Rydberg Equation Calculator
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| Lower level n₁ | 2 |
|---|---|
| Upper level n₂ | 3 |
Rydberg Equation Calculator
Find the wavelength, frequency, and photon energy of a hydrogen spectral line from the two energy levels of the electron transition. Enter the lower and upper principal quantum numbers n₁ and n₂ to get the line in the Lyman, Balmer, or Paschen series.
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Understanding the Rydberg equation
When the single electron in a hydrogen atom drops from a higher energy level to a lower one, it emits a photon whose wavelength depends only on the two levels involved. The Rydberg equation gives that wavelength:
λ1=RH(n121−n221)| Symbol | Quantity | Unit |
|---|---|---|
| λ | Wavelength of the line | m (shown in nm) |
| R_H | Rydberg constant | m⁻¹ |
| n₁ | Lower principal quantum number | — |
| n₂ | Upper principal quantum number | — |
Here and are positive integers with . The lower number fixes which spectral series the line belongs to, while selects the line within that series. Once is known, the photon's frequency follows from and its energy from .
Worked example
The brightest visible line of hydrogen, H-alpha, comes from the transition . With :
λ1=RH(221−321)=1.0973731568×107×(41−91)The bracket evaluates to , so
λ=1.0973731568×107×0.138891≈6.563×10−7 m=656.3 nmThat is the deep-red H-alpha line. Its photon energy is
E=λhc=6.563×10−76.626×10−34×2.998×108≈3.03×10−19 J≈1.89 eVThe spectral series
Each value of produces a whole family of lines, named after the physicists who studied them. The series sit in different parts of the spectrum because the energy gaps shrink as the levels climb.
| Series | n₁ | Spectral region |
|---|---|---|
| Lyman | 1 | Ultraviolet |
| Balmer | 2 | Visible |
| Paschen | 3 | Infrared |
| Brackett | 4 | Infrared |
The Balmer series is the one the eye can see: H-alpha at 656 nm (red), H-beta at 486 nm (blue-green), and H-gamma at 434 nm (violet). As grows, the lines bunch together and converge on a series limit at .
Why the lines are discrete
A hydrogen electron can only occupy specific energy levels, so the gaps between them — and the photons that bridge those gaps — take only specific values. That is why hydrogen glows in sharp lines rather than a continuous rainbow. The Rydberg equation captures this with two integers, and the fact that such a simple rule fits the measured spectrum so well was one of the first hints that energy inside an atom is quantised, a result the Bohr model later explained.
Beyond hydrogen
The same form works for any one-electron ion, such as He⁺ or Li²⁺, if the right-hand side is multiplied by the square of the nuclear charge . For multi-electron atoms the picture is more complicated because the electrons shield one another, and the clean integer pattern breaks down. The idealised constant used here, , assumes an infinitely heavy nucleus; correcting for the proton's finite mass lowers it by about one part in two thousand.
Frequently Asked Questions (FAQ)
What is the Rydberg equation?
The Rydberg equation gives the wavelength of light a hydrogen atom emits or absorbs when its electron jumps between two energy levels: 1/λ = R_H (1/n₁² − 1/n₂²). Here λ is the wavelength, R_H = 1.097 × 10⁷ m⁻¹ is the Rydberg constant, n₁ is the lower principal quantum number, and n₂ is the upper one, with n₂ > n₁. Once λ is known, the frequency follows from f = c/λ and the photon energy from E = hc/λ. The equation reproduces the discrete lines seen in the hydrogen spectrum and was a key clue that led to the Bohr model of the atom.
What are the hydrogen spectral series?
Each value of the lower level n₁ defines a family of spectral lines called a series. The Lyman series (n₁ = 1) lies in the ultraviolet, the Balmer series (n₁ = 2) falls in the visible range and is the one usually seen by eye, and the Paschen series (n₁ = 3) sits in the infrared. Higher series — Brackett (n₁ = 4) and Pfund (n₁ = 5) — lie further into the infrared. Within a series, lines crowd closer together as n₂ rises, converging on a series limit where n₂ → ∞.
What is the Balmer H-alpha line?
The Balmer series consists of transitions ending on the n₁ = 2 level, and its first and brightest member, H-alpha, is the n₂ = 3 → n₁ = 2 transition. The Rydberg equation gives 1/λ = 1.097 × 10⁷ × (1/4 − 1/9), so λ ≈ 656 nm — the deep red line that gives glowing hydrogen and many emission nebulae their characteristic colour. The next Balmer lines, H-beta (n₂ = 4) at 486 nm and H-gamma (n₂ = 5) at 434 nm, appear blue-green and violet.
What is the value of the Rydberg constant?
For an idealised hydrogen atom with an infinitely heavy nucleus the Rydberg constant is R_∞ = 1.0973731568 × 10⁷ m⁻¹, one of the most precisely measured constants in physics. This calculator uses that value. A slightly smaller value, R_H ≈ 1.09678 × 10⁷ m⁻¹, accounts for the finite mass of the proton and matches the hydrogen spectrum even more closely; the difference is about one part in two thousand and shifts the computed wavelengths by well under a nanometre for visible lines.
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