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Foundations: Physics

Photon energy from frequency

Energy of one photon of light or other electromagnetic radiation, equal to Planck’s constant times the frequency.

Ref
FPH-61
Rev
v1
Field
Foundations: Physics
Result
E [J]

Planck’s constant h is 6.626 × 10⁻³⁴ J·s. The photon energy is h times the frequency, in joules; divide by 1.602 × 10⁻¹⁹ to get electronvolts. To start from wavelength instead of frequency, use the photon energy page.

Energy of one photon of light or other electromagnetic radiation, equal to Planck’s constant times the frequency. Higher frequency means more energy per photon.

Calculator

Solve for
J
Hz
1e91e20
  • Energy in electronvolts2.068 eV
  • Wavelength, c/ν599.6 nm

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Photon energy (J) current value 3.313e-19 J
νFrequencyKey driver
6.63e-256.63e-14
  • A photon of 500 THz carries 2.068 eV, which corresponds to a wavelength of 599.6 nm.
  • Energy is directly proportional to frequency: doubling the frequency doubles each photon’s energy.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (J)Local stepChange in result
ν Frequency1e9 to 1e20 Hz6.626e-256.626e-146.626e-14+10%+10 %

Sweep of Frequency

Frequency (Hz)Photon energy (J)
1e96.626e-25
2.371e101.571e-23
5.623e113.726e-22
1.334e138.836e-21
3.162e142.095e-19
7.499e154.969e-18
1.778e171.178e-16
4.217e182.794e-15
1e206.626e-14

Sweep one variable

Photon energy (J)Hover, touch, or use the arrow keys on the chart
02e−144e−146e−14 1G100G10000G1000000G100000000G10000000000G Frequency (Hz), log scale now 3.313e-19

Variables and how each one contributes

E

Photon energy

Result · J

What the equation gives you. Everything else feeds it.

ν

Frequency

Hz · typical 1e9 to 1e20

The frequency of the radiation, in Hz. Raises the photon energy in direct proportion. Raising it by 10 % (to 5.5×1014) moves the result by +10 %, from 3.313×10−19 to 3.644×10−19 J.

Worked example with the values above

  1. h = 6.626 × 10⁻³⁴ J·s.
  2. E = hν = 6.626 × 10⁻³⁴ × 5×1014.
  3. E = 3.31304×10−19 J, which is 2.0678 eV.

Assumptions and limits

  • A single photon of one frequency.
  • Frequency measured in hertz, not angular frequency.

Common mistakes

  • Using angular frequency ω in place of ν (that needs ħ, not h).
  • Mixing joules and electronvolts.
  • Putting a wavelength in the frequency box.

Statistics and uncertainty

The formula is exact. The uncertainty is that of the frequency, which can be measured to extremely high precision for lasers and atomic transitions.

Where it is used

  • Comparing LED, laser and X-ray photon energies.
  • Checking photoelectric and band-gap thresholds.

Maths you need

  • Frequency and wavelength: c = fλ
  • Electronvolt: energy gained by an electron through 1 V

Related formulas

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