ARCHETYPE: Spectral-Series & Rydberg Problems

The arithmetic you just drilled is the daily language of astrophysics. When a star or a distant galaxy sends its light through a spectrograph, the hydrogen Balmer lines you computed — $H_\alpha$ at $656$ nm, $H_\beta$ at $486$ nm — appear as dark absorption notches, and their positions are the most-used ruler in the sky. Slide the whole Balmer pattern to longer wavelengths and you have measured a redshift: the $H_\alpha$ line of a receding galaxy arrives stretched, and the fractional shift gives its recession speed, which is how Hubble first weighed the expansion of the universe. The Lyman-$\alpha$ forest — the thicket of Lyman absorption lines from intervening hydrogen clouds, each notch a genuine Rydberg $2 \to 1$ transition — lets astronomers map neutral hydrogen across billions of light-years, and the $21$-centimetre radio line (a separate hyperfine spin-flip of the ground state, a different mechanism from the Rydberg series) extends the map further. Closer to home, the same formula calibrates hydrogen and deuterium lamps used as wavelength standards, and the hydrogen-like $Z^2$ scaling of example 3 is exactly how the characteristic lines of highly stripped ions are predicted in fusion-plasma and X-ray-source diagnostics. Rydberg's difference of two inverse-squares, written in 1888, is still the first thing a spectrum tells you.

Bookkeeping first, because in a numerical gym the accounting is the physics. A hydrogen atom has a ladder of allowed energies $E_n = -13.6/n^2$ eV, rising from the ground floor $E_1 = -13.6$ eV toward the free-electron ceiling $E_\infty = 0$. When the electron drops from an upper level $n_2$ to a lower level $n_1$, the atom sheds the energy difference as a single photon, and that photon's wavelength is fixed by $\Delta E = 1240/\lambda$. Do the subtraction once and you have the whole spectrum:

where: $n_1$ — the lower (final) level, always the smaller integer; $n_2 > n_1$ — the upper (initial) level; $\Delta E$ — the emitted photon energy; the constant $13.6$ eV is the ionisation energy of hydrogen from its ground state. Read it as a subtraction on the energy ladder: the deeper the electron falls, the more energy the photon carries, and the shorter its wavelength. The lines sort themselves into series by their common landing level $n_1$. Every transition that ends on $n_1 = 1$ is a Lyman line (all in the ultraviolet); every one ending on $n_1 = 2$ is a Balmer line (the visible ones Balmer first fitted); $n_1 = 3$ is Paschen (infrared), $n_1 = 4$ Brackett, $n_1 = 5$ Pfund, and so on into the far infrared. Within a series the individual lines are labelled by the starting level $n_2$: the $\alpha$ line is the shortest jump ($n_2 = n_1 + 1$), then $\beta$, $\gamma$, crowding closer together as $n_2$ climbs.

Two features of this ladder run every problem. First, the levels crowd together as $n$ grows — the gap from $n = 2$ to $n = 3$ is $1.89$ eV, but from $n = 6$ to $n = 7$ it is only about $0.1$ eV. So within any series the lines bunch up toward a short-wavelength edge. Second, that edge — the series limit — is the photon you get when the electron falls all the way from the very top of the ladder ($n_2 \to \infty$) down to $n_1$. Its energy is simply $13.6/n_1^2$, because the $1/n_2^2$ term has died away to nothing.

Three tools run the entire family. The first is the Rydberg formula in its native wavenumber form, the way spectroscopists actually measured these lines:

where: $1/\lambda$ — the wavenumber (inverse wavelength) of the line; $R$ — the Rydberg constant, the same number for every hydrogen line. This is Balmer's 1885 fit generalised: Balmer only had the visible ($n_1 = 2$) lines, and Rydberg saw in 1888 that every series is the same expression with a different $n_1$. Equations (1) and (2) are not two facts but one — they are a single constant apart, because $hcR = 13.6$ eV. Multiply the wavenumber $1/\lambda$ by $hc = 1240$ eV·nm and you recover the energy $\Delta E$; that is the bridge you cross whenever a problem hands you $R$ instead of $13.6$ eV, or the reverse. The second tool is the series limit, the short-wavelength edge of a series. Set $n_2 \to \infty$ so the $1/n_2^2$ term vanishes:

where: $\lambda_{\text{limit}}$ — the wavelength of the series limit; a photon at this energy carries the electron from rest at infinity down to level $n_1$ (or, run backwards, is exactly the energy to ionise an atom sitting in level $n_1$). The Lyman limit is $91.2$ nm (deep UV), the Balmer limit $365$ nm (just past violet), the Paschen limit $821$ nm (near infrared). Beyond the limit — at shorter wavelengths still — the spectrum becomes a faint continuum, because a free electron with any leftover kinetic energy can be captured, and its energy is no longer quantised. The third tool is the $Z^2$ scaling for a hydrogen-like ion — any atom stripped down to a single electron, such as He$^+$ ($Z = 2$) or Li$^{2+}$ ($Z = 3$). Every energy on the ladder deepens by $Z^2$, so:

where: $Z$ — the nuclear charge (the atomic number) of the one-electron ion. The whole spectrum scales rigidly: every wavelength shrinks by the factor $1/Z^2$. So He$^+$ ($Z = 2$) takes hydrogen's Balmer $H_\alpha$ at $656$ nm and moves it to $656/4 = 164$ nm, deep in the ultraviolet. This is only valid for one-electron ions; a neutral or many-electron atom has electron–electron repulsion that the simple $Z^2$ law ignores.

Which series, and where in the spectrum. Given a transition $n_2 \to n_1$, the lower number is the series ($1$ Lyman/UV, $2$ Balmer/visible, $3$ Paschen/IR). Plug into equation (1) for $\Delta E$, then $\lambda = 1240/\Delta E$. A quick sanity flag: Lyman lines are all shorter than $122$ nm, Balmer lines live between $365$ and $656$ nm, Paschen lines are all longer than $821$ nm.

Find the series limit. Use equation (3) directly: $\lambda_{\text{limit}} = n_1^2/R$, or in energy $13.6/n_1^2$. The limit is the shortest wavelength of a series (all its lines are longer), and its energy equals the binding energy of level $n_1$.

Jump to a hydrogen-like ion. Multiply by $Z^2$ (equation 4). The fastest route is to scale a known hydrogen wavelength: $\lambda_{\text{ion}} = \lambda_{\text{H}}/Z^2$. He$^+$ ($Z=2$) divides every hydrogen wavelength by $4$; Li$^{2+}$ ($Z=3$) by $9$.

Count the lines. An atom (or a gas of atoms) excited to level $n$ can de-excite by any downward path, and the number of distinct spectral lines it can emit is the number of pairs of levels at or below $n$, namely $n(n-1)/2$. Level $4$ gives $6$ lines, level $5$ gives $10$. To find the longest and shortest of them, remember: the smallest energy gap (hence longest wavelength) is the top step $n \to n-1$, and the largest gap (shortest wavelength) is the full drop $n \to 1$.

1 · Board level — which series and what wavelength (CBSE and NEET pattern)

A hydrogen atom makes a transition from the level $n = 5$ to the level $n = 2$. Name the spectral series this line belongs to, find its wavelength, and state its colour.

Step 1. Name the series from the lower level. The electron lands on $n_1 = 2$, so this is a Balmer line — expect a visible wavelength.

Step 2. Subtract the rungs, equation (1): $\Delta E = 13.6\left(1/2^2 - 1/5^2\right) = 13.6\left(0.25 - 0.04\right) = 13.6\times0.21 = \boxed{2.86\ \text{eV}}$.

Step 3. Convert to wavelength: $\lambda = 1240/\Delta E = 1240/2.86 = \boxed{434\ \text{nm}}$ — violet light, the Balmer $H_\gamma$ line.

Step 4. Sense-check: $434$ nm sits between the Balmer limit ($365$ nm) and $H_\alpha$ ($656$ nm), exactly where a Balmer line should be, and it is shorter than $H_\alpha$ because the $5 \to 2$ jump is bigger than $3 \to 2$. Consistent.

2 · Board level — the series limit, both series (ICSE and CBSE pattern)

Find the series limit of the Balmer series and of the Lyman series of hydrogen. In each case say what the limit physically represents.

Step 1. Balmer limit, equation (3) with $n_1 = 2$: $\Delta E_{\text{limit}} = 13.6/2^2 = 13.6/4 = 3.40$ eV, so $\lambda_{\text{limit}} = 1240/3.40 = \boxed{365\ \text{nm}}$ (near ultraviolet, just past violet).

Step 2. Cross-check by the Rydberg road: $\lambda_{\text{limit}} = n_1^2/R = 4/(1.097\times10^{7}) = 3.65\times10^{-7}$ m $= 365$ nm. The two roads agree.

Step 3. Lyman limit with $n_1 = 1$: $\Delta E_{\text{limit}} = 13.6/1 = 13.6$ eV, so $\lambda_{\text{limit}} = 1240/13.6 = \boxed{91.2\ \text{nm}}$ (deep ultraviolet).

Step 4. Meaning: each limit is the photon emitted when the electron falls from the very top of the ladder ($n_2 \to \infty$) to $n_1$. Its energy equals the binding energy of level $n_1$ — so the Lyman limit, $13.6$ eV, is exactly the ionisation energy of ground-state hydrogen. All lines of a series are longer in wavelength than its limit.

3 · Competitive level — a hydrogen-like ion (JEE-Advanced and A-Level pattern)

Singly ionised helium (He$^+$, $Z = 2$) is a one-electron ion. Find the wavelength of its line analogous to the Balmer $H_\alpha$ transition ($n = 3 \to n = 2$), and the limit of its Balmer series. Compare each with hydrogen.

Step 1. Scale the known hydrogen line. Every wavelength divides by $Z^2 = 4$. Hydrogen's $H_\alpha$ is $656$ nm, so He$^+$ gives $\lambda = 656/4 = \boxed{164\ \text{nm}}$ — moved from red all the way into the deep ultraviolet.

Step 2. Check by the full formula, equation (4): $\Delta E = 13.6\,Z^2\left(1/4 - 1/9\right) = 13.6\times4\times0.1389 = 7.56$ eV, and $\lambda = 1240/7.56 = 164$ nm. Same answer, so the shortcut is safe.

Step 3. He$^+$ Balmer limit: scale hydrogen's $365$ nm by $1/Z^2$: $\lambda_{\text{limit}} = 365/4 = \boxed{91.2\ \text{nm}}$. Equivalently $\Delta E = 13.6\times4/4 = 13.6$ eV.

Step 4. Compare: the He$^+$ Balmer limit, $91.2$ nm, is identical to hydrogen's Lyman limit. That is no accident — the He$^+$ Balmer limit energy is $13.6\times Z^2/n_1^2 = 13.6\times4/4 = 13.6$ eV, the same as hydrogen's Lyman limit ($13.6/1$). The $Z^2$ and the $n_1^2$ conspire. Timing note: about 75 seconds. The moment you see a one-electron ion, do not restart from the formula — take the matching hydrogen wavelength and divide by $Z^2$. The full equation is only for the cross-check.

4 · Competitive level — counting the lines and finding the edges (JEE-Main and A-Level pattern)

A sample of hydrogen is excited so that its atoms reach the level $n = 4$. How many distinct spectral lines can the gas emit? Sort them by series, and find the longest and shortest wavelengths present.

Step 1. Count the lines: with the top level $n = 4$, the number of distinct transitions is $n(n-1)/2 = 4\times3/2 = \boxed{6}$ lines.

Step 2. Sort by landing level. The six jumps are $4\to1, 3\to1, 2\to1$ (three Lyman lines), $4\to2, 3\to2$ (two Balmer lines), and $4\to3$ (one Paschen line) — three series in one glow.

Step 3. Longest wavelength = smallest energy gap = the top step $4 \to 3$ (Paschen): $\Delta E = 13.6\left(1/9 - 1/16\right) = 13.6\times0.04861 = 0.661$ eV, so $\lambda = 1240/0.661 = \boxed{1876\ \text{nm}}$ (infrared).

Step 4. Shortest wavelength = largest energy gap = the full drop $4 \to 1$ (Lyman): $\Delta E = 13.6\left(1 - 1/16\right) = 13.6\times0.9375 = 12.75$ eV, so $\lambda = 1240/12.75 = \boxed{97.3\ \text{nm}}$ (deep ultraviolet). The whole emission thus spans from $97.3$ nm to $1876$ nm — a factor of nearly twenty in wavelength. Timing note: about 90 seconds. Do not compute all six wavelengths — the examiner only wants the extremes, and those are always the top step (longest) and the full fall to the ground (shortest). Counting is $n(n-1)/2$; sorting is by the landing level.

Reading the wrong integer as the series. The series is the lower level $n_1$, the rung the electron lands on, never the upper level. A $6 \to 2$ line is Balmer, not "the sixth series". Write the smaller integer as $n_1$ every time, and let the colour band confirm it: Lyman UV, Balmer visible, Paschen IR.

Swapping the levels and getting a negative energy. $\Delta E = 13.6\left(1/n_1^2 - 1/n_2^2\right)$ with $n_1 < n_2$ is positive. If you put the bigger level first you get a negative number; that is a bookkeeping sign for absorption, not an error to "fix" by dropping the minus. For an emitted-photon wavelength, always use the smaller level as $n_1$.

Forgetting the $Z^2$ on a hydrogen-like ion. He$^+$, Li$^{2+}$ and Be$^{3+}$ are not hydrogen — every wavelength divides by $Z^2$. The single fastest way in is to scale a known hydrogen line: $\lambda_{\text{ion}} = \lambda_{\text{H}}/Z^2$. And this only works for one-electron ions; neutral helium or lithium does not obey it.

Confusing a series limit with a line. The series limit ($n_2 \to \infty$) is the shortest wavelength of a series and marks the edge of the continuum; it is not one more line in the list. Its energy is $13.6/n_1^2$, which for Lyman equals the ionisation energy. Every real line of the series is longer than the limit.

Miscounting the lines. From level $n$ the number of distinct lines is $n(n-1)/2$, not $n$ and not $n-1$. And those lines are spread across several series — a gas excited to $n = 4$ shows Lyman, Balmer and Paschen lines at once, not just one series.

Board scope. Spectral-series and Rydberg numericals are firmly examined on CBSE, ICSE, JEE-Main, JEE-Advanced and NEET, and on A-Level 9702 (energy levels and line spectra). The hydrogen-like $Z^2$ ion extension is a JEE-Advanced and A-Level flavour rather than a routine CBSE ask. IB treats atomic spectra qualitatively and the Bohr maths only at Higher Level, and AP Physics 2 covers line spectra and energy levels conceptually but does not set this Rydberg-drill family as a numerical objective — do not revise this gym for an AP-2 or IGCSE paper, and note IGCSE 0625 has no quantitative atomic spectra at all.

Texts. NCERT Physics Class 12, Ch. 12 ("Atoms"), the hydrogen-spectrum sections and exercises · H. C. Verma, Concepts of Physics Vol. 2, "Bohr's Model and Physics of the Atom," the spectral-series problems · D. C. Pandey / Irodov atomic-spectra problem sets (for the series-limit, $Z^2$-scaling and line-counting variants) · Cambridge A-Level 9702, atomic-spectra and energy-level problems.

Provenance. Standard exam-archive problems on the hydrogen spectrum. Recompute: master template $3 \to 2$ Balmer $\Rightarrow \Delta E = 1.89$ eV, $\lambda = 656$ nm; $5 \to 2$ Balmer $H_\gamma \Rightarrow 2.86$ eV, $434$ nm; series limits Balmer $3.40$ eV / $365$ nm, Lyman $13.6$ eV / $91.2$ nm, Paschen $1.51$ eV / $821$ nm; He$^+$ ($Z = 2$) analogue of $H_\alpha \Rightarrow 7.56$ eV, $164$ nm, and He$^+$ Balmer limit $91.2$ nm; lines from level $4 = 6$ (Lyman $3$, Balmer $2$, Paschen $1$), with longest $4 \to 3 = 1876$ nm and shortest $4 \to 1 = 97.3$ nm; ladder $E_n = -13.6/n^2 = -13.6, -3.40, -1.51, -0.85, -0.544$ eV; reference lines $H_\beta$ ($4 \to 2$) $486$ nm, $H_\delta$ ($6 \to 2$) $410$ nm, Lyman $\alpha$ ($2 \to 1$) $122$ nm; bridge $hcR = 13.6$ eV; Rydberg $R = 1.097\times10^{7}$ m$^{-1}$.

Prerequisites. Energy Levels and the Hydrogen Spectrum — the concept article that builds the ladder $E_n = -13.6/n^2$, the Rydberg formula and the Lyman/Balmer/Paschen series this gym drills.

Go deeper (UG). The full quantum-mechanical account — the hydrogen wavefunctions, the reduced-mass and relativistic (fine-structure) corrections to the Rydberg constant, and why the simple $Z^2$ law breaks for many-electron atoms — is developed in the university seam The Hydrogen Energy Spectrum and the Rydberg Constant (`aq-qm-hydrogen-energy-spectrum-rydberg`). Enrichment only; nothing here depends on it.