Energy levels and line spectra of hydrogen

Hold a glass prism up to sunlight and you get a smooth, unbroken rainbow — every colour blending into the next with no gaps. Now do the same thing to the light from a glowing hydrogen lamp, and something strange happens: instead of a continuous band, you see a handful of sharp, isolated bright lines on a dark background — a violet line here, a blue-green line there, a red line further along — with nothing but blackness in between. For decades this was one of physics' most stubborn puzzles. Why would an atom, the simplest atom of all, refuse to glow in every colour and instead pick out only a few exact wavelengths, always the same ones, atom after atom, lamp after lamp? The answer turned out to be one of the most important clues in the history of physics, and it comes directly from the energy levels you met in the Bohr model. Because a hydrogen electron is only allowed to sit at certain discrete energies $E_n = -13.6/n^2$ eV — never in between — it can only ever emit a photon whose energy exactly equals the gap between two of those allowed levels. A discrete ladder of energies can only ever produce a discrete set of photon energies, and a discrete set of photon energies is exactly what a line spectrum is. Every coloured line in that hydrogen lamp is the fingerprint of one specific electron jump. This article builds the energy-level diagram of hydrogen from the Bohr formula, shows how transitions between levels are read off it, and organises the resulting spectral lines into named families — Lyman, Balmer, Paschen, Brackett, Pfund — each defined by which floor its transitions land on. We keep the treatment exactly where CBSE Class 12 places it: qualitative for the spectral series (which region of light each family occupies, and why they group the way they do), fully quantitative for the underlying energy arithmetic you already know from the Bohr model. By the end you should be able to sketch the diagram from memory, calculate the photon energy for any named transition, and instantly say which series it belongs to and whether it is visible to the human eye.

Key takeaways

Hydrogen's allowed energies are discrete, given by $E_n = -13.6/n^2$ eV for $n = 1, 2, 3, \ldots$ , and can be drawn as horizontal rungs on an energy-level diagram, converging to $E_\infty = 0$ eV.
A photon is emitted or absorbed only when its energy exactly matches a gap between two levels: $h\nu = E_{n_i} - E_{n_f}$ for emission (higher to lower), the same equation in reverse for absorption.
Discrete levels force a discrete (line) spectrum — this is the entire reason hydrogen glows in isolated coloured lines rather than a smooth rainbow.
Transitions are grouped into series by their common landing level $n_f$ : Lyman ( $n_f=1$ , ultraviolet), Balmer ( $n_f=2$ , the only series with visible lines), Paschen ( $n_f=3$ ), Brackett ( $n_f=4$ ), Pfund ( $n_f=5$ ) — all infrared.
Levels bunch together as $n$ grows because $E_n \propto -1/n^2$ flattens out, so gaps near the ionisation limit ( $E=0$ ) are tiny compared with the huge $n=1 \to n=2$ gap of $10.2$ eV.
This syllabus keeps the series treatment qualitative — know which series lies where and why, but you are not expected to compute exact line wavelengths from the full Rydberg formula.
Key idea: because a hydrogen electron can only occupy the discrete energies $E_n=-13.6/n^2$ eV, every photon it emits or absorbs has one of only finitely many possible energies $h\nu = E_{n_i}-E_{n_f}$ , and grouping these transitions by their common lower level $n_f$ produces the named Lyman/Balmer/Paschen/Brackett/Pfund spectral series.

The core insight

The discreteness of hydrogen's energy levels $E_n=-13.6/n^2$ eV forces every emitted or absorbed photon to carry one of only finitely many energies $h\nu=E_{n_i}-E_{n_f}$ , and sorting these transitions by their shared lower level $n_f$ produces the named line-spectrum series, with only Balmer ( $n_f=2$ ) falling in visible light.

Where it fits in the syllabus

This topic sits directly after the Bohr model within CBSE Class 12 Physics Unit IX ("Atoms and Nuclei"), and it is the payoff of everything the Bohr postulates set up: once you accept that orbit energies are quantised, the line spectrum follows almost for free. The board expects you to draw and read a hydrogen energy-level diagram, calculate the energy and identify the series of a photon emitted or absorbed in a named transition, and state qualitatively which series lies in which region of the electromagnetic spectrum. The full quantitative Rydberg formula for computing exact series wavelengths is explicitly kept out of scope at this level — series membership and general trends are what matter, not numeric line positions to several significant figures.

The physical picture

Every atom of hydrogen anywhere in the universe has access to exactly the same ladder of allowed energies, because that ladder comes from the same physics — an electron bound to a single proton. From the Bohr model, the energy of the $n$ th allowed state is

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

where: $n$ is the principal quantum number (which orbit/level the electron occupies) and $13.6$ eV is the ionisation energy of hydrogen — the energy needed to pull the ground-state electron ( $n=1$ ) completely free. The physical reading is that $E_n$ is always negative for a bound electron (negative energy means "bound," the electron cannot escape without extra energy fed in), it becomes less negative as $n$ grows (the electron is more loosely held the farther out it sits), and it approaches zero as $n \to \infty$ — the point at which the electron is no longer bound at all, i.e. the atom is ionised. Because $n$ can only take whole-number values, $E_n$ can only take the specific values $-13.6, -3.4, -1.51, -0.85, -0.54, \ldots$ eV (for $n=1,2,3,4,5,\ldots$ ) and nothing in between. Draw these as horizontal lines stacked one above another and you get the energy-level diagram — the single most useful picture in this entire topic.

Figure

A hydrogen energy level diagram with horizontal rungs at n equals 1 through 5 and the ionisation limit, rungs bunching closer together as n increases, with an upward arrow for absorption and a downward arrow for emission between two levels

An electron does not glide smoothly between levels; it makes an instantaneous jump, and every jump has an energy cost or payout fixed exactly by the two levels involved:

h\nu = E_{n_i} - E_{n_f}

where: $h = 6.63\times10^{-34}\ \text{J s}$ is Planck's constant, $\nu$ is the frequency of the photon emitted (if $n_i > n_f$ , a downward jump) or absorbed (if $n_i < n_f$ , an upward jump), $E_{n_i}$ is the energy of the initial level and $E_{n_f}$ the energy of the final level. The physical reading: when the electron falls from a higher level to a lower one, it must shed exactly that much energy, and it does so by emitting one photon of that exact energy — no more, no less, no partial photons. When the electron is pushed from a lower level to a higher one by an incoming photon, that photon must carry exactly the energy of the gap; a photon of the wrong energy simply cannot be absorbed by that transition, however intense the light source is. Since only whole-number values of $n$ exist, only a finite, discrete set of gaps $E_{n_i}-E_{n_f}$ exist, and therefore only a discrete set of photon energies — hence wavelengths — can ever appear in hydrogen's spectrum. This is precisely why a hydrogen lamp shows isolated bright lines rather than a smooth rainbow: nature has no photon to offer at any energy that does not correspond to some pair of levels.

Spectral series. All the transitions that share the same lower level $n_f$ are grouped into one family, named after the person who first studied it:

Lyman series — transitions ending at $n_f = 1$ (any $n_i > 1 \to 1$ ); lies entirely in the ultraviolet, since these are the largest energy gaps in the whole diagram.
Balmer series — transitions ending at $n_f = 2$ ; the only series with any lines in visible light, and historically the first one observed, which is exactly why it carries the most famous name.
Paschen series — transitions ending at $n_f = 3$ ; infrared.
Brackett series — transitions ending at $n_f = 4$ ; infrared.
Pfund series — transitions ending at $n_f = 5$ ; infrared, even longer wavelengths than Brackett.

Within one series, as the starting level $n_i$ climbs higher and higher, the photon energy creeps up toward a ceiling called the series limit — the transition from $n_i \to \infty$ down to that same $n_f$ — and consecutive lines in the series crowd closer and closer together, exactly mirroring the way the energy levels themselves bunch up near the top of the diagram. This syllabus asks only for this qualitative picture — which series occupies which part of the spectrum and why the lines within a series converge — not the full Rydberg-formula arithmetic for exact wavelengths.

Derivation AWhy hydrogen's spectrum must be made of discrete lines

The goal here is not to compute a number but to build the logical chain, in plain algebra, from "the allowed energies are discrete" to "the emitted light must be discrete lines, never a continuous rainbow" — the single idea that makes the entire line-spectrum topic make sense rather than being a list of names to memorise.

Step 1. Start from the fact established by the Bohr model: the electron's energy can only take the values $E_n = -13.6/n^2$ eV for whole-number $n$ . There is no rule, no mechanism, and no orbit that allows the electron to sit at, say, $-5.0$ eV — that value simply does not appear anywhere on the list $-13.6, -3.4, -1.51, -0.85, \ldots$ eV. The set of allowed energies is a discrete, countable list, not a continuous range.

Step 2. Any transition the electron makes is a jump directly from one item on that list, $E_{n_i}$ , to another item on that list, $E_{n_f}$ . Because both endpoints are drawn from the same discrete list, the difference $E_{n_i} - E_{n_f}$ can also only take a discrete, countable set of values — one value for every possible pair $(n_i, n_f)$ you could choose, and no values in between those.

Step 3. Conservation of energy then fixes the photon completely: whatever energy the electron loses (emission) or gains (absorption) must be carried away by, or supplied by, exactly one photon, so $h\nu = E_{n_i}-E_{n_f}$ . Since $h$ is a fixed constant, a discrete set of allowed values for $E_{n_i}-E_{n_f}$ forces $\nu$ — and therefore the wavelength $\lambda = c/\nu$ — to take only a discrete set of values too.

Step 4. A light source that can only emit photons at a discrete set of frequencies, when passed through a prism or diffraction grating, cannot paint a smooth rainbow; it can only light up a spectrograph screen at the specific positions corresponding to those specific frequencies, leaving every position in between completely dark. Those lit positions are exactly the bright lines you see in a hydrogen line spectrum, and the dark gaps between them are the direct, visible proof that the underlying energy levels are quantised rather than continuous.

Conclusion. The chain is airtight and runs in one direction only: discrete energy levels imply discrete transition energies imply discrete photon frequencies imply a line spectrum. Historically the logic was actually run backwards — physicists saw the discrete lines first, decades before Bohr, and it was the observed line spectrum of hydrogen that Bohr's model was built to explain. Seeing the derivation forwards, as done here, is what makes the model's success feel inevitable rather than lucky.

Worked examples

Board level - identifying a transition's series and energy

Question: An electron in a hydrogen atom falls from the $n=4$ level to the $n=1$ level. Find the energy of the photon emitted, and state which spectral series this transition belongs to.

Step 1. Compute the two level energies from $E_n = -13.6/n^2$ eV: $E_4 = -13.6/16 = -0.85$ eV, and $E_1 = -13.6/1 = -13.6$ eV.

Step 2. Apply $h\nu = E_{n_i} - E_{n_f}$ with $n_i = 4$ and $n_f = 1$ : photon energy $= E_4 - E_1 = (-0.85) - (-13.6) = 12.75$ eV.

Step 3. The transition lands on $n_f = 1$ , so by definition it belongs to the Lyman series, which lies entirely in the ultraviolet — invisible to the human eye, so this particular jump could never be one of the coloured lines seen by simply looking at a hydrogen lamp.

Answer: The emitted photon carries $12.75$ eV of energy, and the transition belongs to the Lyman series (ultraviolet).

Competitive level - comparing two transitions and locating them on the spectrum

Question: Two transitions occur in a hydrogen sample: Transition A, from $n=3$ to $n=2$ , and Transition B, from $n=5$ to $n=2$ . Find the photon energy for each, state which series each belongs to, and determine which transition releases the more energetic photon.

Step 1. List the level energies needed: $E_2 = -13.6/4 = -3.4$ eV, $E_3 = -13.6/9 = -1.51$ eV, $E_5 = -13.6/25 = -0.54$ eV.

Step 2. Transition A energy $= E_3 - E_2 = (-1.51) - (-3.4) = 1.89$ eV. Transition B energy $= E_5 - E_2 = (-0.54) - (-3.4) = 2.86$ eV.

Step 3. Both transitions land on $n_f = 2$ , so both belong to the Balmer series — the only series that reaches into visible light, so both of these lines could in principle be seen directly with the eye (or a simple spectrometer) as coloured lines in the hydrogen lamp's spectrum.

Step 4. Compare the two energies directly: $2.86$ eV (Transition B) is larger than $1.89$ eV (Transition A), so Transition B releases the more energetic photon. This also matches the general trend within a series: a jump that starts from farther away ( $n_i=5$ versus $n_i=3$ ) releases more energy, because the initial level sits higher up (less negative) on the diagram, making the gap down to $n_f=2$ bigger.

Answer: Transition A emits $1.89$ eV (Balmer series), Transition B emits $2.86$ eV (Balmer series), and Transition B is the more energetic photon of the two. Timing note: This is a two-line-item comparison, not a single lookup — in an exam, compute both energies first using the same $E_n$ formula before comparing, rather than trying to eyeball which gap looks bigger from a rough sketch; budget about 3 minutes for a question with two transitions like this, since it is really just the Board-level calculation done twice plus one comparison step.

Common traps

Thinking any pair of levels can define a series — a series is fixed only by its common lower level $n_f$ ; many different $n_i$ values feed into the same series, but a series is never defined by the upper level.
Assuming the Balmer series is somehow the "main" or the only real series — it is simply the one series with lines in visible light, which is why it was discovered and named first; Lyman, Paschen, Brackett and Pfund are equally real, just outside the range the human eye can see.
Forgetting that absorption uses the identical energy-level diagram as emission, just read in the opposite direction — an atom does not need a separate "absorption diagram"; the same rungs and the same gaps apply both ways.
Trying to memorise or derive exact line wavelengths using the full Rydberg formula for a CBSE answer — this syllabus explicitly keeps hydrogen spectral series at the qualitative level (existence, grouping, which region of the spectrum); quoting the full numeric Rydberg calculation is not what is being tested here.
Sign errors when subtracting negative level energies — always compute $E_{n_i}-E_{n_f}$ carefully with the negative signs kept in (e.g. $(-0.85)-(-13.6)=+12.75$ , not $-12.75$ or $-14.45$ ); a photon energy must always come out positive for an emission.

You should be able to…

Sketch a hydrogen energy-level diagram from memory, including the ground state, several excited states, and the ionisation limit at $E=0$ .
Calculate the photon energy $h\nu = E_{n_i}-E_{n_f}$ for any named transition between two given levels, keeping the signs of negative energies straight.
Identify which spectral series (Lyman, Balmer, Paschen, Brackett, Pfund) a given transition belongs to, based purely on its lower level $n_f$ .
State which region of the electromagnetic spectrum each series occupies, and explain why only Balmer reaches into visible light.
Explain qualitatively why the energy levels — and hence the spectral lines within a series — bunch closer together as $n$ increases.
Distinguish an emission transition (electron falls, photon released) from an absorption transition (electron rises, photon consumed) on the same diagram.

Ask the tutor: Want to work through a full energy-level diagram with three or four transitions marked at once, or check which series a tricky transition (like $n=7\to3$ ) falls into — ask the AI tutor to walk it through with you.

Why is the Balmer series the only one we can actually see with our eyes?

Visible light is only a narrow window of the full electromagnetic spectrum, roughly $1.8$ eV to $3.1$ eV in photon energy. The Lyman series, ending on $n_f=1$ , involves the biggest energy gaps in the whole hydrogen diagram — every one of those photons carries more energy than violet light, putting the whole series in the ultraviolet. Paschen, Brackett and Pfund, ending on $n_f=3,4,5$ , involve much smaller gaps near the top of the ladder where the levels have bunched together, putting all of them below red light in the infrared. Only the Balmer series, sitting in between, happens to straddle the visible window — which is really just a coincidence of where $n_f=2$ sits relative to how human eyes evolved to see.

If the electron can only be at discrete energies, how does it get from n=1 all the way to n=4 to start with?

It absorbs a photon (or gets a kick from a collision) whose energy exactly matches the $n=1$ to $n=4$ gap, which is $12.75$ eV — the same number as the Board-level example above, just run in reverse. If the incoming photon has too little or too much energy, the electron generally cannot use it for that particular jump at all; the transition simply does not happen. This is why hydrogen gas is transparent to most wavelengths of light passing through it, but absorbs sharply at exactly the wavelengths matching its own energy gaps, producing dark absorption lines in an otherwise continuous background spectrum — the mirror image of the bright emission lines you get from a glowing hydrogen lamp.

Do all the electrons in a sample of hydrogen gas jump the same way at the same time, producing only one spectral line?

No — a real sample contains an enormous number of hydrogen atoms, and at any moment their electrons sit in a whole mixture of different levels, having been excited by heat, electric sparks or collisions to varying degrees. Because so many atoms are present, essentially every possible transition is happening somewhere in the sample at every instant, which is exactly why a real hydrogen spectrum shows the entire Lyman series, entire Balmer series, and so on, all at once, rather than just a single line. A single isolated hydrogen atom making one specific jump would indeed emit only one photon of one specific energy at that instant, but a lamp full of gas is really billions of atoms doing this simultaneously and independently.

Is the energy-level diagram exactly the same for every hydrogen atom in the universe, or does it depend on the atom's temperature or motion?

The set of allowed energies $E_n = -13.6/n^2$ eV depends only on the fixed physical constants of the electron-proton system (charge, mass, Planck's constant), so yes — the ladder of levels is identical for every hydrogen atom everywhere, regardless of temperature, pressure or motion. Temperature and collisions only affect which levels the electrons happen to occupy at a given moment (a hotter sample has more electrons pushed into higher, excited levels), not the positions of the rungs themselves. This is actually how astronomers identify hydrogen in distant stars and galaxies: the pattern of spectral lines is a fixed fingerprint, though the whole pattern can appear shifted in wavelength due to the object's motion relative to us (the Doppler effect), which is a separate topic from the energy-level structure itself.