Hydrogen Spectrum and Spectral Series

In short

When hydrogen is excited (in a discharge tube, in a star, in a nebula) and the light passes through a prism, the spectrum is a set of discrete lines at very specific wavelengths. Every one of these wavelengths is given by the Rydberg formula:

\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \qquad R = 1.0974 \times 10^7\ \text{m}^{-1},\ n_2 > n_1.

Each choice of n_1 defines a series. n_1 = 1 is the Lyman series (ultraviolet). n_1 = 2 is the Balmer series (visible — the red H-alpha line at 656.3 nm sits in here). n_1 = 3 is Paschen, n_1 = 4 is Brackett, n_1 = 5 is Pfund — all in the infrared. The series limit (n_2 \to \infty) is the shortest wavelength in each series; for Lyman it is 91.2 nm, the ionisation wavelength of hydrogen.

The Rydberg formula is not an empirical fit. It falls out of the Bohr model: hc/\lambda = E_{n_2} - E_{n_1} = 13.6\ \text{eV}\cdot(1/n_1^2 - 1/n_2^2), and the coefficient R = 13.6\ \text{eV}/(hc) is just the Bohr ionisation energy expressed as a reciprocal wavelength. Every spectral line is one electron in one atom falling from n_2 to n_1 and emitting one photon.

Limit of the Bohr picture. The formula works with uncanny precision for hydrogen and hydrogen-like ions (He^+, Li^{2+}). It fails for every other atom. Helium has two electrons that repel each other; the spectrum is far richer than Bohr's single-electron picture allows. The modern fix — solving the Schrödinger equation with full electron-electron interactions — is in Energy levels and quantum numbers. Hydrogen is the one atom where Bohr got the answer exactly right.

At the Kodaikanal Solar Observatory in Tamil Nadu, every clear morning since 1904, astronomers have photographed the sun through a filter that lets through only one colour of red — the wavelength 656.3 nanometres. The pictures show a boiling, fibrous surface that has looked nothing like a smooth disc for over a century. The filter is tuned to a single spectral line called H-alpha, and the reason that line exists at exactly 656.3 nm — not 655, not 657, not a smear — is the subject of this article.

If you take a glass tube, fill it with hydrogen gas, seal it, and run a few thousand volts across it, the gas glows a pale pink. Pass that pink light through a prism. What comes out is not a continuous rainbow. It is four coloured lines in the visible range — a red one, a turquoise, a blue, a deep violet — and nothing in between. No orange, no yellow, no green. Just four narrow bright lines on a dark background, always at the same wavelengths, always in the same positions. Every hydrogen atom in the universe emits these exact four lines when you excite it. The red one is H-alpha, the very same line the Kodaikanal telescope is pointed at.

Three decades before Bohr wrote down his atomic model, a Swiss schoolteacher named Johann Balmer stared at the measured wavelengths of these four lines — 656.3, 486.1, 434.0, 410.2 nm — and found, after much hunting, that they fit the tidy formula

\lambda = 364.5\ \text{nm}\cdot\frac{n^2}{n^2 - 4}, \qquad n = 3, 4, 5, 6.

No one at the time knew why. The formula was a fact staring up from the data, with no theory underneath. Within a decade the Swedish physicist Johannes Rydberg had extended it into a general formula — the Rydberg formula — that correctly predicted spectral lines in the ultraviolet and infrared as well, lines that were found after the formula was proposed. The number 364.5 nm became an empirical constant, baptised the Rydberg constant. For thirty years it was a number with no explanation.

Then Bohr, in 1913, wrote down three postulates about the hydrogen atom and — as a consequence, not as a fit — the Rydberg formula and the number 364.5 nm fell out. That is the subject of this article: the spectrum, the formula, the series, and how every last wavelength is one electron in one atom dropping from one orbit to another.

Why a hot gas emits discrete lines — the Bohr logic

The Bohr model fixes the electron in hydrogen to a ladder of allowed energies,

E_n = -\frac{13.6\ \text{eV}}{n^2}, \qquad n = 1, 2, 3, \ldots

The ground state is n = 1 (E_1 = -13.6 eV). Excited states are n = 2 (E_2 = -3.40 eV), n = 3 (E_3 = -1.51 eV), and so on. The levels crowd together as n grows, with E_\infty = 0 — the energy of an electron just barely escaping the atom.

In a discharge tube the collisions between electrons and gas molecules keep pumping atoms into excited states. An excited atom is unstable: after a time of order 10^{-8} s, the electron jumps down to a lower level, and the energy difference is carried away as one photon. Bohr's third postulate says:

hf = E_{n_2} - E_{n_1} \qquad (n_2 > n_1).

The photon's frequency f, and therefore its wavelength \lambda = c/f, is determined entirely by the two levels involved. Because the levels are discrete, the photon wavelengths are discrete too. You see lines, not a continuum.

The full machinery of the Rydberg formula is now just one line of substitution.

Deriving the Rydberg formula

Step 1. Write the Bohr energy gap for a transition n_2 \to n_1, with n_2 > n_1.

\Delta E = E_{n_2} - E_{n_1} = \left(-\frac{13.6\ \text{eV}}{n_2^2}\right) - \left(-\frac{13.6\ \text{eV}}{n_1^2}\right) = 13.6\ \text{eV}\cdot\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right).

Why: both E_{n_1} and E_{n_2} are negative, with E_{n_2} closer to zero. Their difference is positive — it is the energy released when the electron drops from the higher orbit to the lower one. The 1/n_1^2 - 1/n_2^2 combination is positive as long as n_2 > n_1.

Step 2. Set this equal to the photon energy hf = hc/\lambda.

\frac{hc}{\lambda} = 13.6\ \text{eV}\cdot\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right).

Why: postulate 3 — one photon carries exactly the energy difference, no more and no less. Use f = c/\lambda to rewrite energy in terms of the wavelength that a prism or a grating will actually measure.

Step 3. Solve for 1/\lambda.

\boxed{\frac{1}{\lambda} = \frac{13.6\ \text{eV}}{hc}\cdot\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \equiv R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right).}

The coefficient R \equiv 13.6\ \text{eV}/(hc) is the Rydberg constant.

Step 4. Plug in numbers. 13.6\ \text{eV} = 13.6 \times 1.602\times 10^{-19}\ \text{J} = 2.179 \times 10^{-18}\ \text{J}. hc = (6.626\times 10^{-34}\ \text{J·s}) \cdot (3.00\times 10^8\ \text{m/s}) = 1.988 \times 10^{-25}\ \text{J·m}. Dividing,

R = \frac{2.179\times 10^{-18}}{1.988\times 10^{-25}}\ \text{m}^{-1} = 1.0963\times 10^7\ \text{m}^{-1}.

Why: the modern high-precision value is R = 1.0973731568\times 10^7 m^{-1} — one of the most accurately known constants in physics. The small difference from our estimate comes from the reduced-mass correction (the proton is not truly infinitely heavy) and the rounding we did on 13.6 eV. The agreement to four significant figures is already extraordinary for a theory with one free postulate.

The units are reciprocal length, which is a quirk of how spectroscopists think. You can always convert back: \lambda = 1/[R(1/n_1^2 - 1/n_2^2)].

The series — one value of n_1 each

Fix n_1 and let n_2 run through all integers greater than it. Each choice gives an infinite family of lines — a series named after the physicist who measured it first.

Energy-level diagram. The horizontal lines are the Bohr levels; the arrows are the downward transitions that emit photons. Each series is the family of all transitions ending on the same lower level. Series limits ($n_2 \to \infty$) correspond to arrows that start from the dashed $n = \infty$ line — the shortest-wavelength line in each series.

Lyman series — ultraviolet (n_1 = 1)

Transitions that end at the ground state. Because the gap from any excited level down to n = 1 is the largest possible gap in hydrogen, the photons are the most energetic, with the shortest wavelengths. All Lyman lines lie in the ultraviolet part of the spectrum (the visible band is 400–700 nm; Lyman lines are 91–122 nm).

The Lyman-alpha line — n = 2 \to 1 — is the single most important spectral feature in the hydrogen spectrum. Its wavelength:

\frac{1}{\lambda_{\text{Ly}\alpha}} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R\cdot\frac{3}{4}.

\lambda_{\text{Ly}\alpha} = \frac{4}{3R} = \frac{4}{3 \times 1.097\times 10^7}\ \text{m} = 1.216\times 10^{-7}\ \text{m} = 121.6\ \text{nm}.

The series limit of Lyman (n_2 \to \infty, called Lyman-beta, Lyman-gamma, …, up to the continuum edge) is \lambda_\infty = 1/R = 91.2 nm. A photon shorter than 91.2 nm doesn't just excite the atom — it ionises it, tearing the electron off entirely. This is the ionisation wavelength of hydrogen.

ISRO's ASTROSAT UVIT instrument operates in the far-UV window from about 130 to 180 nm and can image the Lyman-alpha emission from galaxies. When astronomers talk about "seeing the universe in Lyman-alpha", they are tracking the light of hydrogen atoms falling from n = 2 back to n = 1, anywhere in the observable universe.

Balmer series — visible (n_1 = 2)

Transitions ending at n = 2. These are the four lines the schoolteacher Balmer stared at, plus all the higher ones. The first four:

n = 3 \to 2: H-\alpha at 656.3 nm (red)
n = 4 \to 2: H-\beta at 486.1 nm (turquoise)
n = 5 \to 2: H-\gamma at 434.0 nm (blue)
n = 6 \to 2: H-\delta at 410.2 nm (deep violet)

Verify the first: 1/\lambda = R(1/4 - 1/9) = R \cdot 5/36, so \lambda = 36/(5R) = 6.565\times 10^{-7} m = 656.5 nm. Within the precision of the Bohr model (i.e. before you include the reduced-mass correction), that is the H-\alpha line.

The Balmer series limit is \lambda = 4/R = 364.5 nm — just into the near-UV. That is the mysterious number Balmer found in his 1885 fit.

The red of H-\alpha is why astronomers' photographs of the sun taken at Kodaikanal (and of nebulae like the Orion Nebula) look vivid red. The light is coming from hydrogen atoms that were ionised in a stellar environment and are now recombining — and one of the transitions they pass through on the way back to the ground state is n = 3 \to 2.

Paschen series — near infrared (n_1 = 3)

Transitions ending at n = 3. First line: n = 4 \to 3 at 1875 nm (near infrared). Series limit: \lambda = 9/R = 820 nm (barely into the IR). The Paschen lines are invisible to the eye but easily detected by IR astronomy cameras; they are used to probe cool hydrogen regions in the interstellar medium.

Brackett series — mid infrared (n_1 = 4) and Pfund series (n_1 = 5)

Transitions ending at n = 4 and n = 5 respectively. These lines lie from about 2.2 to 7.5 micrometres — deep infrared, well beyond the eye's reach. Historically they were the last of the named series to be measured. Today they are used (along with Paschen) for hydrogen thermometry in high-temperature plasmas and in the atmospheres of late-type stars. Higher series (n_1 \geq 6) exist but do not carry personal names.

Explore the Rydberg formula

Drag the slider for n_2 below and watch the emitted wavelength. The slider for n_1 picks which series you are in. The readout shows the wavelength in nanometres and places the line in the electromagnetic spectrum.

The red slider sets $n_1$ (lower level — which series you are in). The dark slider sets $n_2$ (upper level). The emitted photon's wavelength and energy are shown. Try $n_1 = 2, n_2 = 3$ — you should get 656 nm, the H-alpha line.

The series limit and the ionisation energy

Let n_2 \to \infty in the Rydberg formula. The 1/n_2^2 term vanishes and

\frac{1}{\lambda_\infty} = \frac{R}{n_1^2}, \qquad \lambda_\infty = \frac{n_1^2}{R}.

For n_1 = 1: \lambda_\infty = 1/R = 91.2 nm. This is the series limit of the Lyman series and also the ionisation wavelength of hydrogen. A photon of exactly 91.2 nm has enough energy to tear an electron from an atom in the ground state and leave it with zero kinetic energy at infinity. The corresponding photon energy is hc/\lambda_\infty = 13.6 eV — the ionisation energy of hydrogen, as expected.

Photons shorter than 91.2 nm ionise with energy to spare — the extra goes into kinetic energy of the freed electron. This is photoionisation, and it is the dominant mechanism by which hydrogen is ionised in the atmospheres of hot stars and in planetary ionospheres.

The named hydrogen series placed on the electromagnetic spectrum (logarithmic wavelength axis). The three Balmer lines H-$\alpha$, H-$\beta$, H-$\gamma$ lie in the visible band; the Lyman series is entirely in the ultraviolet; the Paschen, Brackett and Pfund series are infrared. Each series converges to its series limit on the short-wavelength side.

Emission and absorption — the same lines, different direction

The discussion so far has been about emission: an excited atom drops to a lower level and emits a photon. Run the process in reverse and you get absorption: an atom in a lower level swallows a photon of exactly the right energy and jumps to a higher level.

The rule is the same. The photon energy must match the energy gap:

E_\gamma = E_{n_2} - E_{n_1}.

A photon of slightly different wavelength passes through the gas unaffected. The gas is transparent to every wavelength except the Rydberg wavelengths, which it absorbs.

This is why, when you look at the sun's spectrum with a good spectrograph, you see dark lines at precisely the Balmer wavelengths — the Fraunhofer lines. The sun's interior produces a continuous spectrum (a thermal blackbody); as the light passes through the cooler outer atmosphere, hydrogen atoms there absorb the Balmer wavelengths on their way out. The dark H-\alpha, H-\beta, H-\gamma, H-\delta absorption lines are one of the standard ways astronomers identify hydrogen in distant objects. The Kodaikanal Solar Observatory's H-\alpha filter images both: emission from excited chromospheric gas, and absorption in the solar atmosphere.

Each element has a unique fingerprint

The Rydberg pattern — 1/\lambda = R\cdot(1/n_1^2 - 1/n_2^2) — is special to hydrogen. Every other element has its own distinctive set of spectral lines, and those patterns do not fit a simple Rydberg formula. Helium has hundreds of lines from its two-electron configurations. Sodium famously has a pair of yellow D-lines at 589.0 and 589.6 nm (these are the yellow you see in street lamps in Indian cities). Iron, which is commonly found in stars and in terrestrial industrial settings, has thousands of lines.

The pattern is still the same in spirit — each line is one electron dropping between two energy levels — but the energy levels themselves are different for every element, because different nuclei have different charges (Z) and different numbers of electrons that screen that charge. Two atoms of sodium, anywhere in the universe, have identical energy levels; one atom of sodium and one of iron do not.

This is the basis of atomic spectroscopy. When astronomers look at the light from a distant star and decompose it through a grating, the set of absorption lines acts as a fingerprint telling them exactly which elements are in that star's atmosphere. When Indian industrial labs use inductively-coupled-plasma optical emission spectroscopy (ICP-OES) to check the purity of a steel alloy, they are measuring the same fingerprint for every element in the sample.

Worked examples

Example 1: Wavelength of H-alpha

H-\alpha is the n = 3 \to n = 2 transition in hydrogen — the red Balmer line. Compute its wavelength from the Rydberg formula and check that it matches what you would measure in a school discharge-tube experiment.

The $n = 3 \to 2$ transition in hydrogen. The emitted photon has energy 1.89 eV and wavelength 656 nm — deep red.

Step 1. Identify n_1 and n_2.

n_1 = 2 (final, lower level — this is the Balmer series). n_2 = 3 (initial, upper level).

Why: convention is n_1 < n_2. The atom drops from n_2 to n_1. In Balmer, every transition ends at 2.

Step 2. Evaluate the bracket in the Rydberg formula.

\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}.

Why: common denominator 36 so the subtraction is clean. The bracket is always a proper fraction between 0 and 1.

Step 3. Apply the formula.

\frac{1}{\lambda} = R\cdot\frac{5}{36} = 1.097\times 10^7\ \text{m}^{-1}\cdot\frac{5}{36} = 1.524\times 10^6\ \text{m}^{-1}.

Step 4. Invert to get the wavelength.

\lambda = \frac{1}{1.524\times 10^6\ \text{m}^{-1}} = 6.562\times 10^{-7}\ \text{m} = 656.2\ \text{nm}.

Why: the simple Bohr model, with no corrections, predicts 656.2 nm. The actual measured wavelength of H-alpha in air is 656.28 nm. The agreement is the reason everyone believed Bohr immediately.

Step 5. Cross-check via photon energy.

E_\gamma = 13.6\ \text{eV}\cdot\frac{5}{36} = 1.889\ \text{eV}.

Convert: \lambda = hc/E = (1240\ \text{eV·nm})/1.889 = 656.4 nm.

Why: the convenient rule hc = 1240 eV·nm lets you check \lambda in nanometres by dividing 1240 by the photon energy in eV. Both routes (Rydberg and energy) must give the same answer because they are the same equation rearranged.

Result: \lambda_{\text{H}\alpha} = 656.2 nm — red light, right at the edge of visible into deep red. This is the line the Kodaikanal telescope is tuned to.

What this shows: Starting from a single postulate about angular momentum, the Bohr model predicts a wavelength that matches a four-significant-figure measurement. This is not a fit. The coefficient R comes from fundamental constants only; the integers n_1 and n_2 are small positive integers. Everything else is arithmetic.

Example 2: Ultraviolet photons ionising hydrogen

A hydrogen atom in the ground state absorbs an ultraviolet photon of wavelength \lambda = 80 nm. Does the atom get excited to a higher bound level, or does it ionise? If it ionises, what is the kinetic energy of the ejected electron?

The 80-nm photon has 15.5 eV — more than the 13.6 eV needed to liberate the electron. The excess 1.9 eV becomes kinetic energy of the freed electron.

Step 1. Convert the photon's wavelength to energy.

E_\gamma = \frac{hc}{\lambda} = \frac{1240\ \text{eV·nm}}{80\ \text{nm}} = 15.50\ \text{eV}.

Why: use the hc = 1240 eV·nm shortcut. A 100-nm UV photon is almost exactly 12.4 eV; an 80-nm photon scales up proportionally.

Step 2. Compare with the ionisation energy of hydrogen from the ground state.

Ionisation energy: |E_1| = 13.6 eV (the energy to move an electron from n=1 to n=\infty).

E_\gamma = 15.50 eV > 13.6 eV, so the photon ionises the atom.

Why: there is no bound-state level 15.50 eV above n = 1 — the deepest possible transition from n = 1 tops out at exactly 13.6 eV (to n = \infty). Anything more energetic cannot be absorbed into a bound transition; it must ionise, because the continuum is available and can soak up any excess.

Step 3. Apply energy conservation to find the ejected electron's kinetic energy.

E_\gamma = |E_1| + KE_e

KE_e = E_\gamma - |E_1| = 15.50 - 13.6 = 1.90\ \text{eV}.

Why: the photon's energy splits into two pieces — the fixed amount 13.6 eV needed to free the electron, and whatever is left over becomes the freed electron's kinetic energy. This is Einstein's photoelectric equation applied to an atom instead of a metal surface.

Step 4. Convert the electron's kinetic energy to a speed for intuition.

KE_e = \tfrac{1}{2}m_e v^2 \quad\Rightarrow\quad v = \sqrt{\frac{2\cdot KE_e}{m_e}}.

With KE_e = 1.90\ \text{eV} = 3.04\times 10^{-19} J and m_e = 9.11\times 10^{-31} kg:

v = \sqrt{\frac{2 \times 3.04\times 10^{-19}}{9.11\times 10^{-31}}} = \sqrt{6.68\times 10^{11}} \approx 8.17\times 10^5\ \text{m/s}.

Why: this is about 0.003c — non-relativistic, so the Newtonian formula is fine. No need for the relativistic E^2 = (pc)^2 + (mc^2)^2.

Result: The photon ionises the ground-state atom and ejects the electron with kinetic energy 1.90 eV and speed 8.17 × 10⁵ m/s.

What this shows: The same energy-balance logic that handled bound-to-bound transitions (the Rydberg formula) extends directly to bound-to-free transitions (photoionisation). The only difference is that instead of landing in a discrete level n_2, the electron lands in the continuum above n = \infty with positive kinetic energy. This is the basis of the photoelectric effect.

Example 3: Identifying an unknown line

In a lab experiment you measure a spectral line from hydrogen at \lambda = 1875 nm. Which transition does it correspond to?

The $n = 4 \to 3$ transition — the first line in the Paschen series, known as Paschen-$\alpha$.

Step 1. Locate the line in the EM spectrum. 1875 nm is deep in the infrared — well beyond the visible (ends near 700 nm) and past the Paschen series limit at 820 nm. So this is Paschen or beyond.

Step 2. Try the Paschen series (n_1 = 3). The first line is n_2 = 4:

\frac{1}{\lambda} = R\left(\frac{1}{9} - \frac{1}{16}\right) = R\cdot\frac{16-9}{144} = R\cdot\frac{7}{144}.

Why: working out common denominators first keeps the algebra clean. For Paschen, 1/n_1^2 = 1/9 is fixed.

Step 3. Compute.

\lambda = \frac{144}{7R} = \frac{144}{7 \times 1.097\times 10^7}\ \text{m} = \frac{144}{7.679\times 10^7}\ \text{m} = 1.875\times 10^{-6}\ \text{m} = 1875\ \text{nm}.

Why: matches the observed wavelength almost exactly. No other simple transition gives 1875 nm — this must be n = 4 \to 3.

Result: The 1875-nm line is Paschen-\alpha, the n = 4 \to 3 transition.

What this shows: In principle, every spectral line is a candidate for every possible (n_1, n_2) pair. In practice, one pair usually dominates: the wavelength constrains the series (via the EM region) and then a short search through small integers finds the exact transition. This is what astronomers and spectroscopists do every day.

Common confusions

"Why are the lines so sharp? Nothing in nature is infinitely sharp." The lines are not infinitely sharp — they have a finite width set by the finite lifetime of the excited state (Heisenberg: \Delta E \cdot \Delta t \sim \hbar), Doppler broadening from thermal motion, and pressure broadening from collisions. But the widths are small fractions of a nanometre — much narrower than any dye filter — so to the naked eye the lines look infinitely narrow. High-resolution spectroscopy resolves the line profile.
"Isn't the Rydberg formula empirical? Bohr's model is outdated." The Rydberg formula was discovered empirically (by Balmer and Rydberg from spectral measurements) before Bohr. Bohr then derived it from first principles. Modern quantum mechanics (Schrödinger equation, QED) derives the same formula with additional small corrections (fine structure, Lamb shift, hyperfine). The formula is correct to excellent approximation for hydrogen; it is only "outdated" in the sense that better theories exist with finer corrections.
"Why doesn't the Rydberg formula work for helium?" Helium has two electrons. Each electron feels the nucleus' pull plus a repulsion from the other electron. That repulsion is not a central force and breaks the single-particle structure on which Bohr's L = n\hbar argument rests. Helium has energy levels and spectral lines, but they come from solving the two-electron Schrödinger equation, not from any simple 1/n^2 formula.
"Spectral lines are different for different atoms, so the universe must have different physics in different places." Exactly the opposite. The fact that hydrogen lines measured in an Indian physics lab are at the same wavelengths as hydrogen lines measured from a galaxy 10 billion light years away is one of the strongest pieces of evidence that physics is the same everywhere. Redshift moves the lines; the pattern stays identical.
"The Lyman series ends at 91.2 nm. What is at wavelengths shorter than that?" The continuum — free electrons returning to the nucleus from positive-energy states above n = \infty. A free electron can have any kinetic energy, so the continuum is, well, continuous. Below 91.2 nm the spectrum is a smooth unstructured background, not a line pattern. The same is true at the short-wavelength limit of every series.
"Ultraviolet photons from the sun should destroy all the hydrogen." Most ionising UV (< 91.2 nm) is absorbed high in the Earth's atmosphere — in the ionosphere, ironically by ionising the hydrogen and oxygen that is already there. Very little reaches the ground. The hydrogen at Earth's surface (mostly bound in water) is safe.

If you came here to know what the hydrogen spectrum is, how it gives the Balmer lines, and what the Rydberg formula is for, stop here — you have the whole story. What follows is for readers who want the precision corrections, the Z-dependence for hydrogen-like ions, and the connection to modern atomic physics.

The Rydberg constant and reduced mass

The formula above used the electron mass m_e. The proton is not infinitely heavy — it also orbits the common centre of mass. The correct mass to use is the reduced mass

\mu = \frac{m_e m_p}{m_e + m_p} = m_e \cdot \frac{1}{1 + m_e/m_p}.

Since m_e/m_p \approx 1/1836, the reduced mass is about 0.0544% smaller than m_e. Bohr's derivation is unchanged except that m_e is everywhere replaced by \mu, which means R \to R_H where

R_H = R_\infty\cdot\frac{1}{1 + m_e/m_p} = 1.09677\times 10^7\ \text{m}^{-1}.

Here R_\infty = 1.09737\times 10^7 m^{-1} is the "infinite-nuclear-mass" Rydberg constant (the limit m_p \to \infty) and R_H is the value measured for hydrogen. Deuterium, with a nucleus about twice as massive, has R_D slightly closer to R_\infty — the observed wavelength difference between hydrogen and deuterium lines (about 0.18 nm for H-\alpha) was the first evidence of deuterium's existence, discovered in 1931.

Hydrogen-like ions: the Z^2 scaling

Any one-electron atom with nuclear charge Z (e.g. He^+ with Z = 2, Li^{2+} with Z = 3) has the same Bohr structure with modified energies:

E_n = -\frac{Z^2 \cdot 13.6\ \text{eV}}{n^2}, \qquad r_n = \frac{n^2 a_0}{Z}.

The Rydberg formula becomes

\frac{1}{\lambda} = R Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right).

For He^+, the wavelength of the analogue of H-\alpha (n = 3 \to 2) is 656.3/4 = 164.1 nm — far ultraviolet. He^+ spectra are seen in the hot atmospheres of O and B stars. Fe^{25+} (iron stripped of all but one electron) has Z = 26; its Lyman-\alpha line is at 121.6/26^2 = 0.18 nm — a hard X-ray, routinely detected by X-ray astronomy missions observing supernova remnants and tokamak plasmas.

Fine structure and the Lamb shift

Bohr's formula predicts that every level n is a single sharp state. High-resolution spectroscopy shows each line is a close multiplet:

Fine structure. The electron's relativistic kinetic energy correction and its spin-orbit coupling split each level into sublevels labelled by the total angular momentum j. The splitting is of order \alpha^2 E_n \sim 10^{-4} of the Bohr energy. Dirac's relativistic theory (1928) gets the fine structure right.
Lamb shift. In 1947, Willis Lamb measured a tiny splitting between the 2s_{1/2} and 2p_{1/2} states of hydrogen — which Dirac theory predicted should be exactly degenerate. The splitting is \sim 1000 MHz (about 4\times 10^{-6} eV) and arises from the electron's interaction with its own quantum-fluctuation-filled electromagnetic environment — quantum electrodynamics (QED). The Lamb shift was one of the first triumphs of QED.
Hyperfine structure. The interaction between the electron's spin and the proton's spin splits every level by a tiny amount. The ground-state hyperfine splitting of hydrogen corresponds to a photon of wavelength 21 cm — a radio wavelength. The 21-cm line is how radio astronomers map neutral hydrogen throughout the Milky Way and in other galaxies. The Giant Metrewave Radio Telescope (GMRT) in Pune is one of the world's most sensitive instruments for observing this line.

The Zeeman effect — spectral lines in a magnetic field

Put a hydrogen discharge tube in a strong magnetic field (say, from a laboratory electromagnet or a sunspot's magnetic field). Each spectral line splits into several closely spaced components. The energy of a level of magnetic quantum number m_\ell shifts by

\Delta E = m_\ell \mu_B B,

where \mu_B = e\hbar/(2m_e) = 9.27\times 10^{-24} J/T is the Bohr magneton. Different sublevels have different m_\ell and therefore different energies, so transitions between them split the original line into three (for a normal Zeeman effect) or more (for an anomalous Zeeman effect, which requires spin). Astronomers use the Zeeman splitting of spectral lines in sunspots to measure the strength of solar magnetic fields.

Why the electron does not fall in spite of accelerating

Classically, an orbiting electron must radiate and spiral in — this was Rutherford's problem and Bohr's ad-hoc fix was just to declare that stationary orbits are stable. The modern resolution is that the electron's ground state is not a circular orbit at all; it is a spherically symmetric standing wave — the 1s orbital — that has no net accelerating charge distribution. There is a nonzero expectation value of |\vec r|^2, but the centre of charge stays at the nucleus. No net dipole moment, no radiation. The classical "accelerating charge radiates" argument needs a current with a time-varying dipole moment; the 1s state has none. Bohr's postulate 3 is not a fudge; it is the first peek at the fact that electrons in atoms are waves, not particles.

The series converges but does not stop

You might worry that there are infinitely many lines in each series, crowding near the series limit. They become impossibly close to one another at the limit — in principle, any spectroscope of finite resolution will see a continuum of lines near the limit merging into a smooth blur. In high-resolution measurements, this is exactly what is seen: the discrete Rydberg lines visible as individual lines at low n, merging into a blur and then a true ionisation continuum as n \to \infty. At n = 50, the atom is a metre-scale electron cloud — a Rydberg atom — with its own rich behaviour, exploited in atomic-physics labs for quantum information experiments.

Where this leads next

Bohr's model of the hydrogen atom — the postulates that give the energy levels from which every Rydberg wavelength is computed.
Energy levels and quantum numbers — the full quantum-mechanical picture that replaces n with the four quantum numbers (n, \ell, m_\ell, m_s) and explains why Bohr works for hydrogen but not for helium.
Photons and wave-particle duality of light — the energy quanta that carry each spectral line's energy.
Photoelectric effect — Einstein's explanation of photon absorption by bound electrons, identical in spirit to photoionisation at the series limit.
X-rays — production and properties — the hydrogen-like transitions in heavy-ion cores that produce the characteristic X-ray lines.
de Broglie hypothesis — why Bohr's angular-momentum quantisation is the standing-wave condition on the electron's matter wave.