Chalk−1

Chemistry · General chemistry I · Worked example

Find a photon’s energy from its wavelength

Find the frequency, the energy of one photon and the energy of a mole of photons for light of wavelength 425 nm.

E=h⁢cλ

Convert the wavelength

425 nm = 425 × 10⁻⁹ m = 4.25 × 10⁻⁷ m.

λ=4.25×10−7 m

Find the frequency

ν = c/λ: divide the speed of light by the wavelength.

ν=2.998×108 m/s4.25×10−7 m=7.05×1014 s−1
ν=2.998×108 m/s4.25×10−7 m=7.05×1014 s−1

Find the energy of one photon

E = hc/λ. The meters cancel, leaving joules.

E=(6.626×10−34 Js)⁢(2.998×108 m/s)4.25×10−7 m=4.67×10−19 J
E=(6.626×10−34)⁢(2.998×108)4.25×10−7=4.67×10−19 J

Scale up to a mole of photons

Multiply by Avogadro’s number: 2.81 × 10⁵ J/mol, or 281 kJ/mol.

(4.67×10−19 J)⁢(6.022×1023 mol−1)=281 kJ/mol
(4.67×10−19)⁢(6.022×1023)=2.81×105 J/mol=281 kJ/mol

Result

ν = 7.05 × 10¹⁴ Hz; E = 4.67 × 10⁻¹⁹ J per photon, or 281 kJ per mole of photons.

Your turn

What is the energy of one photon of light with frequency 5.00 × 10¹⁴ Hz?

Show the answer and explanation

3.31 × 10⁻¹⁹ J.

E = hν = (6.626 × 10⁻³⁴ J·s)(5.00 × 10¹⁴ s⁻¹) = 3.31 × 10⁻¹⁹ J.

E=(6.626×10−34 Js)⁢(5.00×1014 s−1)=3.31×10−19 J
E=(6.626×10−34)⁢(5.00×1014) J=3.31×10−19 J

Keep exploring

In Photons & spectrophotometry, halve the wavelength to 212.5 nm. The energy per photon doubles, to 9.35 × 10⁻¹⁹ J.

Return to the concept →
Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Make it concrete

Try in the workspace

Open the example inputs, change a value and keep a useful result on your board.

Convert it in Photons & spectrophotometry Check the energy in a Chemistry box Open worked example on a board Chemistry formulas: atomic structure

Your existing work stays on this device. Examples open as editable copies.