Energy Levels & the Hydrogen Spectrum

The hydrogen spectrum is not a museum piece; it is a working instrument. Because every element prints its own set of lines at its own fixed wavelengths, spectroscopy lets us read the composition of anything that glows without touching it. Astronomers identify the elements in distant stars and galaxies purely from their line patterns, and measure how fast those objects recede by how far the whole barcode is Doppler-shifted — the redshift that revealed the expanding universe rests on comparing observed hydrogen lines against their laboratory wavelengths. The $21$-centimetre radio line of hydrogen, from an even finer splitting of the ground state, maps the cold hydrogen gas threaded through our own galaxy and others. Closer to home, the Rydberg constant is measured so precisely — to twelve significant figures — that hydrogen spectroscopy is one of the sharpest tests of quantum electrodynamics, and discrepancies in it (the "proton radius puzzle") have sent physicists back to the drawing board more than once. The tidy ladder Bohr sketched in 1913 is still where fundamental constants get pinned down.

Start from what the Bohr model handed us in the previous article. The electron in a hydrogen atom cannot orbit at any radius it likes; it is restricted to a set of quantised orbits labelled by a whole number $n = 1, 2, 3, \dots$, the principal quantum number. Each orbit has its own fixed total energy — kinetic plus the (negative) electrical potential energy of an electron bound to a proton — and Bohr's calculation gives that energy in one compact line: $E_n = -13.6/n^2$ electronvolts. That single formula is the spine of this whole article, so read it slowly. The energies are negative because the electron is bound: you would have to add energy to tear it away from the proton, and we agree to call a free electron at rest, infinitely far away, the zero of energy. Anything trapped in the atom therefore sits below zero. Because $n$ can only be a whole number, the allowed energies are not a smooth ramp but a set of discrete rungs — a ladder. The lowest rung, $n=1$, is the ground state at $E_1 = -13.6$ eV: the deepest, most tightly bound, most stable state, where an undisturbed hydrogen atom spends essentially all its time. Above it sit the excited states: $E_2 = -3.4$ eV, $E_3 = -1.51$ eV, $E_4 = -0.85$ eV, and so on. Notice how the rungs crowd together as $n$ grows — the $1 \to 2$ gap is a huge $10.2$ eV, but the $4 \to 5$ gap is under a tenth of an electronvolt. They pile up against the ceiling at $E = 0$, which is the atom's ionisation threshold: the electron finally free of the proton.

Now the crucial move, Bohr's third postulate. An electron sitting in an allowed level does not radiate — that is why the atom is stable and does not spiral into the nucleus. Light is emitted only when the electron makes a transition, jumping from a higher level $n_2$ to a lower level $n_1$. Energy must be conserved, so the energy the atom loses reappears as a single photon, a lump of light whose energy is exactly the gap between the two rungs. A definite energy gap means a definite photon energy, which means a definite frequency and a definite wavelength — a single sharp spectral line. Run the same jump backwards — the atom absorbs a photon of exactly that energy and the electron climbs from $n_1$ to $n_2$ — and you get a dark absorption line at the identical wavelength. The spectrum is discrete for one reason only: the ladder is discrete.

Everything begins with the one formula the Bohr model earns for hydrogen. For a single electron bound to a single proton, the allowed total energies are

where: $n$ is the principal quantum number (a positive whole number), and $13.6$ eV is the ground-state binding energy of hydrogen, a number Bohr built out of the electron mass, its charge, Planck's constant and the permittivity of free space. The physical reading is threefold. The energy is quantised — only $n = 1, 2, 3, \dots$ are permitted, nothing between. The energy is negative — the electron is trapped in the proton's electrical well, and zero is reserved for a free electron at rest. And the spacing shrinks like $1/n^2$: consecutive rungs get closer and closer, converging on $E = 0$. Put in the numbers and the ladder appears: $E_1 = -13.6$ eV, $E_2 = -13.6/4 = -3.4$ eV, $E_3 = -13.6/9 = -1.51$ eV, $E_4 = -13.6/16 = -0.85$ eV, $E_5 = -13.6/25 = -0.54$ eV. The gap from $n=1$ to $n=2$ is $10.2$ eV — a colossal jump that lands in the ultraviolet. From $n=2$ to $n=3$ the gap is only $1.89$ eV, a visible-light photon. Higher up the gaps become tiny. This uneven spacing is why hydrogen's three families of lines live in three different parts of the spectrum, as we will see. The top of the ladder, $E = 0$ at $n \to \infty$, is where the electron is just barely free. The energy required to lift the electron from the ground state all the way to freedom is the ionisation energy of hydrogen: $E_\infty - E_1 = 0 - (-13.6) = 13.6$ eV. This is one of the most-quoted numbers in atomic physics, and it drops straight out of equation (1). Ionise from an excited state and it costs less — from $n=2$ only $3.4$ eV, from $n=3$ only $1.51$ eV — because those electrons are already part-way out of the well.

An atom radiates only when its electron falls from a higher rung $n_2$ down to a lower rung $n_1$. Conservation of energy fixes the photon exactly: the atom loses $E_{n_2} - E_{n_1}$, and that energy leaves as one photon. Writing the drop out with equation (1),

where: $n_1 < n_2$, so the bracket is positive and $\Delta E > 0$ — the emitted photon carries real, positive energy. Because $n_1$ is the lower level, note the order inside the bracket: $1/n_1^2$ (the bigger term) comes first. This is the single most-tested piece of bookkeeping in the topic. The same $\Delta E$ read as an absorption tells you the exact photon a ground-or-excited atom will swallow to climb from $n_1$ to $n_2$. A photon of energy $\Delta E$ has a wavelength you get from the photon shortcut $E[\mathrm{eV}] = 1240/\lambda[\mathrm{nm}]$, so

where: $1240$ eV·nm is the product $hc$ in convenient units. Bigger energy drops make shorter (bluer) wavelengths; the modest drops high on the ladder make long (redder or infrared) light. As the test of the whole scheme, take the $3 \to 2$ jump: $\Delta E = 13.6\left(1/4 - 1/9\right) = 13.6 \times 0.1389 = 1.89$ eV, so $\lambda = 1240/1.89 = 656$ nm — the famous red H$\alpha$ line, the deep red glow of every hydrogen discharge tube and of the Sun's chromosphere. It matches Ångström's 1868 measurement to the digit.

Sort every possible jump by the level it lands on, and the whole spectrum falls into families. Each family is named after the person who first identified it as a series, and each occupies its own band of the spectrum because the landing level fixes how big the energy drops can be. The Lyman series is every jump ending on $n_1 = 1$ (that is, $2\to1, 3\to1, 4\to1, \dots$). These land on the deepest rung, so the drops are the largest — $10.2$ eV or more — and every Lyman line is in the ultraviolet. The first line, Lyman-$\alpha$ ($2\to1$), has $\Delta E = 13.6(1 - 1/4) = 10.2$ eV and $\lambda = 122$ nm. The Balmer series is every jump ending on $n_1 = 2$ ($3\to2, 4\to2, 5\to2, \dots$). These are the only hydrogen lines the human eye can see, which is exactly why this series was the first to be recognised and fitted to a formula, by Balmer in 1885 (using Ångström's 1868 measurements). H$\alpha$ ($3\to2$) is $656$ nm red; H$\beta$ ($4\to2$) is $486$ nm blue-green; H$\gamma$ ($5\to2$) is $434$ nm violet; H$\delta$ ($6\to2$) is $410$ nm deep violet. The Paschen series ends on $n_1 = 3$ ($4\to3, 5\to3, \dots$). The rungs above $n=3$ are close together, so the drops are small and every Paschen line is in the infrared. Each series has a series limit — the shortest wavelength in the family, reached when the electron falls from $n_2 = \infty$ (a just-free electron) down to the series' landing level. That is the biggest possible drop for the family. For Balmer, the limit is $\Delta E = 13.6(1/4 - 0) = 3.4$ eV, $\lambda = 365$ nm — just into the ultraviolet, which is why the visible Balmer lines bunch up and stop near $410$ nm. The Lyman limit is $91.2$ nm and the Paschen limit is $821$ nm. Beyond the limit the spectrum becomes continuous, because a free electron with any leftover kinetic energy can be captured, contributing a photon of any energy above the limit.

Divide equation (2) by $hc$ and you convert an energy statement into the wavelength statement that spectroscopists actually use. Since photon energy $\Delta E = hc/\lambda$, we have $1/\lambda = \Delta E / hc$, and equation (2) becomes

where: $R$ is the Rydberg constant, one of the most precisely measured numbers in all of physics. Equation (4) is Rydberg's 1888 generalisation of Balmer's formula, and it predicts every line hydrogen can emit from the two integers $n_1$ and $n_2$ alone. The reading is that the reciprocal wavelength (the "wavenumber") of any hydrogen line is a difference of two terms $R/n^2$, one for each level — spectra are made of differences, which is the deep hint Bohr chased. That $R$ is nothing more than the ionisation energy $13.6$ eV divided by $hc$ is the whole point: the Bohr ladder and the Rydberg formula are the same physics wearing two different unit systems.

Board level — the first two Balmer lines and the series limit (CBSE and ICSE pattern)

For atomic hydrogen take $E_n = -13.6/n^2$ eV and the photon shortcut $\lambda = 1240/\Delta E$ nm. Find the wavelength of (a) the H$\beta$ line, the $4\to2$ Balmer transition, and (b) the Balmer series limit, and state the region of the spectrum each lies in.

Step 1. Energy of the $4\to2$ jump. Using $\Delta E = 13.6\left(1/n_1^2 - 1/n_2^2\right)$ with $n_1 = 2, n_2 = 4$: $\Delta E = 13.6\left(1/4 - 1/16\right) = 13.6\left(0.250 - 0.0625\right) = 13.6\times0.1875 = 2.55$ eV.

Step 2. Its wavelength: $\lambda = 1240/2.55 = \boxed{486\ \mathrm{nm}}$ — a blue-green line in the visible. This is H$\beta$, the second Balmer line.

Step 3. The series limit is the largest Balmer drop, from $n_2 = \infty$ to $n_1 = 2$: $\Delta E = 13.6\left(1/4 - 0\right) = 3.40$ eV.

Step 4. Its wavelength: $\lambda = 1240/3.40 = \boxed{365\ \mathrm{nm}}$ — just past the violet edge, in the near ultraviolet. Every Balmer line lies between H$\alpha$ ($656$ nm) and this $365$ nm limit, bunching up toward the limit.

Board level — ionisation energy from the ground state and from an excited state (CBSE and NEET pattern)

Using $E_n = -13.6/n^2$ eV, find the energy needed to ionise a hydrogen atom (a) from its ground state and (b) from the first excited state. (c) What wavelength photon would ionise a ground-state atom exactly?

Step 1. Ionisation means lifting the electron to $E = 0$ (free, at rest). From the ground state $n=1$: energy needed $= 0 - E_1 = 0 - (-13.6) = \boxed{13.6\ \mathrm{eV}}$.

Step 2. From the first excited state $n=2$, where $E_2 = -13.6/4 = -3.4$ eV: energy needed $= 0 - (-3.4) = \boxed{3.4\ \mathrm{eV}}$. An already-excited atom is far cheaper to ionise, because the electron is part-way out of the well.

Step 3. A photon that just ionises the ground state carries exactly $13.6$ eV: $\lambda = 1240/13.6 = \boxed{91.2\ \mathrm{nm}}$ — deep ultraviolet, and exactly the Lyman series limit, since ionisation from $n=1$ is the $\infty\to1$ "jump".

Competitive level — how many lines and which is reddest (JEE and NEET pattern)

A sample of atomic hydrogen is excited so that every atom reaches the $n = 4$ level. As the atoms return to the ground state, (a) how many distinct spectral lines appear, (b) which transition gives the longest wavelength, and (c) what is that wavelength?

Step 1. Count the lines. From a level $n$, the number of distinct downward transitions (and hence distinct lines) is $n(n-1)/2$. For $n = 4$: $4\times3/2 = \boxed{6}$ lines. They are $4\to3, 4\to2, 4\to1, 3\to2, 3\to1, 2\to1$.

Step 2. Longest wavelength means smallest energy drop. Scanning the six, the smallest gap is between the two closest rungs in play, $4\to3$: $\Delta E = 13.6\left(1/9 - 1/16\right) = 13.6\times0.04861 = 0.661$ eV.

Step 3. That wavelength: $\lambda = 1240/0.661 = \boxed{1876\ \mathrm{nm}}$ — deep infrared (this is the first Paschen line). The shortest wavelength, by contrast, is the biggest drop $4\to1$, a Lyman line at $\lambda = 1240/[13.6(1 - 1/16)] = 1240/12.75 = 97.3$ nm in the ultraviolet. Timing note: about 90 seconds. The line count is instant once you know $n(n-1)/2$; the trap is (b) — reach for "smallest energy gap = longest wavelength" and pick the two adjacent top rungs $4\to3$, not the huge $4\to1$ drop. Do not compute all six wavelengths; you only need the extremes.

Competitive level — a hydrogen-like ion and the Z-squared scaling (JEE Advanced pattern)

Singly ionised helium (He⁺) has one electron orbiting a nucleus of charge $Z = 2$. Its energy levels are $E_n = -13.6\,Z^2/n^2$ eV. Find (a) the ground-state energy and ionisation energy of He⁺, and (b) the wavelength of its "H$\alpha$-analogue" $3\to2$ line, comparing with hydrogen's $656$ nm.

Step 1. Ground state: $E_1 = -13.6\times2^2/1^2 = -13.6\times4 = \boxed{-54.4\ \mathrm{eV}}$. The ionisation energy of He⁺ is therefore $54.4$ eV — four times hydrogen's, because every level is pulled down by the factor $Z^2 = 4$.

Step 2. The $3\to2$ transition energy scales the same way: $\Delta E = 13.6\,Z^2\left(1/4 - 1/9\right) = 4\times1.89 = 7.56$ eV, four times the hydrogen H$\alpha$ energy.

Step 3. Its wavelength: $\lambda = 1240/7.56 = \boxed{164\ \mathrm{nm}}$. Equivalently, since $\lambda \propto 1/\Delta E \propto 1/Z^2$, just divide hydrogen's $656$ nm by $Z^2 = 4$: $656/4 = 164$ nm. The helium line sits at a quarter of the wavelength, deep in the ultraviolet. Timing note: about 60 seconds. Do not redo the whole Rydberg calculation — recognise that every hydrogen-like energy and every transition energy scales as $Z^2$, so every wavelength scales as $1/Z^2$. One division, $656/4$, gets you the answer.

Which integer names the series. The series is fixed by the lower level $n_1$, the one the electron falls to: $n_1 = 1$ Lyman (UV), $n_1 = 2$ Balmer (visible), $n_1 = 3$ Paschen (IR). Students routinely name the series after the starting level and lose the mark. On CBSE, ICSE, JEE and NEET this is asked almost every year; it appears on AP Physics 2, IB HL and A-Level as well, but the named-series taxonomy is background there rather than a routine exam item.

The order of terms in the bracket. Always $13.6\left(1/n_1^2 - 1/n_2^2\right)$ with the smaller integer first, so the answer is positive. Writing it the other way round gives a negative $\Delta E$ and a negative wavelength — physically impossible, and an instant sign-error red flag.

Ionisation energy is not the ground-state energy's value with the minus sign ignored by accident. It equals $0 - E_n$. From the ground state that is $+13.6$ eV; from $n=2$ it is $+3.4$ eV, not $13.6$ eV. Always ask "ionise from which level?" — many marks are lost by assuming ground state when the question specifies an excited one.

Series limit means $n_2 = \infty$, not $n_2 =$ the next integer. The limit is the family's shortest wavelength, reached in the $\infty \to n_1$ transition. Balmer limit $365$ nm, Lyman $91.2$ nm, Paschen $821$ nm. Do not confuse the limit with the first line of the series.

Wavelength inequality flips against energy. Bigger energy drop means shorter wavelength. So Lyman lines (biggest drops) are the shortest-wavelength UV, and Paschen lines (smallest drops) the longest-wavelength IR. Sanity-check every answer: a "656 nm Lyman line" is nonsense.

Board scope. The hydrogen energy levels, the Bohr-model spectrum, the named Lyman/Balmer/Paschen series and the Rydberg formula are examined quantitatively on CBSE, ICSE, JEE Main, JEE Advanced and NEET. AP Physics 2, IB HL (Theme E) and A-Level (CIE 9702) carry the energy-level ladder and line spectra (the photon relation $hf = E_2 - E_1$), but treat the Rydberg formula and the named-series taxonomy as background rather than examined content. IB Standard Level carries line spectra qualitatively but not the Bohr-model maths, and IGCSE 0625 has none of this quantitative content — an IGCSE student should not rely on this page for their syllabus.

Texts. NCERT Physics Class 12, Chapter 12 ("Atoms"), the sections on the hydrogen spectrum, energy levels and the Rydberg formula · H. C. Verma, Concepts of Physics Vol. 2, the "Bohr's Model and Physics of the Atom" chapter · Halliday, Resnick and Walker, "All About Atoms", the hydrogen-spectrum section.

Provenance. Locked hydrogen anchors. Recompute: $E_2 = -13.6/4 = -3.4$ eV, $E_3 = -13.6/9 = -1.51$ eV, $E_4 = -0.85$ eV, $E_5 = -0.54$ eV; H$\alpha$ ($3\to2$) $\Delta E = 1.89$ eV, $\lambda = 656$ nm; H$\beta$ ($4\to2$) $486$ nm; H$\gamma$ ($5\to2$) $434$ nm; H$\delta$ ($6\to2$) $410$ nm; Balmer limit $365$ nm; Lyman-$\alpha$ ($2\to1$) $10.2$ eV, $122$ nm; Lyman limit $91.2$ nm; Paschen limit $821$ nm; $4\to3$ first Paschen line $0.661$ eV, $1876$ nm; $4\to1$ line $97.3$ nm; ionisation from ground $13.6$ eV, from $n=2$ is $3.4$ eV; $R = 1.097\times10^7$ m⁻¹ $= 13.6\ \mathrm{eV}/hc$; Balmer $B = 4/R = 3646$ Å; number of lines from level $n$ is $n(n-1)/2$ (six from $n=4$, three from $n=3$); He⁺ ground state $-54.4$ eV, $3\to2$ line $164$ nm.

Prerequisites. The Bohr Model of the Atom — where the quantised orbits, the angular-momentum rule and the formula $E_n = -13.6/n^2$ are derived. This article takes that ladder as given and reads the spectrum off it. Leads to. the university seams The Hydrogen Spectrum Series and The Hydrogen Energy Spectrum and the Rydberg Constant (go-deeper enrichment, never an exam dependency here), which redo this with the full Schrödinger treatment — including the fine structure the Bohr ladder cannot show.