10,973,731.5682
m⁻¹
8,230,298.6761
m⁻¹
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nm
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Hz
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eV
10,973,731.5682
m⁻¹
8,230,298.6761
m⁻¹
—
nm
—
Hz
—
eV
The Rydberg Constant Calculator computes the wavelengths, frequencies, and energies of spectral lines for hydrogen-like atoms using the generalized Rydberg formula $$\frac{1}{\lambda} = R_\infty Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ where $$R_\infty = 1.097373 \times 10^7\;\text{m}^{-1}$$ is the Rydberg constant, Z is the atomic number, and $$n_1, n_2$$ are the principal quantum numbers of the lower and upper energy levels.
The Rydberg constant is the most precisely measured fundamental constant in physics, known to 12 significant figures. Originally discovered empirically by Johannes Rydberg in 1888 to describe hydrogen spectral lines, it was later derived from first principles by Niels Bohr's atomic model and ultimately from quantum electrodynamics. This calculator predicts spectral line positions for any hydrogen-like (single-electron) atom or ion.
The Rydberg formula arises from the energy difference between two atomic orbitals. For a hydrogen-like atom with nuclear charge Z, the energy of level n is:
$$E_n = -\frac{Z^2 R_\infty h c}{n^2} = -\frac{13.606\;Z^2}{n^2}\;\text{eV}$$
When an electron transitions from level $$n_2$$ to $$n_1$$ (with $$n_2 > n_1$$), it emits a photon with energy $$\Delta E = E_{n_1} - E_{n_2}$$. The corresponding wavenumber (inverse wavelength) is:
$$\frac{1}{\lambda} = R_\infty Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$
The Rydberg constant itself is composed of fundamental constants:
$$R_\infty = \frac{m_e e^4}{8\varepsilon_0^2 h^3 c}$$
The frequency and photon energy are obtained from $$\nu = c/\lambda$$ and $$E = h\nu$$. This calculator works for any pair of quantum numbers with $$n_2 > n_1 \geq 1$$ and any atomic number Z from 1 to 118.
The wavelength tells you where the spectral line falls: below 400 nm is ultraviolet, 400–700 nm is visible, above 700 nm is infrared. The photon energy in eV gives the energy released (emission) or required (absorption) for the transition. Higher Z or larger jumps between levels produce shorter wavelengths and higher energies. The Rydberg constant itself is displayed as a reference value.
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The n = 3 → 2 transition in hydrogen produces the famous red Hα line at 656.3 nm, the most prominent line in the Balmer series and the characteristic color of hydrogen emission nebulae.
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He⁺ (Z = 2) has Z² = 4 times the wavenumber of hydrogen. The 2→1 transition gives a 30.4 nm extreme UV photon with 40.8 eV energy, important in solar physics.
The Rydberg constant $$R_\infty = 1.0973731568160 \times 10^7\;\text{m}^{-1}$$ is a fundamental physical constant that appears in the formula for spectral line wavelengths of hydrogen-like atoms. It equals $$m_e e^4/(8\varepsilon_0^2 h^3 c)$$ and is the most precisely measured constant in physics.
The series are named by the lower level $$n_1$$: Lyman ($$n_1=1$$, UV), Balmer ($$n_1=2$$, visible/UV), Paschen ($$n_1=3$$, near-IR), Brackett ($$n_1=4$$, IR), Pfund ($$n_1=5$$, far-IR). Each series converges to a series limit at $$\lambda = n_1^2 / (R_\infty Z^2)$$.
No. The Rydberg formula applies only to hydrogen-like (one-electron) systems: H, He⁺, Li²⁺, etc. Multi-electron atoms have electron-electron repulsion and shielding effects that make the energy levels much more complex. Approximate methods like Hartree-Fock or density functional theory are needed.
$$R_\infty$$ assumes an infinitely heavy nucleus (theoretical limit). The hydrogen-specific constant $$R_H = R_\infty/(1 + m_e/m_p) = 10\,967\,758\;\text{m}^{-1}$$ accounts for the proton's finite mass. The correction is ~0.054%, small but measurable in precision spectroscopy.
Atomic hydrogen spectroscopy, particularly the 1S–2S transition measured by laser frequency combs, achieves extraordinary precision. Combined with QED calculations that include relativistic, radiative, and nuclear-size corrections, $$R_\infty$$ is determined to a relative uncertainty of $$1.9 \times 10^{-12}$$.
The ionization energy of hydrogen (from ground state to $$n = \infty$$) is $$E_i = R_\infty h c = 13.606\;\text{eV}$$. For hydrogen-like ions, the ionization energy scales as $$Z^2 \times 13.606\;\text{eV}$$. So He⁺ requires 54.4 eV and Li²⁺ requires 122.4 eV to ionize.
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