Photon Energy Calculator
Calculate the energy of a photon from its wavelength or frequency. Enter either value and get the energy per photon in joules and electronvolts, plus the energy per mole of photons in kJ/mol.
Photon Energy from Wavelength or Frequency
Photon Energy
Where:
- \(E\) — energy of one photon (J)
- \(h\) — Planck constant, \(6.62607015 \times 10^{-34}\) J·s (exact)
- \(\nu\) — frequency of the light (Hz)
- \(c\) — speed of light in vacuum, \(2.99792458 \times 10^{8}\) m/s (exact)
- \(\lambda\) — wavelength in vacuum (m)
- \(N_A\) — Avogadro constant, \(6.02214076 \times 10^{23}\) mol⁻¹ (exact)
- \(E_{\text{mol}}\) — energy per mole of photons, also called one einstein (J/mol)
This is the Planck relation: the energy of a single photon is fixed entirely by the frequency (or equivalently the wavelength) of the light, because \(\lambda \nu = c\).
Photon Energy – Calculation Example
Example: green light with a wavelength of 500 nm (the calculator's default value).
\[ \lambda = 500\ \text{nm} = 5.00 \times 10^{-7}\ \text{m} \]
\[ \nu = \frac{c}{\lambda} = \frac{2.99792458 \times 10^{8}}{5.00 \times 10^{-7}} = 5.996 \times 10^{14}\ \text{Hz} \approx 599.6\ \text{THz} \]
\[ E = h\nu = 6.62607015 \times 10^{-34} \times 5.996 \times 10^{14} = 3.973 \times 10^{-19}\ \text{J} \]
\[ E = \frac{3.973 \times 10^{-19}}{1.602176634 \times 10^{-19}} \approx 2.48\ \text{eV} \qquad E_{\text{mol}} = 3.973 \times 10^{-19} \times 6.02214076 \times 10^{23} \approx 239\ \text{kJ/mol} \]
Result: each photon of 500 nm light carries \(3.973 \times 10^{-19}\) J ≈ 2.48 eV, and one mole of such photons carries about 239 kJ. This lies in the green region of the visible spectrum.
How to Use the Calculator
- In the Input Type dropdown, select Wavelength (λ) or Frequency (ν), depending on which quantity you know.
- If you chose wavelength, enter the value in the Wavelength (λ) field and pick the unit (nm, µm, mm, cm or m). If you chose frequency, enter the value in the Frequency (ν) field and pick the unit (Hz, kHz, MHz, GHz or THz).
- The result panel shows the Energy per Photon in joules and electronvolts, the Energy per Mole of Photons in kJ/mol, plus the complementary wavelength or frequency and the wavenumber in cm⁻¹.
- Check the Calculation Details panel for the step-by-step computation and the About Photon Energy panel for context on the spectral region and typical benchmarks.
What Is Photon Energy?
Light behaves both as a wave and as a stream of particles called photons. Each photon carries a tiny, discrete packet of energy whose size depends only on the light's frequency: higher frequency (shorter wavelength) means more energetic photons. This is the essence of the Planck relation \(E = h\nu\), the equation that launched quantum physics in 1900 and later earned Einstein's explanation of the photoelectric effect a Nobel Prize.
A single photon's energy is extremely small — around \(10^{-19}\) J for visible light — so it is usually quoted in electronvolts (eV), where 1 eV is the energy an electron gains crossing a 1-volt potential (\(1\ \text{eV} = 1.602 \times 10^{-19}\) J). Visible photons span roughly 1.6 eV (750 nm, red) to 3.3 eV (380 nm, violet); UV photons are more energetic still, and gamma-ray photons reach millions of eV.
In chemistry and photochemistry the energy per mole of photons (one einstein) is often the more useful figure, because it lets you compare photon energy directly with bond energies and activation energies in kJ/mol. For example, 254 nm UV light delivers about 471 kJ/mol — enough to break many covalent bonds — while 700 nm red light delivers only about 171 kJ/mol, below even the weakest O–O single bonds. This is why UV light causes sunburn and photodegradation while red light generally does not.
Note that photon energy depends only on the color (wavelength) of the light, never on its brightness. Intensity determines how many photons arrive per second; each individual photon always carries exactly \(h\nu\). If you are combining light energies with bond energies to estimate reaction enthalpies, our Reaction Enthalpy from Bond Energies calculator is a natural companion tool.
Reference Values
| Wavelength | Region | Energy (eV) | Energy (kJ/mol) |
|---|---|---|---|
| 200 nm | Far UV | 6.20 | 598 |
| 254 nm | UV (Hg lamp) | 4.89 | 471 |
| 350 nm | Near UV | 3.54 | 342 |
| 450 nm | Visible (blue) | 2.76 | 266 |
| 500 nm | Visible (green) | 2.48 | 239 |
| 700 nm | Visible (red edge) | 1.77 | 171 |
Assumptions and Limitations
- The calculation assumes the light travels in vacuum. For air the difference is negligible — the speed of light in air is only about 0.03% lower than in vacuum, which changes the wavelength-to-frequency mapping imperceptibly for most purposes.
- The constants \(h\), \(c\) and \(N_A\) are the exact SI 2019 values, so the only source of imprecision is the input itself.
- Wavelength and frequency must refer to the same medium; mixing a vacuum wavelength with an in-medium frequency would introduce an error equal to the refractive index.
- The calculator treats light as single photons. Effects such as multiphoton absorption (where two lower-energy photons are absorbed together) are nonlinear processes not covered by the simple Planck relation.
- When comparing with bond energies, remember that a photon must supply at least the bond dissociation energy in a single absorption event; surplus energy appears as kinetic energy, heat or fluorescence.
If you work with the wave nature of light instead — for example diffraction by crystals — you may also find our Bragg's Law calculator useful for relating wavelength to diffraction angles, or the Standard Cell Potential calculator when converting between electrochemical voltages and electron energies in eV.
Frequently Asked Questions
How do I calculate photon energy by hand?
Convert the wavelength to metres first, then divide the speed of light by it to get the frequency, and multiply by Planck's constant. For example, for 400 nm: \(\nu = 3\times10^8 / 4\times10^{-7} = 7.5\times10^{14}\) Hz, so \(E = 6.626\times10^{-34} \times 7.5\times10^{14} \approx 4.97\times10^{-19}\) J. Forgetting to convert nanometres to metres is by far the most common mistake and gives answers that are wrong by a factor of \(10^9\).
Is there a quick shortcut for converting wavelength in nm to eV?
Yes. The product \(E \cdot \lambda = hc \approx 1240\) eV·nm, so \(E(\text{eV}) \approx 1240 / \lambda(\text{nm})\). For example, 620 nm gives 2.0 eV, and 310 nm gives 4.0 eV. This handy rule is exact enough for everyday spectroscopy work. Similarly, for kJ/mol you can use \(E \approx 119{,}600 / \lambda(\text{nm})\) kJ/mol.
Does photon energy change when light passes through glass or water?
No. Inside a medium the wavelength shortens by the refractive index, but the frequency — and therefore the energy \(E = h\nu\) — stays exactly the same. A blue photon remains a blue photon underwater. This is why spectroscopists prefer frequency or wavenumber over wavelength: those quantities are medium-independent, whereas the wavelength value depends on what the light is travelling through.
Why doesn't a brighter light have more energetic photons?
Intensity and photon energy are independent. A dim laser pointer and a blinding spotlight of the same colour deliver identical photons; the bright source just sends more of them per second. Only changing the colour (wavelength) changes the per-photon energy. This follows directly from the photoelectric effect: below a threshold frequency, no amount of red light can eject electrons, while even weak UV light can.
How do I convert between joules, electronvolts and kJ/mol?
Divide joules by \(1.602\times10^{-19}\) to get eV, and multiply by Avogadro's constant then divide by 1000 to get kJ/mol. Handy equivalences: 1 eV ≈ \(1.602\times10^{-19}\) J ≈ 96.49 kJ/mol, and 1 kJ/mol ≈ 0.01036 eV per photon. So a 3 eV photon corresponds to roughly 290 kJ/mol — comparable to a strong chemical bond.
What is the wavenumber and why do spectroscopists use it?
The wavenumber is the number of wavelengths per centimetre, \(\tilde\nu = 1/\lambda\), usually quoted in cm⁻¹. It is proportional to photon energy: \(E = hc\tilde\nu\), with 1 cm⁻¹ ≈ \(1.986\times10^{-23}\) J ≈ 0.01196 kJ/mol. IR and Raman spectroscopists prefer it because it avoids medium-dependent wavelength values and gives a linear energy scale, e.g. green 500 nm light corresponds to 20,000 cm⁻¹.
Can two low-energy photons break a bond that one photon cannot?
Yes, but only in special cases. In ordinary photochemistry each molecule absorbs one photon, so a single photon must carry at least the bond energy. However, in multiphoton absorption — possible with intense pulsed lasers — two or more lower-energy photons are absorbed nearly simultaneously and their energies add up. This is a nonlinear optical effect that the simple Planck relation does not describe, so do not rely on it for everyday light sources.