ABC Tools

Photon Energy, Explained

After reading this you will be able to convert freely between a photon's wavelength, frequency and energy, quote the answer in both joules and electronvolts, and say which part of the spectrum a given light belongs to.

What a photon's energy means

Light comes in discrete packets called photons. Each photon carries a fixed amount of energy set by its frequency, and nothing else. A single green photon at 550 nm carries about 2.25 eV, which is 3.61 \times 10^{-19} joules. That is a tiny amount. A 1-watt green laser pushes out roughly 2.8 \times 10^{18} such photons every second.

Here is the hook. Radio waves, visible light and gamma rays are all the same thing, electromagnetic waves, differing only in frequency. Yet a gamma photon can knock electrons out of atoms and a radio photon cannot, no matter how bright the radio signal. The reason is entirely about energy per photon, and that is what this page teaches you to compute.

When to use E = hf, and when not

Use E = hf whenever you care about a single quantum of light: whether one photon can eject an electron (the photoelectric effect), excite a molecule, or break a chemical bond. The threshold behavior of these processes depends on energy per photon, not total power.

You do not need it when you only care about total energy flow. A heat lamp warming your hand is described well by power in watts and intensity in watts per square meter. The photon picture still holds underneath, but counting individual photons adds nothing to the bookkeeping.

The formula treats light as massless quanta moving at c. It does not apply to massive particles. For the wavelength of a moving electron or proton, use the de Broglie wavelength calculator instead, which uses momentum rather than frequency.

The formula and where it comes from

Two facts combine. First, Planck and Einstein found that a photon's energy is proportional to its frequency:

E = hf

Here E is energy in joules, f is frequency in hertz (cycles per second), and h is Planck's constant, 6.62607 \times 10^{-34} joule-seconds. Planck's constant is small, which is why a single photon carries so little energy.

Second, every electromagnetic wave in vacuum travels at the same speed c, and speed equals frequency times wavelength:

c = f\lambda

with c = 2.99792 \times 10^{8} meters per second and \lambda the wavelength in meters. Solve for f = c/\lambda and substitute into the first equation:

E = \frac{hc}{\lambda}

This is the working form. Energy is inversely proportional to wavelength. Halve the wavelength and you double the energy per photon. That single fact explains why ultraviolet light at 300 nm damages skin while infrared at 1000 nm mostly warms it: the UV photon carries more than three times the energy.

The product hc is worth memorizing in convenient units. In electronvolt-nanometers it is 1239.84 eV\cdot nm, so E(\text{eV}) = 1239.84 / \lambda(\text{nm}). That one line replaces most of the arithmetic.

A worked example with the demo numbers

Green light at 550 nm

Start from the tool's default wavelength of 550 nm and reproduce every output field.

  1. Convert the wavelength to meters: \lambda = 550 \times 10^{-9} = 5.50 \times 10^{-7} m.
  2. Frequency: f = c/\lambda = (2.998 \times 10^{8}) / (5.50 \times 10^{-7}) = 5.451 \times 10^{14} Hz, which is 545.1 THz. This matches the demo frequency of 545 THz.
  3. Energy in joules: E = hf = (6.626 \times 10^{-34})(5.451 \times 10^{14}) = 3.612 \times 10^{-19} J.
  4. Energy in electronvolts: divide by the charge on one electron, 1.602 \times 10^{-19} C, giving E = 3.612 \times 10^{-19} / 1.602 \times 10^{-19} = 2.255 eV. This matches the demo energy of 2.25 eV.
  5. The shortcut agrees: 1239.84 / 550 = 2.254 eV.

So a 550 nm photon sits in the green, at 545 THz and 2.25 eV. All three descriptions are the same photon.

As wavelength rises from 380 nm to 750 nm across the visible band, frequency falls from about 789 THz to 400 THz and energy per photon falls from about 3.26 eV to 1.65 eV. Energy and wavelength are inversely related: a 380 nm violet photon carries nearly twice the energy of a 750 nm red photon.

Reading and interpreting the result

The calculator names the spectral band because the number alone rarely tells you what matters. Use these rough boundaries, quoted in both wavelength and photon energy.

Spectral bands with representative photon energy
BandWavelengthFrequencyPhoton energy
Radio> 1 m< 300 MHz< 1.24 µeV
Microwave1 mm to 1 m0.3 to 300 GHz1.24 µeV to 1.24 meV
Infrared750 nm to 1 mm0.3 to 400 THz1.24 meV to 1.65 eV
Visible380 to 750 nm400 to 789 THz1.65 to 3.26 eV
Ultraviolet10 to 380 nm0.789 to 30 PHz3.26 to 124 eV
X-ray and gamma< 10 nm> 30 PHz> 124 eV

The energy scale is what predicts effects. Breaking a typical carbon-carbon bond needs about 3.6 eV, which corresponds to 345 nm, deep in the UV. That is why UV light drives photochemistry and sunburn while visible light at 2.25 eV mostly gets absorbed and re-emitted as heat.

Energy falls as 1/wavelength, so the curve is a hyperbola, not a line. The shaded band is visible light; the marker sits at 550 nm, 2.25 eV.

Common mistakes

The errors here are almost all unit errors. The physics is one line, so watch the numbers.

  • Forgetting to convert nanometers to meters. Using 550 instead of 5.50 \times 10^{-7} in c/\lambda gives a frequency off by a factor of a billion. The 1239.84 eV\cdot nm shortcut sidesteps this because it expects nanometers directly.
  • Mixing joules and electronvolts. One electronvolt is 1.602 \times 10^{-19} J. Report both, but never add them.
  • Confusing THz and Hz. The demo frequency 545 THz is 5.45 \times 10^{14} Hz. Slotting 545 into E = hf gives an energy 12 orders of magnitude too small.
  • Assuming the wave speed is always c. The relation c = f\lambda uses the vacuum speed. Inside glass, light slows and its wavelength shortens while frequency stays fixed. If your problem involves light entering a medium, handle the bending with the Snell's law calculator and remember that the photon energy does not change.

Frequency, not wavelength, is the invariant when light crosses into a new medium. A 550 nm photon in air becomes about 367 nm inside glass with refractive index 1.5, yet it is still a green, 2.25 eV photon because f is unchanged. Compute energy from frequency, or from the vacuum wavelength, never from the wavelength inside a material.

Related tools

Photon energy connects to several nearby topics. If a moving source shifts the frequency you observe, the Doppler effect calculator gives the shifted frequency, which you then feed into E = hf. When photons are gathered by a lens or mirror, the lens and mirror equation calculator tells you where the image forms. If you are estimating a measured wavelength with an instrument uncertainty and want the resulting spread in energy, run it through the error propagation calculator, or check a single value against a reference with the percent error calculator. For high-energy photons emitted during nuclear transitions, the radioactive decay calculator handles the timing of the source.

Frequently asked questions

Why do UV rays burn but radio waves do not?

A UV photon at 300 nm carries about 4.1 eV, enough to break bonds and damage DNA. A radio photon at 1 m carries about 1.24 µeV, a factor of three million less. Turning up the radio power sends more photons, but each one is still far too weak to disrupt a molecule.

How do I convert eV to joules?

Multiply by 1.602 \times 10^{-19}. So 2.25 eV equals 2.25 \times 1.602 \times 10^{-19} = 3.60 \times 10^{-19} J. To go the other way, divide joules by the same number.

What is the energy of a single visible photon in everyday terms?

About 3.6 \times 10^{-19} J for green light. Lifting a single grain of sand one millimeter takes roughly 10^{-8} J, tens of billions of times more. One photon is a very small energy budget.

Does a brighter light have more energetic photons?

No. Brightness at a fixed color means more photons per second, each carrying the same E = hf. To raise the energy per photon you must raise the frequency, meaning shorten the wavelength.

Can I use this formula for X-rays and gamma rays?

Yes. The relation E = hc/\lambda holds across the entire spectrum. A gamma photon at 0.01 nm carries 1239.84 / 0.01 = 123984 eV, about 124 keV, which is why such photons ionize atoms readily.

Try it yourself in the tool

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