²⁵ ÷ 3.03 × 10⁻¹⁹ = 6.56 × 10⁻⁷ m = 656 nm → red

This is hydrogen’s H-alpha line. See the hydrogen emission spectrum.

Worked example 6: molar energy → wavelength

The bond energy of Cl–Cl is 242 kJ/mol. What is the longest wavelength of light that can break a Cl–Cl bond?

Energy per bond = 242,000 ÷ 6.022 × 10²³ = 4.02 × 10⁻¹⁹ J

λ = hc ÷ E = 1.988 × 10⁻²⁵ ÷ 4.02 × 10⁻¹⁹ = 4.95 × 10⁻⁷ m = 495 nm

Any light with wavelength shorter than 495 nm (blue, violet and ultraviolet) carries enough energy per photon to break a Cl–Cl bond. This is why a mixture of chlorine and methane reacts in sunlight but not in the dark: light starts a radical chain reaction.

Worked example 7: can this photon break a C–H bond?

The C–H bond energy is about 413 kJ/mol. Can a photon of visible light at 400 nm break it?

Energy of a 400 nm photon per mole: (1.988 × 10⁻²⁵ ÷ 4.00 × 10⁻⁷) × 6.022 × 10²³ = 299 kJ/mol

299 < 413, so no. Breaking C–H bonds needs ultraviolet light of wavelength shorter than about 290 nm. This is why many organic substances are stable in visible light but degrade in strong UV.

Worked example 8: identifying the region of the spectrum

λ	Region
below 10 nm	X-rays / gamma rays
10–400 nm	ultraviolet
400–700 nm	visible
700 nm – 1 mm	infrared
1 mm – 1 m	microwaves
above 1 m	radio waves

A photon has energy 1.50 × 10⁻²⁰ J. Identify the region.

λ = 1.988 × 10⁻²⁵ ÷ 1.50 × 10⁻²⁰ = 1.33 × 10⁻⁵ m = 13.3 μm → infrared

Infrared photons of this energy make molecular bonds vibrate, which is the basis of infrared spectroscopy.

Worked example 9: counting photons

A 5.00 mW green laser pointer emits light at 532 nm. How many photons does it emit per second?

Power = 5.00 × 10⁻³ W = 5.00 × 10⁻³ J per second

Energy per photon = 1.988 × 10⁻²⁵ ÷ 5.32 × 10⁻⁷ = 3.74 × 10⁻¹⁹ J

Photons per second = 5.00 × 10⁻³ ÷ 3.74 × 10⁻¹⁹ = 1.34 × 10¹⁶ photons

Over ten thousand trillion photons every second, from a pocket-sized laser.

Worked example 10: wavenumbers

Spectroscopists often use wavenumber, ν̃ = 1/λ, usually in cm⁻¹.

An IR absorption appears at 1715 cm⁻¹. Calculate the wavelength and photon energy.

λ = 1 ÷ 1715 cm⁻¹ = 5.83 × 10⁻⁴ cm = 5.83 × 10⁻⁶ m (5.83 μm)

E = hc ÷ λ = 1.988 × 10⁻²⁵ ÷ 5.83 × 10⁻⁶ = 3.41 × 10⁻²⁰ J (about 20.5 kJ/mol)

That’s the typical absorption of a C=O bond in a ketone.

Relationships at a glance

• Wavelength and frequency are inversely proportional: double λ, halve f.
• Energy and frequency are directly proportional: double f, double E.
• Energy and wavelength are inversely proportional: double λ, halve E.

So: shorter wavelength → higher frequency → more energy.

Common mistakes

1. Not converting nm to m. Using λ = 589 instead of 5.89 × 10⁻⁷ gives answers wrong by a factor of 10⁹.
2. Confusing energy per photon with energy per mole. Multiply or divide by Avogadro’s constant.
3. Forgetting to convert kJ to J before dividing by N_A.
4. Using E = hc/λ with λ in cm or nm. h and c are in SI units; λ must be in metres.
5. Mixing up direction: longer wavelength means less energy, not more.

Practice questions

1. Calculate the frequency of light with wavelength 450 nm.
2. Calculate the energy per mole of photons at 300 nm.
3. The first ionisation energy of sodium is 496 kJ/mol. Calculate the longest wavelength that could ionise a sodium atom.
4. A microwave oven uses 2.45 GHz radiation. Calculate the wavelength.

Answers:

1. f = 3.00 × 10⁸ ÷ 4.50 × 10⁻⁷ = 6.67 × 10¹⁴ Hz
2. E = (1.988 × 10⁻²⁵ ÷ 3.00 × 10⁻⁷) × 6.022 × 10²³ = 399 kJ/mol
3. E per atom = 496,000 ÷ 6.022 × 10²³ = 8.24 × 10⁻¹⁹ J; λ = 1.988 × 10⁻²⁵ ÷ 8.24 × 10⁻¹⁹ = 241 nm (UV)
4. λ = 3.00 × 10⁸ ÷ 2.45 × 10⁹ = 0.122 m (12.2 cm)

Where these calculations are used

These three equations appear throughout chemistry and beyond. Analytical chemists use them to choose wavelengths for spectroscopy; atmospheric chemists use them to work out which sunlight can break ozone or pollutant molecules; medical physicists use them to understand the energy delivered by X-rays and UV lamps; and engineers use them to design LEDs, lasers and solar cells. Getting comfortable with the unit conversions now makes all of these topics easier later.

Key takeaways

• c = fλ links wavelength and frequency; E = hf and E = hc/λ give photon energy.
• Always use SI units: λ in metres, f in Hz, E in joules.
• Multiply by Avogadro’s constant to convert energy per photon to energy per mole.
• Shorter wavelength means higher frequency and more energetic photons.
• Comparing photon energies with bond or ionisation energies shows which light can drive chemical change. See photons and energy levels.