The Bohr Model of the Atom

Rutherford had just handed physics a beautiful disaster. His 1911 scattering data proved the atom is almost all empty space, with a tiny, dense, positive nucleus and electrons somewhere outside — a miniature solar system. The trouble is that a solar system made of charges cannot exist. An electron circling a nucleus is an accelerating charge, and Maxwell's own electromagnetism says an accelerating charge must radiate. Radiating away its energy, the electron should spiral into the nucleus in about a hundred-billionth of a second, sweeping through every frequency as it falls. Every atom in your body should have collapsed before you finished reading this sentence, emitting a smear of all colours on the way down. Atoms plainly do not collapse, and — even stranger — when they do emit light, they emit it not as a smear but as a handful of razor-sharp, fixed colours: the line spectra of the previous article. Hydrogen's visible glow is four crisp lines, always at the same wavelengths, the same for every hydrogen atom in the universe. Any correct model of the atom has to explain two miracles at once: why atoms are stable, and why they radiate only a discrete menu of wavelengths. In 1913 a 27-year-old Dane named Niels Bohr cut the knot with three bold postulates that flatly contradicted classical physics — and, in return, computed the exact wavelengths of hydrogen's spectrum from nothing but the electron's charge, its mass, Planck's constant, and the speed of light. This article builds his model from those postulates, derives the orbit radius, speed and energy with school algebra, and then watches Rydberg's mysterious empirical constant fall out of it. We also mark honestly where the model breaks.

Key takeaways

The classical catastrophe: a Rutherford atom is unstable — an orbiting electron radiates, spirals in, and would emit a continuous smear, contradicting both stability and the observed sharp lines.
Bohr's three postulates: electrons occupy only certain stationary circular orbits and do not radiate in them; the allowed orbits are those with angular momentum $L = n\hbar = nh/2\pi$ ; light is emitted or absorbed only when an electron jumps between orbits, with $\Delta E = hf$ .
Orbit radius: $r_n = 0.529\,\dfrac{n^2}{Z}\ \text{Å}$ — the ground state of hydrogen sits at the Bohr radius $a_0 = 0.529\ \text{Å}$ ; radii grow as $n^2$ .
Orbit speed: $v_n = \dfrac{2.19\times10^{6}\,Z}{n}\ \mathrm{m\,s^{-1}}$ — the ground-state electron moves at about $1/137$ of the speed of light.
Orbit energy: $E_n = -13.6\,\dfrac{Z^2}{n^2}\ \text{eV}$ — negative (bound), deepest at $n=1$ ( $-13.6\ \text{eV}$ ), climbing toward zero as $n\to\infty$ ; ionising hydrogen from the ground state costs $13.6\ \text{eV}$ .
Bohr's triumph: the energies reproduce the empirical Rydberg constant $R = 1.097\times10^{7}\ \mathrm{m^{-1}}$ and the hydrogen wavelengths (H $\alpha$ $656\ \text{nm}$ ) from first principles.
Key idea: an atom is stable because only certain quantised orbits are allowed, and it radiates a sharp line only when an electron jumps between two of them, the photon carrying exactly the energy gap $\Delta E = hf$ .

The core insight

An electron in an atom may occupy only discrete orbits fixed by the quantisation of angular momentum $L = nh/2\pi$ ; in each such orbit it has a definite radius $r_n \propto n^2/Z$ , speed $v_n \propto Z/n$ and negative energy $E_n = -13.6\,Z^2/n^2\ \text{eV}$ , and it radiates a single sharp photon of energy $hf = E_i - E_f$ only when it jumps from a higher orbit to a lower one.

Where this lives today

The Bohr model is wrong in its picture and right in its arithmetic, which is exactly why it survives. The full quantum atom of Schrödinger (1926) replaced the little circular orbits with three-dimensional probability clouds and needs calculus well beyond this article — we only name it — yet for hydrogen it returns the same energy ladder $-13.6/n^2\ \text{eV}$ that Bohr wrote down. So the numbers on this page are not a historical curiosity; they are the correct hydrogen energies, still used daily. Astronomers read the Balmer and Lyman lines predicted here to measure the composition, temperature and velocity of stars and galaxies — the redshift of hydrogen's lines is how we know the universe is expanding. The $21\ \text{cm}$ line of hydrogen, from an even finer split of the ground state, maps neutral gas across the whole Milky Way. And the electronvolt scale set by $13.6\ \text{eV}$ is the natural currency of all of chemistry, since chemical bonds are made and broken by rearranging exactly these outer electrons. Every gas-discharge lamp, every sodium street light, every neon sign is a Bohr transition made visible.

The physical picture

Keep Rutherford's picture of the atom — a heavy positive nucleus with light electrons around it, held by the electrostatic (Coulomb) attraction that plays the role gravity plays for planets — but strip away the classical assumption that any orbit is allowed. Bohr's radical move was to permit only a discrete set of orbits, like rungs on a ladder, and to declare by decree that an electron sitting in one of these special orbits simply does not radiate, Maxwell notwithstanding. He could not justify the decree from deeper principles; he justified it by its results. (The why — an electron wave that must close on itself around the ring — waited for de Broglie a decade later, and is the next-but-one article's job.) In this picture the hydrogen atom is a single electron on one of a set of nested circular tracks labelled by a whole number $n = 1, 2, 3, \dots$ , the principal quantum number. The innermost track, $n=1$ , is the ground state: the smallest, fastest, most tightly bound orbit, the one the atom settles into and stays in forever if left alone. The outer tracks, $n = 2, 3, \dots$ , are excited states, reached only if the atom is given energy. Crucially the tracks are not evenly spaced: their radii balloon as $n^2$ , so the second orbit is four times the first, the third nine times, and so on.

Figure

The Bohr hydrogen atom — an electron occupies only the quantised orbits n equals one two three whose radii grow as n squared; a jump from an outer orbit to an inner one releases a single photon carrying the exact energy gap

The electron does not radiate while it circles — that is the postulate that buys stability. It radiates only when it jumps from an outer orbit to an inner one, and it absorbs light only when a photon of exactly the right energy lifts it from an inner orbit to an outer one. Because the orbits are discrete, the jumps come in discrete sizes, and so the emitted and absorbed photons come in discrete energies — sharp spectral lines. The barcode of the previous article is nothing but the ladder of allowed orbits, read out in light.

Bohr's three postulates

Everything follows from three statements. Bohr grafted Planck's and Einstein's quantum ideas onto Rutherford's atom and accepted the contradictions with classical physics as the price of a working model.

Postulate 1 — stationary states. An electron in an atom can move only in certain allowed circular orbits, called stationary states, and while in such an orbit it does not radiate energy, even though it is accelerating. This is the flat denial of Maxwell that rescues the atom from collapse.

Postulate 2 — quantised angular momentum. The allowed orbits are exactly those for which the electron's orbital angular momentum is an integer multiple of $h/2\pi$ :

L = m v_n r_n = n\,\frac{h}{2\pi} = n\hbar, \qquad n = 1, 2, 3, \dots \qquad (1)

where: $m = 9.11\times10^{-31}\ \text{kg}$ is the electron mass, $v_n$ and $r_n$ the speed and radius of the $n$ th orbit, $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ Planck's constant, and $\hbar = h/2\pi = 1.05\times10^{-34}\ \mathrm{J\,s}$ the "reduced" Planck constant. The physical reading is that angular momentum, like Planck's energy, comes only in whole lumps of size $\hbar$ ; the integer $n$ is what labels the ladder.

Postulate 3 — the frequency condition. An atom emits or absorbs a photon only when an electron jumps between two stationary states, and the photon's energy equals the difference between the two orbit energies:

hf = E_i - E_f, \qquad f = \frac{E_i - E_f}{h} \qquad (2)

where: $E_i$ and $E_f$ are the energies of the initial and final orbits and $f$ is the emitted (if $E_i > E_f$ ) or absorbed (if $E_i < E_f$ ) light's frequency. This is where the sharp lines come from: fixed energy levels give fixed energy gaps, and fixed gaps give fixed frequencies. Note what postulate 3 does not say — it says nothing about the frequency at which the electron physically circles the nucleus. The light's frequency is set by the gap between orbits, not by the motion within one, a clean break from the classical idea that an orbiting charge radiates at its orbital frequency.

The size, speed and energy of an orbit

Two pieces of physics fix everything. The first is Newton's second law for circular motion: the Coulomb attraction between the nucleus (charge $+Ze$ ) and the electron (charge $-e$ ) supplies exactly the centripetal force needed to bend the electron around its circle. The second is Bohr's quantisation rule, equation (1). Put them together and $r_n$ , $v_n$ and $E_n$ all drop out with nothing harder than algebra. The force balance reads

\frac{k Z e^2}{r_n^{\,2}} = \frac{m v_n^{\,2}}{r_n} \quad\Longrightarrow\quad m v_n^{\,2} = \frac{k Z e^2}{r_n} \qquad (3)

where: $k = 1/4\pi\varepsilon_0 = 8.99\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}$ is the Coulomb constant, $e = 1.60\times10^{-19}\ \text{C}$ the elementary charge, and $Z$ the number of protons in the nucleus ( $Z=1$ for hydrogen). Equation (3) says the kinetic energy $\tfrac12 m v_n^2$ is exactly half the size of the Coulomb potential energy — the virial relation for an inverse-square force — a fact we cash in below. The energy of an orbit is the sum of the electron's kinetic energy and its Coulomb potential energy, $E_n = \tfrac12 m v_n^2 - kZe^2/r_n$ . The potential energy is negative because the electron is bound — you would have to do work to pull it away — and it is this negativity that makes every $E_n$ negative, with $E = 0$ reserved for a free electron infinitely far out and at rest.

Derivation ARadius, speed and energy of the nth Bohr orbit

The goal is to obtain closed formulas for the radius $r_n$ , speed $v_n$ and energy $E_n$ of the $n$ th allowed orbit of a one-electron atom of nuclear charge $Ze$ , using only the two ingredients already on the table — the Coulomb-centripetal force balance of equation (3) and the angular-momentum quantisation of equation (1) — and nothing beyond school algebra. The strategy is to eliminate the speed between the two equations to get the radius, back-substitute for the speed, and finally assemble the energy from the virial relation.

Step 1. From the quantisation rule (1), solve for the speed: $v_n = nh/(2\pi m r_n)$ . Square it: $v_n^2 = n^2 h^2 / (4\pi^2 m^2 r_n^2)$ .

Step 2. Substitute this into the force balance (3), $m v_n^2 = kZe^2/r_n$ , giving $m \cdot n^2 h^2/(4\pi^2 m^2 r_n^2) = kZe^2/r_n$ , that is $n^2 h^2/(4\pi^2 m r_n^2) = kZe^2/r_n$ . Multiply both sides by $r_n^2$ and divide by $kZe^2$ :

r_n = \frac{n^2 h^2}{4\pi^2 m k Z e^2} = \frac{n^2}{Z}\,a_0, \qquad a_0 \equiv \frac{h^2}{4\pi^2 m k e^2} = \frac{\hbar^2}{m k e^2}. \qquad (4)

Putting in the constants, the Bohr radius $a_0 = 0.529\ \text{Å} = 0.529\times10^{-10}\ \text{m}$ . So the radii scale as $n^2/Z$ : for hydrogen $r_1 = 0.529\ \text{Å}$ , $r_2 = 2.12\ \text{Å}$ , $r_3 = 4.76\ \text{Å}$ .

Step 3. Back-substitute $r_n$ into $v_n = nh/(2\pi m r_n)$ . The $n^2$ in $r_n$ cancels one power of $n$ , and one $h$ cancels, leaving

v_n = \frac{2\pi k Z e^2}{n h} = \frac{Z}{n}\,(2.19\times10^{6}\ \mathrm{m\,s^{-1}}). \qquad (5)

The ground-state speed $v_1 = 2.19\times10^{6}\ \mathrm{m\,s^{-1}}$ is about $2.19\times10^6 / 3\times10^8 = 0.0073 = 1/137$ of the speed of light — the famous fine-structure ratio, comfortably non-relativistic, which is why the non-relativistic model works so well for hydrogen.

Step 4. Assemble the energy. By the virial relation from (3), the kinetic energy is $\tfrac12 m v_n^2 = kZe^2/(2r_n)$ , and the potential energy is $-kZe^2/r_n$ , so their sum is

E_n = \frac{kZe^2}{2r_n} - \frac{kZe^2}{r_n} = -\frac{kZe^2}{2r_n} = -\frac{k Z e^2}{2}\cdot\frac{Z}{n^2 a_0} = -\frac{Z^2}{n^2}\,\frac{ke^2}{2a_0}. \qquad (6)

The constant $ke^2/(2a_0)$ evaluates to $2.18\times10^{-18}\ \text{J} = 13.6\ \text{eV}$ , so

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

For hydrogen this gives the ladder $E_1 = -13.6\ \text{eV}$ , $E_2 = -3.40\ \text{eV}$ , $E_3 = -1.51\ \text{eV}$ , $E_4 = -0.85\ \text{eV}$ , climbing toward $0$ . The energy is negative because the electron is bound; the least negative rung ( $n\to\infty$ , $E\to 0$ ) is a barely-free electron. Verified: $a_0 = \hbar^2/(mke^2) = 0.529\ \text{Å}$ , $v_1 = 2\pi ke^2/h = 2.19\times10^6\ \mathrm{m\,s^{-1}} = c/137$ , $ke^2/(2a_0) = 13.6\ \text{eV}$ , and the ladder $-13.6, -3.40, -1.51, -0.85\ \text{eV}$ — recomputed in verify_hs2-bohr-model.py.

Three formulas worth memorising, all for a one-electron (hydrogen-like) atom of nuclear charge $Z$ :

r_n = 0.529\,\frac{n^2}{Z}\ \text{Å}, \qquad v_n = \frac{2.19\times10^{6}\,Z}{n}\ \mathrm{m\,s^{-1}}, \qquad E_n = -\frac{13.6\,Z^2}{n^2}\ \text{eV}. \qquad (7)

Two useful energy relations fall straight out of the virial split in Step 4: the kinetic energy equals the positive of the total energy, $KE_n = +13.6\,Z^2/n^2\ \text{eV} = -E_n$ , and the potential energy is twice the total, $PE_n = -27.2\,Z^2/n^2\ \text{eV} = 2E_n$ . So for ground-state hydrogen the electron has $+13.6\ \text{eV}$ of kinetic energy and $-27.2\ \text{eV}$ of potential energy, summing to $-13.6\ \text{eV}$ .

Bohr's triumph: computing the Rydberg constant

For thirty years spectroscopists had a formula but no reason for it. In 1885 Johann Balmer noticed that hydrogen's four visible lines fit a simple numerical pattern, and in 1888 Johannes Rydberg generalised every hydrogen series into a single equation with one empirical constant $R$ :

\frac{1}{\lambda} = R\left(\frac{1}{n_1^{\,2}} - \frac{1}{n_2^{\,2}}\right), \qquad R = 1.097\times10^{7}\ \mathrm{m^{-1}}, \quad n_2 > n_1. \qquad (8)

Nobody knew where the number $R = 1.097\times10^{7}\ \mathrm{m^{-1}}$ came from — it was measured, not explained. Bohr's model explains it: combine the frequency condition (2) with the energy ladder (6), and Rydberg's constant is forced to be a specific combination of $e$ , $m$ , $h$ and $c$ , whose value the model then predicts to four figures. That agreement, more than any postulate, is what convinced physicists Bohr was onto something.

Figure

The hydrogen energy-level ladder — levels crowd toward zero as one over n squared and every downward jump emits a photon; jumps ending on n equals one form the ultraviolet Lyman series, on n equals two the visible Balmer series, on n equals three the infrared Paschen series

Go deeper · Derivation B · 1 minThe Rydberg constant from e, m, h and c

Derivation BThe Rydberg constant from e, m, h and c

The goal is to show that Bohr's energy ladder reproduces Rydberg's empirical equation (8) and predicts the value of his constant $R$ from fundamental constants alone, using only the frequency condition and the energy formula already derived — no new physics, pure algebra. The strategy is to write the photon energy of a jump $n_2 \to n_1$ two ways and match them.

Step 1. A jump from an upper level $n_2$ to a lower level $n_1$ in hydrogen ( $Z=1$ ) releases a photon whose energy, by the frequency condition (2) and the ladder (6), is $\Delta E = E_{n_2} - E_{n_1} = -13.6/n_2^2 - (-13.6/n_1^2) = 13.6\left(1/n_1^2 - 1/n_2^2\right)$ electronvolts, a positive number since $n_2 > n_1$ .

Step 2. A photon of energy $\Delta E$ has wavelength $\lambda = hc/\Delta E$ , so $1/\lambda = \Delta E/(hc)$ . Using the classroom shortcut $hc = 1240\ \text{eV}\,\cdot\,\text{nm}$ ,

\frac{1}{\lambda} = \frac{\Delta E}{hc} = \frac{13.6}{1240}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\ \text{nm}^{-1}.

Step 3. This already has exactly Rydberg's form (8), with $R$ identified as $R = (13.6\ \text{eV})/(hc)$ . Evaluate: $13.6/1240 = 1.097\times10^{-2}\ \text{nm}^{-1} = 1.097\times10^{7}\ \text{m}^{-1}$ . The empirical Rydberg constant is reproduced to four significant figures — and, tracing $13.6\ \text{eV} = ke^2/2a_0 = 2\pi^2 m k^2 e^4/h^2$ back through the derivation, it is built entirely from $e$ , $m$ , $h$ and $c$ . Bohr computed a number that spectroscopists had only ever measured.

Step 4. Check one line. The Balmer H $\alpha$ line is the jump $n_2=3 \to n_1=2$ : $\Delta E = 13.6(1/4 - 1/9) = 13.6\times0.1389 = 1.89\ \text{eV}$ , so $\lambda = 1240/1.89 = 656\ \text{nm}$ — the deep-red line you see through a spectroscope pointed at a hydrogen lamp, predicted to the nanometre. Verified: $R = 13.6/1240\ \text{nm}^{-1} = 1.097\times10^{7}\ \text{m}^{-1}$ ; H $\alpha$ $\Delta E = 1.89\ \text{eV}$ , $\lambda = 656\ \text{nm}$ — recomputed in verify_hs2-bohr-model.py.

Limiting cases

case	result	meaning
ground state $n=1$	$r_1 = 0.529\ \text{Å}$ , $E_1 = -13.6\ \text{eV}$	smallest, most tightly bound orbit — where the atom rests
$n \to \infty$	$r \to \infty$ , $E \to 0$	electron barely bound then free — the ionisation limit
raise $Z$ (He $^+$ , Li $^{2+}$ )	$r \propto 1/Z$ , $E \propto Z^2$	a heavier nucleus pulls the orbit in tighter and deeper
large $n$	levels crowd, $\Delta E \to 0$	classical continuum recovered — Bohr's correspondence principle
$n_2 \to \infty$ jump to fixed $n_1$	series limit wavelength	shortest wavelength of a spectral series
ionisation from $n$	$\|E_n\| = 13.6\,Z^2/n^2\ \text{eV}$	energy to lift the electron from level $n$ to free

Common trapHigher energy level means less negative energy, not "more energy you can point to"

The commonest slip is with the sign of the energy. The levels are all negative, and "higher up the ladder" means closer to zero, i.e. less negative. So $E_2 = -3.40\ \text{eV}$ is a higher energy than $E_1 = -13.6\ \text{eV}$ , even though $3.40 < 13.6$ — because $-3.40 > -13.6$ . An atom absorbs energy to climb from $n=1$ to $n=2$ and releases energy falling back. The second trap hiding here is treating a Bohr orbit as a literal planetary path with a definite position and velocity: it is a bookkeeping device for the allowed energies, and quantum mechanics later showed the electron has no such sharp trajectory. Use the model for the energies and wavelengths it gets right; do not over-read the little circles.

Worked examples

Board level — the second orbit of hydrogen and the H-alpha line (CBSE and ICSE pattern)

For a hydrogen atom take $a_0 = 0.529\ \text{Å}$ , ground-state energy $-13.6\ \text{eV}$ , and $v_1 = 2.19\times10^{6}\ \mathrm{m\,s^{-1}}$ . Find, for the $n=2$ orbit, (a) its radius, (b) the electron's speed, and (c) its energy. Then (d) find the wavelength of the light emitted when the electron falls from $n=3$ to $n=2$ .

Step 1. Radius scales as $n^2/Z$ with $Z=1$ : $r_2 = 0.529\times 2^2 = 0.529\times 4 = \boxed{2.12\ \text{Å}}$ .

Step 2. Speed scales as $Z/n$ : $v_2 = 2.19\times10^{6}/2 = \boxed{1.10\times10^{6}\ \mathrm{m\,s^{-1}}}$ .

Step 3. Energy scales as $Z^2/n^2$ : $E_2 = -13.6/2^2 = -13.6/4 = \boxed{-3.40\ \text{eV}}$ .

Step 4. The jump $n=3 \to n=2$ releases $\Delta E = E_3 - E_2 = -1.51 - (-3.40) = 1.89\ \text{eV}$ . Its wavelength is $\lambda = 1240/\Delta E = 1240/1.89 = \boxed{656\ \text{nm}}$ , the red Balmer H $\alpha$ line. (Cross-check with the ladder: $E_3 = -13.6/9 = -1.51\ \text{eV}$ .)

Competitive level — the He-plus ion (JEE and NEET pattern)

Singly-ionised helium He $^+$ is hydrogen-like with $Z=2$ . Find (a) the energy needed to ionise it from its ground state, (b) its ground-state orbit radius, and (c) the wavelength of its $n=3 \to n=2$ line. Comment on how these compare with hydrogen.

Step 1. Ground-state energy scales as $Z^2$ : $E_1 = -13.6\times Z^2/1^2 = -13.6\times 4 = -54.4\ \text{eV}$ . Ionisation energy is the energy to reach $E=0$ , so it is $\boxed{54.4\ \text{eV}}$ — four times hydrogen's $13.6\ \text{eV}$ , because a $Z=2$ nucleus binds twice as hard and energy goes as $Z^2$ .

Step 2. Radius scales as $1/Z$ : $r_1 = 0.529/Z = 0.529/2 = \boxed{0.264\ \text{Å}}$ — half of hydrogen's, the orbit pulled in by the stronger pull.

Step 3. The transition energy carries the same $Z^2$ factor: $\Delta E = 13.6\,Z^2(1/2^2 - 1/3^2) = 13.6\times 4\times(0.25 - 0.1111) = 13.6\times 4\times 0.1389 = 7.56\ \text{eV}$ . So $\lambda = 1240/7.56 = \boxed{164\ \text{nm}}$ — the same "H $\alpha$ " transition, but pushed from red $656\ \text{nm}$ into the ultraviolet, shorter by a factor of $Z^2 = 4$ ( $656/4 = 164\ \text{nm}$ ). Timing note: about 90 seconds once the $Z$ -scalings are reflexes — $E\propto Z^2$ , $r\propto 1/Z$ , transition $\lambda \propto 1/Z^2$ . The whole item is three one-line scalings off the hydrogen anchors; the only arithmetic is the final $1240/7.56$ . Spotting that the He $^+$ line is just the hydrogen line divided by four saves the last division entirely.

Competitive level — orbital frequency and the atomic current (JEE Advanced pattern)

In the ground state of hydrogen the electron circles at $v_1 = 2.19\times10^{6}\ \mathrm{m\,s^{-1}}$ on an orbit of radius $r_1 = 0.529\ \text{Å}$ . Find (a) how many times per second it goes round (its orbital frequency), and (b) the effective electric current this circulating charge represents.

Step 1. The orbital frequency is speed over circumference: $f_1 = v_1/(2\pi r_1) = (2.19\times10^{6})/(2\pi\times 0.529\times10^{-10}) = (2.19\times10^{6})/(3.32\times10^{-10}) = \boxed{6.59\times10^{15}\ \text{Hz}}$ . The electron laps the nucleus about six and a half thousand-trillion times a second; equivalently the period is $T = 1/f_1 = 1.52\times10^{-16}\ \text{s}$ .

Step 2. A charge $e$ passing a point $f_1$ times per second is a current $I = e f_1 = (1.60\times10^{-19})(6.59\times10^{15}) = 1.05\times10^{-3}\ \text{A} = \boxed{1.05\ \text{mA}}$ . A single hydrogen atom is a milliamp current loop — which, incidentally, is the origin of its magnetic moment. Timing note: about one minute. Note the orbital frequency $6.59\times10^{15}\ \text{Hz}$ is not the frequency of any emitted light — postulate 3 forbids the confusion. Emitted light comes from the gap between levels; this number is how fast the electron physically circles, a different thing entirely, and a favourite trap.

Exam traps

The sign of the energy levels. Levels are negative and get less negative going up. $E_2 = -3.40\ \text{eV}$ is above $E_1 = -13.6\ \text{eV}$ . Writing " $E_2$ is smaller than $E_1$ " and concluding the atom emits when it goes $1\to 2$ is the classic sign error — climbing the ladder absorbs energy. Appears on CBSE, NEET, JEE, ICSE and A-Level every year.

Orbital frequency is not the emitted-light frequency. Postulate 3 sets the light frequency by $hf = E_i - E_f$ , the gap between orbits — never by how fast the electron circles. A question that computes $v/2\pi r$ and asks for the emitted wavelength is baiting exactly this error.

Getting the $Z$ -scalings backwards. For hydrogen-like ions, energy goes up as $Z^2$ (deeper binding), radius goes down as $1/Z$ , and transition wavelengths go down as $1/Z^2$ . Students routinely put $Z$ on the wrong side. Anchor it to He $^+$ : $E_1 = -54.4\ \text{eV}$ ( $=4\times 13.6$ ), $r_1 = 0.264\ \text{Å}$ ( $=0.529/2$ ).

Confusing excitation with ionisation. Exciting hydrogen from $n=1$ to $n=2$ costs $10.2\ \text{eV}$ ; ionising it from the ground state costs the full $13.6\ \text{eV}$ (to reach $E=0$ ). They are different numbers — do not report $10.2\ \text{eV}$ as the ionisation energy.

Over-claiming the model. Bohr's model is exact only for one-electron systems (H, He $^+$ , Li $^{2+}$ ). It does not predict multi-electron spectra, line intensities, or fine structure. Do not use $-13.6/n^2$ for the levels of helium or lithium atoms — only for their singly-electron ions.

Board scope. The Bohr model and hydrogen energy levels are core on CBSE, ICSE, JEE Main, JEE Advanced, NEET, AP Physics 2, IB HL (Theme E) and A-Level 9702. IB Standard Level carries the line-spectra idea but not the Bohr orbit algebra — so the $r_n$ , $v_n$ derivations here are HL-only for IB. IGCSE 0625 has none of this quantitative content.

Origins · 1 minNiels Bohr and how this was discovered

Niels Bohr

1885–1962 · Copenhagen

Niels Bohr was the Danish physicist who, in a three-part paper of 1913 written soon after working in Rutherford's Manchester laboratory, fused the new quantum idea with Rutherford's nuclear atom to produce the model in this article. By quantising angular momentum and forbidding radiation from stationary states, he explained at one stroke why atoms are stable and why hydrogen emits its sharp Balmer lines, deriving Rydberg's constant from first principles — the calculation that won over a sceptical physics community and earned him the 1922 Nobel Prize. Bohr went on to found his Institute for Theoretical Physics in Copenhagen, a magnet for a whole generation of quantum physicists, and to shape the interpretation of quantum mechanics through his principle of complementarity. His model, though later superseded, remains the doorway through which every student first enters the quantum atom.

History

1885

— Johann Balmer finds an empirical formula fitting hydrogen's four visible lines.

1888

— Johannes Rydberg generalises it to a single constant R covering every hydrogen series.

1911

— Ernest Rutherford's alpha-scattering reveals the tiny nuclear atom — which classical physics says cannot be stable.

1913

— Niels Bohr's three-part paper quantises the orbits, rescues stability, and computes R from e, m and h.

1914

— James Franck and Gustav Hertz measure a sharp 4.9 eV threshold for energy loss when electrons collide with mercury atoms — soon recognised (though they first read it as an ionisation potential) as direct evidence for discrete, Bohr-type energy levels.

1922

— Bohr receives the Nobel Prize in Physics for his work on atomic structure.

Going further

Texts. NCERT Physics Class 12, Chapter 12 ("Atoms"), the Bohr-model and hydrogen-spectrum sections · H. C. Verma, Concepts of Physics Vol. 2, "Bohr's Model and Physics of the Atom" · Halliday, Resnick and Walker, "All About Atoms" · Eisberg and Resnick, Quantum Physics, for the historical route.

Provenance. Standard hydrogen constants. Recompute: $a_0 = \hbar^2/(mke^2) = 0.529\ \text{Å}$ ; $v_1 = 2\pi ke^2/h = 2.19\times10^6\ \mathrm{m\,s^{-1}} = c/137$ ; $ke^2/(2a_0) = 13.6\ \text{eV} = 2.18\times10^{-18}\ \text{J}$ ; ladder $E_2=-3.40$ , $E_3=-1.51$ , $E_4=-0.85\ \text{eV}$ ; radii $r_2=2.12$ , $r_3=4.76\ \text{Å}$ ; $R = 13.6/1240\ \text{nm}^{-1} = 1.097\times10^7\ \text{m}^{-1}$ ; Balmer H $\alpha$ $3\to2$ $=1.89\ \text{eV}=656\ \text{nm}$ , H $\beta$ $4\to2=2.55\ \text{eV}=486\ \text{nm}$ , Lyman $\alpha$ $2\to1=10.2\ \text{eV}=122\ \text{nm}$ , Balmer limit $=365\ \text{nm}$ ; first excitation $10.2\ \text{eV}$ ; He $^+$ $E_1=-54.4\ \text{eV}$ , $r_1=0.264\ \text{Å}$ , $3\to2=164\ \text{nm}$ ; ground-state orbital frequency $f_1 = v_1/2\pi r_1 = 6.59\times10^{15}\ \text{Hz}$ , period $1.52\times10^{-16}\ \text{s}$ , current $I=ef_1=1.05\ \text{mA}$ ; $KE_1=+13.6\ \text{eV}$ , $PE_1=-27.2\ \text{eV}$ .

Prerequisites. Atomic Spectra: Emission & Absorption Lines — the sharp lines this model exists to explain · and Planck's Quantum & the Photon, for $E=hf$ and the quantum of energy that postulate 3 rests on. Leads to. Energy Levels & the Hydrogen Spectrum, which runs the full Lyman–Balmer–Paschen ladder off these energies, and de Broglie's Explanation of Bohr Quantization, which finally supplies the reason for the quantised orbits. For enrichment, the university seam The Bohr Model of Hydrogen (go-deeper only, never an exam dependency here).

Why doesn't the orbiting electron spiral into the nucleus as classical physics predicts?

Classically it should: an accelerating charge radiates energy, so a circling electron should lose energy continuously and spiral in within about $10^{-11}\ \text{s}$ . Bohr's answer was not to explain this away but to forbid it by postulate — he declared that in the special allowed (stationary) orbits the electron simply does not radiate, full stop. It was an admission that classical electromagnetism does not apply inside the atom. The deeper reason came a decade later from de Broglie: the electron behaves as a standing wave that must fit a whole number of wavelengths around the orbit, and a standing wave does not radiate. That is the subject of a later article; here we accept the stability as Bohr's founding assumption.

Why is the energy of the electron negative?

The zero of energy is chosen, by universal convention, to be a free electron at rest infinitely far from the nucleus. Any electron actually bound in the atom has less energy than that free state — you must add energy to free it — so its total energy is negative. Concretely, the electron's kinetic energy is positive ( $+13.6\ \text{eV}$ in the ground state) but its Coulomb potential energy is more negative ( $-27.2\ \text{eV}$ ), and the sum is $-13.6\ \text{eV}$ . The size of that negative number, $13.6\ \text{eV}$ , is precisely the energy you must supply to ionise the atom — to lift the electron from its bound state up to the zero of free space.

What is the difference between exciting and ionising a hydrogen atom?

Exciting means lifting the electron from a lower orbit to a higher but still bound orbit — for example $n=1 \to n=2$ , which costs $E_2 - E_1 = -3.40 - (-13.6) = 10.2\ \text{eV}$ . The atom stays intact; the electron will soon fall back and emit a photon. Ionising means removing the electron entirely, lifting it all the way to $E=0$ and beyond; from the ground state that costs the full $13.6\ \text{eV}$ . A frequent exam trap is to quote the first-excitation energy $10.2\ \text{eV}$ as the ionisation energy — they are different. Ionisation from an excited level $n$ costs only $13.6/n^2\ \text{eV}$ , since the electron starts higher up.

Does the Bohr model work for atoms other than hydrogen?

Only for hydrogen-like atoms — those with a single electron, whatever the nuclear charge: neutral hydrogen ( $Z=1$ ), He $^+$ ( $Z=2$ ), Li $^{2+}$ ( $Z=3$ ), and so on. For these the formulas hold with the $Z$ -scalings $E\propto Z^2$ , $r\propto 1/Z$ . The moment there is a second electron — neutral helium, lithium, or anything heavier — the model fails, because it ignores the repulsion between electrons and their mutual shielding of the nucleus. It cannot predict the spectrum of neutral helium at all. This is one of the model's central limitations and a favourite examiner distinction: use $-13.6\,Z^2/n^2$ for one-electron systems only.

Is the electron really moving in a circle like a tiny planet?

No, and this is the model's biggest conceptual debt. Bohr's circular orbits are a bookkeeping picture that happens to give the right energies for hydrogen, but the full quantum theory (Schrödinger, 1926 — a calculus-level theory we only name here) shows the electron does not have a definite position-and-velocity trajectory at all; it is described by a probability cloud, densest where Bohr's orbit sat but with no sharp path. Trying to pin down both where the electron is and how fast it moves violates Heisenberg's uncertainty principle. So take the orbits as a scaffold for the energy levels — which are genuinely correct — and not as literal little rings. The energies survive the upgrade to modern quantum mechanics; the circular paths do not.

Where does the number 13.6 eV actually come from?

It is the combination $ke^2/(2a_0)$ of fundamental constants, and equivalently $2\pi^2 m k^2 e^4/h^2$ — built from the electron's charge $e$ , its mass $m$ , the Coulomb constant $k$ and Planck's constant $h$ . Numerically it works out to $2.18\times10^{-18}\ \text{J}$ , which is $13.6\ \text{eV}$ . It is simultaneously the ground-state binding energy of hydrogen, its ionisation energy, and (divided by $hc$ ) the Rydberg constant $R$ . That one number ties together the size of the atom, the energy to ionise it, and the wavelengths of its entire spectrum — which is why Bohr's ability to compute it from first principles was so persuasive.

If the Bohr model is superseded, why is it still taught?

Because it is the simplest thing that gets the central facts right. With one integer $n$ and a page of algebra it explains atomic stability, the existence of discrete energy levels, the hydrogen spectrum to four figures, ionisation energy, and the $Z$ -scaling of hydrogen-like ions. The full quantum-mechanical treatment needs calculus and gives, for hydrogen, the same energy ladder $-13.6/n^2\ \text{eV}$ . So the Bohr model is both a faithful stepping-stone and a correct calculator for the one-electron problem. Every physicist meets it first for exactly this reason — and every board exam tests it because its reasoning, not just its answers, is the gateway to quantum physics.