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Joules to Electronvolts Converter

↔ Convert eV to J instead

Common Conversions

J eV
1.602e-19 1
3.204e-19 2
4.806e-19 3
8.01e-19 5
1.602e-18 10
2.18e-18 13.6
1.602e-17 100
1.602e-16 1000

Why this conversion matters in chemistry

Photochemistry reports pulse energies in joules — a typical 248 nm excimer laser pulse might deliver 500 mJ — but what matters for molecules is the energy per photon, which lives at the eV scale. 500 mJ of 248 nm light contains about 6.24 × 10¹⁷ photons at 5.00 eV each, calculated from hc/λ. Dividing the total pulse energy by the per-photon energy (in joules, via 1 eV = 1.602 × 10⁻¹⁹ J) gives you the photon count — the input for any quantum-yield calculation in a photochemistry experiment. The conversion bridges two natural scales: joules for bulk energy delivered, eV for what individual molecules absorb.

Formula

eV = J ÷ 1.602176634 × 10⁻¹⁹

Where the factor comes from

An electronvolt is the work done moving one elementary charge through a potential difference of one volt: E = qV, and since a volt is one joule per coulomb, 1 eV equals e joules with e expressed in coulombs. That used to make the conversion experimental. The 2019 SI revision fixed the elementary charge at 1.602176634 × 10⁻¹⁹ C exactly, and the factor became a definition overnight — before that it carried a relative uncertainty near 10⁻⁸ and reference tables revised it every few years. Inverting the fixed value gives 1 J = 6.241509074 × 10¹⁸ eV. Nothing about the electron is being measured here: the number is the defined charge written out in joules per electronvolt, and it will not move again.

Precision and significant figures

Ten exact digits are available and almost nobody needs more than four. Ionization energies are tabulated to a few decimals in eV — hydrogen at 13.598 eV — and the joule equivalents inherit exactly that precision, since the conversion adds none of its own. Digits actually get lost upstream: a photon energy computed as hc/λ is only as good as the wavelength, so a monochromator set to 248 nm with a 1 nm bandpass gives 5.00 eV, not 4.999363 eV. Photoelectron work quotes binding energies to about 0.1 eV because of analyzer resolution and sample charging. Convert with the full factor, then round to what the spectrometer earned.

Worked Examples

1.602 × 10⁻¹⁹ J = 1 eV

The defining equivalence — the energy one electron gains falling through a 1 V potential, expressed in SI.

2.18 × 10⁻¹⁸ J = 13.6 eV

Hydrogen's ionization energy. The Rydberg — arguably the most memorable per-particle energy in atomic chemistry.

3.0 × 10⁻¹⁹ J = 1.87 eV

A red photon near 660 nm. In the visible range, this is the low-energy end.

5.0 × 10⁻¹⁹ J = 3.12 eV

A violet photon near 400 nm — right at the visible/UV boundary. Enough energy to break some weak bonds.

Common mistakes

Treating eV as a molar energy

An electronvolt is per particle. A 5 eV bond energy does not compare with a 400 kJ/mol table entry until Avogadro's number has been applied — 5 eV per bond works out to 482 kJ/mol. Setting the two numbers side by side without the factor of 96.485 makes bond strengths look about a hundred times too small.

Pulse energy divided into photon energy

Dividing a 500 mJ laser pulse by 1.602 × 10⁻¹⁹ gives the number of electronvolts in the whole pulse, not the energy of one photon. Per-photon energy comes from hc/λ and nothing else. Confusing the two turns a quantum yield calculation into nonsense while leaving every number on the page looking entirely plausible.

Hartrees reported as electronvolts

Quantum chemistry output defaults to atomic units, where one hartree is 27.211 eV or 4.3597 × 10⁻¹⁸ J. Reading a raw energy difference of 0.05 as electronvolts rather than hartrees understates it by a factor of twenty-seven. Check the units block near the top of the output file before converting anything downstream.

Frequently Asked Questions

How do I convert joules to electronvolts?
Divide by 1.602176634 × 10⁻¹⁹. So 2.18 × 10⁻¹⁸ J becomes 13.6 eV, which is hydrogen's ionization energy. The factor is the elementary charge in coulombs — not a coincidence, but the definition of the electronvolt.
Why do electronvolts show up in chemistry?
Because atomic and molecular energies land naturally in the 1–25 eV range. Ionization energies, electron affinities, per-bond dissociation energies, photon energies across the visible and UV — all fit into single or low double digits in eV. The same values in joules come out as unwieldy 10⁻¹⁹ numbers that are much harder to compare at a glance.
How do eV relate to kJ/mol?
Multiply eV by 96.485 to get kJ/mol. The factor is Faraday's constant divided by 1000 — essentially Avogadro's number times the elementary charge. A 5 eV bond energy works out to 482 kJ/mol, which lands it in the typical range for single covalent bonds.
What's the energy range of visible light in eV?
About 1.77 eV at the red end (700 nm) to 3.1 eV at the violet end (400 nm). UV photons are above 3.1 eV, which is why UV light can break some weaker chemical bonds that visible light can't touch.