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Hydrogen-Like Vacuum Emission Line

Find vacuum photon wavelength, frequency and energy for a downward transition in an ideal one-electron ion, with inverse state and charge controls.

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How to use this calculator

  1. Enter the known values in the units shown. Results update as you type.
  2. Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
  3. Use the worked example to check the method. Reset restores the starting fields.

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Formula and method

For an ideal Z = 1 transition from n = 3 to n = 2, the vacuum emitted wavelength is approximately 656.112276 nm and photon energy is approximately 1.889679600 eV. A 2 eV photon alone corresponds to about 619.920992 nm in vacuum.

1/λ = R∞Z²(1/nf² − 1/ni²), with ni > nf. ν = c/λ and Eγ = hν, using 2022 CODATA R∞ and exact SI h and c.

Worked example

Enter these known values and leave the other values blank.

Emitted photon energy
2 eV
Vacuum emitted wavelength
619.9 nm
Emitted photon frequency
483.6 THz

Assumptions and limitations

  • This is emission from a higher to a lower positive principal quantum state in a one-electron ion, with an infinite-mass point nucleus and a nonrelativistic Coulomb spectrum. Initial n must exceed final n.
  • Wavelength is in vacuum. No air refractive-index correction, isotope/reduced-mass adjustment, fine structure, external field or relativistic correction is included. It is not the precise observed wavelength of a particular isotope or fine-structure line.
  • Photon energy is positive, unlike the negative bound-state energies. The photon carries the difference between the ideal initial and final energy levels.
  • The supported numerical range for both state numbers and Z is positive exact integers from 1 to 1,000,000. Many-electron screening is not modeled. At large Z the nonrelativistic approximation is inaccurate.
  • All six quantities are editable. A known wavelength, frequency or photon energy can determine the other two without supplying states or Z. State/charge inverses require enough known quantities and a physically allowed integer result.
  • Equal states produce no finite emitted line. An initial-state inverse at the series limit is unbounded; wavelengths beyond the finite-state inverse domain produce an incomplete or invalid result, not an invented state.
  • Only computed integer inverses are recovered within an absolute 10⁻⁵⁰ arithmetic remainder. This is not a measurement tolerance and does not round manually entered fractional quantum numbers. R∞ has measured uncertainty, which is not propagated.

Common questions

Is this an air wavelength?

No. It is a vacuum wavelength. Comparing to measured air lines requires a suitable refractive-index conversion and isotope/spectral corrections.

Why must the initial state exceed the final state?

A downward transition releases positive photon energy. Equal or upward states do not represent the emission branch solved here.

Can I convert a wavelength without knowing the atom?

Yes. The photon relations ν = c/λ and E = hν need only one positive wavelength, frequency or energy.

Why might an inverse state be invalid?

The given line may lie outside the emission domain or may not yield an integer state. A series-limit wavelength corresponds to an unbounded initial state rather than a finite integer.

References

Use 2022 CODATA Rydberg and exact SI photon constants for the ideal infinite-nucleus model. An inferred integer state or charge is recovered only for a numerical remainder within 10⁻⁵⁰; ordinary fractional inverses and manual fractions remain invalid. Bound-state energy is strictly negative relative to ionization. The state and nuclear charge are exact positive integers from 1 to 1,000,000; inverse square roots require negative bound energy and positive known state or charge. The ideal model omits isotope and relativistic corrections.

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