Electron orbits

Rutherford's nuclear atom had a fatal flaw. Classical electromagnetism insists that any accelerating charge radiates energy, and an electron circling a nucleus is accelerating every instant — so it should spiral inward, losing energy continuously, and crash into the nucleus in a fraction of a second. Atoms plainly do not do this. Every hydrogen atom in the universe is stable, and every one of them, when excited, glows at exactly the same handful of colours rather than a smear of every possible wavelength. Something was stopping the electron from occupying just any orbit at any energy — and classical physics had no idea what. Niels Bohr's answer, in 1913, was audacious: throw out one piece of classical physics and keep the rest. An electron, he proposed, can move only in certain special circular paths — electron orbits — chosen not by continuous freedom but by a strict quantisation rule on angular momentum. While parked in one of these allowed orbits, the electron simply does not radiate, no matter how classical theory insists it must. Only when it jumps between orbits does it exchange energy with the outside world, as a single photon. This one rule is enough to fix, for the first time, exactly how big a hydrogen atom is and how fast its electron moves. It predicts a smallest possible orbit — the Bohr radius, $0.529$ Å — and a whole ladder of larger orbits above it, spaced $1, 4, 9, 16,\dots$ times as wide. It tells you the electron in the outermost orbits crawls compared to the innermost one. And by fixing the orbit, it fixes an energy for each orbit too, setting up the machinery that later explains why hydrogen's light comes out in sharp spectral lines rather than a continuous rainbow. This article builds that ladder of orbits from Bohr's two postulates and shows how to compute radius, speed, and orbit number for hydrogen and hydrogen-like ions.

Key takeaways

Bohr rescued the atom from classical collapse by proposing stationary orbits: special orbits in which an accelerating electron does not radiate energy.
Only orbits obeying the angular momentum quantisation condition $mvr = nh/2\pi$ ( $n = 1,2,3,\dots$ ) are allowed; every other radius is forbidden.
Combining this rule with the Coulomb force acting as the centripetal force gives the orbit radius $r_n = n^2 a_0$ , with Bohr radius $a_0 = 0.529$ Å for hydrogen.
Orbital speed falls as $v_n \propto 1/n$ — electrons in outer orbits move more slowly, not faster.
The energy of the $n$ th orbit is $E_n = -13.6/n^2$ eV for hydrogen; both formulas may be quoted and used without re-deriving them from scratch.
For a hydrogen-like ion of atomic number $Z$ (one electron, charge $+Ze$ nucleus), radii shrink and speeds rise: $r_n \propto n^2/Z$ and $v_n \propto Z/n$ .
Key idea: an electron can occupy only the discrete orbits satisfying $mvr = nh/2\pi$ , and combining that rule with Coulomb's law forces the orbit radius to grow as $n^2$ while the orbital speed falls as $1/n$ , giving hydrogen a smallest stable orbit of radius $0.529$ Å.

The core insight

Bohr's angular-momentum quantisation condition $mvr = nh/2\pi$ , combined with Coulomb attraction supplying the centripetal force, restricts the electron to discrete orbits of radius $r_n = n^2 a_0$ and speed $v_n \propto 1/n$ , so only a countable ladder of orbits — not a continuum — is physically allowed.

Where it fits in the syllabus

This topic is the mathematical heart of the "Bohr model of the hydrogen atom" section inside the CBSE Class 12 Physics chapter Atoms (Unit IX, Atoms and Nuclei), sitting right after Rutherford's scattering experiment and nuclear model. The syllabus asks you to know Bohr's two postulates, the angular momentum condition, and to be able to quote and use the results $r_n = n^2 a_0$ , $v_n \propto 1/n$ , and $E_n = -13.6/n^2$ eV — a full first-principles derivation with every constant is explicitly not the expected spine of the topic, so this article walks the algebra once, lightly, and then leans on the formulas. The quantitative Rydberg-formula treatment of hydrogen's spectral lines belongs to the separate spectra topic and is only touched on qualitatively here.

The physical picture

Bohr kept Rutherford's basic geometry — a tiny, massive, positively charged nucleus with an electron circling it under Coulomb attraction — but overruled classical electrodynamics with two new postulates. The first, the stationary-orbit postulate, simply asserts that certain orbits are radiation-free: the electron can revolve in them indefinitely without losing energy, even though it is constantly accelerating toward the nucleus. The second, the quantisation postulate, is the rule that decides which orbits qualify. It requires the electron's orbital angular momentum to be an integer multiple of $h/2\pi$ :

L = m v r = \frac{nh}{2\pi}, \qquad n = 1, 2, 3, \dots

where: $L$ is orbital angular momentum, $m$ is the electron's mass, $v$ is its orbital speed, $r$ is the orbit radius, $h$ is Planck's constant ( $6.63\times10^{-34}$ J·s), and $n$ — a positive whole number — is the principal quantum number that labels the orbit. Physically this says angular momentum comes only in whole packets of $h/2\pi$ ; a value like $1.5\,h/2\pi$ simply cannot occur. Because $h$ is so tiny, this restriction is utterly unnoticeable for a spinning ball or an orbiting planet — it only becomes the dominant fact of life at atomic size. The second ingredient is ordinary circular-motion mechanics: whatever orbit the electron is in, the Coulomb force of attraction toward the nucleus must exactly supply the centripetal force needed to keep it moving in a circle of that radius,

\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2} = \frac{mv^2}{r}

where: $Z$ is the atomic number of the nucleus (so the nuclear charge is $+Ze$ ; $Z=1$ for hydrogen), $e$ is the electronic charge, and $\varepsilon_0$ is the permittivity of free space. This is the same balance Rutherford's own model would have used — the new physics is entirely in the quantisation condition above, which turns the two unknowns $v$ and $r$ (previously free to take any values at all) into a discrete set indexed by $n$ .

Figure

Three concentric circular orbits around a nucleus labelled n equals 1 2 and 3 with radii growing as 1 4 and 9 times the smallest radius and an arrow showing the electron slowing down as the orbit number increases

Solving the two equations together (eliminating $v$ ) gives the orbit radius as a function of $n$ and $Z$ ; carrying the elimination the other way gives the speed. Both may be quoted directly:

r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2} = \frac{n^2}{Z}\,a_0, \qquad a_0 = 0.529\ \text{Å} = 5.29\times10^{-11}\ \text{m}

v_n = \frac{Z e^2}{2\varepsilon_0 n h} \;\propto\; \frac{Z}{n}

where: $a_0$ , the Bohr radius, is the radius of hydrogen's smallest orbit ( $Z=1$ , $n=1$ ) and serves as the natural length scale of the atom. Physically, $r_n \propto n^2/Z$ says allowed orbits fan out rapidly for higher $n$ (the second orbit is four times as wide as the first, not twice) and shrink for a more highly charged nucleus, while $v_n \propto Z/n$ says the electron moves slower the farther out it sits, but faster if the nucleus pulls harder. Combining these with the mechanical energy of the orbit gives the further quotable result

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

the total energy of the $n$ th orbit for a hydrogen-like ion, negative because the electron is bound to the nucleus; larger $|E_n|$ means more tightly bound. These three formulas — $r_n$ , $v_n$ , $E_n$ — are the working toolkit for every Bohr-model calculation in this syllabus.

Derivation AGetting the orbit radius from Bohr's quantisation rule and Coulomb's law

The goal is to combine Bohr's angular momentum condition with the requirement that the Coulomb force supplies the centripetal force, and eliminate the unknown speed $v$ to leave the orbit radius $r_n$ purely in terms of $n$ and known constants — showing exactly why $r_n$ must grow as $n^2$ rather than in any other way, using nothing beyond algebraic substitution (the syllabus does not expect the full derivation to be reproduced from memory, but seeing it once makes the formula trustworthy rather than a rule to be memorised blindly).

Step 1. Write down the two governing equations. Quantisation gives $mvr = nh/2\pi$ , so $v = \dfrac{nh}{2\pi m r}$ . Force balance gives $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r^2} = \dfrac{mv^2}{r}$ , which rearranges to $v^2 = \dfrac{Ze^2}{4\pi\varepsilon_0 m r}$ .

Step 2. Substitute the first expression for $v$ into the second. Squaring $v = \dfrac{nh}{2\pi m r}$ gives $v^2 = \dfrac{n^2h^2}{4\pi^2 m^2 r^2}$ . Setting this equal to the force-balance expression for $v^2$ : $\dfrac{n^2h^2}{4\pi^2 m^2 r^2} = \dfrac{Ze^2}{4\pi\varepsilon_0 m r}$ .

Step 3. Solve for $r$ . Cross-multiplying and cancelling one power of $r$ and one power of $m$ from both sides leaves $r = \dfrac{n^2h^2 \times 4\pi\varepsilon_0}{4\pi^2 m \times Ze^2} = \dfrac{n^2h^2\varepsilon_0}{\pi m Z e^2}$ . Every quantity on the right except $n$ is a fixed constant of nature ( $h$ , $\varepsilon_0$ , $m$ , $e$ ) or the fixed nuclear charge number $Z$ , so the whole radius scales exactly as $n^2$ .

Step 4. Identify the Bohr radius and check the scaling. Setting $n=1$ and $Z=1$ (hydrogen's innermost orbit) gives $r_1 = \dfrac{h^2\varepsilon_0}{\pi m e^2} = a_0 = 0.529$ Å — this single number, once fixed by the known values of $h$ , $m$ , $e$ , $\varepsilon_0$ , becomes the yardstick for every other orbit, since $r_n = n^2 a_0/Z$ . Doubling $n$ quadruples $r$ ; doubling $Z$ (a more charged nucleus) halves $r$ — both consistent with a more strongly bound, more tightly pulled-in electron for higher $Z$ . Result used throughout this article: $r_n = n^2 a_0/Z$ with $a_0 = 0.529\ \text{Å}$ , obtained by eliminating $v$ between $mvr = nh/2\pi$ and $\dfrac{Ze^2}{4\pi\varepsilon_0 r^2} = \dfrac{mv^2}{r}$ .

Worked examples

Board level - radius and speed of the second Bohr orbit of hydrogen

Question: Find the radius and orbital speed of the second orbit ( $n=2$ ) of a hydrogen atom, given the Bohr radius $a_0 = 0.529$ Å and the first-orbit speed $v_1 = 2.19\times10^6$ m/s.

Step 1. Use $r_n = n^2 a_0$ for hydrogen ( $Z=1$ ). With $n=2$ : $r_2 = (2)^2 \times 0.529\ \text{Å} = 4 \times 0.529\ \text{Å}$ .

Step 2. Calculate the radius: $r_2 = 2.116\ \text{Å} = 2.116\times10^{-10}\ \text{m}$ — four times the ground-state radius, exactly as the $n^2$ scaling predicts.

Step 3. Use the proportionality $v_n \propto 1/n$ , so $v_n = v_1/n$ . With $n=2$ : $v_2 = \dfrac{2.19\times10^6}{2} = 1.095\times10^6\ \text{m/s}$ .

Step 4. Sanity-check the trend: the electron in the wider $n=2$ orbit moves at roughly half the ground-state speed, consistent with $r$ growing while $v$ shrinks as the electron sits farther from the attracting nucleus.

Answer: $r_2 \approx 2.12\times10^{-10}$ m and $v_2 \approx 1.10\times10^6$ m/s.

Competitive level - orbit of a doubly-ionised lithium ion

Question: Li²⁺ (lithium stripped of two electrons, $Z=3$ , one electron left) behaves as a hydrogen-like ion. Find the radius, orbital speed, and orbit energy of its electron in the $n=2$ orbit, given $a_0 = 0.529$ Å, $v_1(\text{H}) = 2.19\times10^6$ m/s (hydrogen's ground-state speed, the reference constant for $v_n$ ), and hydrogen's ground-state energy $-13.6$ eV.

Step 1. Radius: use $r_n = \dfrac{n^2}{Z}a_0$ with $n=2$ , $Z=3$ : $r_2 = \dfrac{(2)^2}{3}\times0.529\ \text{Å} = \dfrac{4}{3}\times0.529\ \text{Å} = 0.7053\ \text{Å}$ .

Step 2. Convert: $r_2 = 0.7053\times10^{-10}\ \text{m} \approx 7.05\times10^{-11}\ \text{m}$ — noticeably smaller than hydrogen's $n=2$ orbit ( $2.116\times10^{-10}$ m), because the triply-charged nucleus pulls the electron in far more tightly.

Step 3. Speed: use $v_n = \dfrac{Z}{n}v_1(\text{H})$ with $Z=3$ , $n=2$ : $v_2 = \dfrac{3}{2}\times2.19\times10^6 = 3.285\times10^6\ \text{m/s}$ — faster than even hydrogen's ground-state electron, because the stronger Coulomb pull demands a higher orbital speed for force balance.

Step 4. Energy: use $E_n = -13.6\dfrac{Z^2}{n^2}\ \text{eV}$ with $Z=3$ , $n=2$ : $E_2 = -13.6\times\dfrac{9}{4} = -13.6\times2.25 = -30.6\ \text{eV}$ — about twice as deeply bound as hydrogen's own ground state ( $-13.6$ eV), despite being a higher orbit number.

Answer: $r_2 \approx 7.05\times10^{-11}$ m, $v_2 \approx 3.29\times10^6$ m/s, $E_2 = -30.6$ eV. Timing note: in a competitive exam, do not re-derive $r_n$ or $v_n$ from Coulomb's law under time pressure — plug straight into $r_n = n^2a_0/Z$ , $v_n = Zv_1/n$ , and $E_n=-13.6Z^2/n^2$ eV, which together should take under ninety seconds for a three-part question like this once the formulas are memorised.

Common traps

Assuming any circular orbit is allowed, the way it would be in pure Rutherford/classical mechanics — only orbits satisfying $mvr = nh/2\pi$ for a whole number $n$ exist; every intermediate radius is simply forbidden by the quantisation rule.
Thinking a stationary orbit means the electron has stopped accelerating — it has not; it is still under constant centripetal acceleration toward the nucleus. Bohr's postulate is that it does not radiate while doing so, a deliberate departure from classical electromagnetism, not a claim about the electron's motion itself.
Mixing up how $r_n$ and $v_n$ scale with $n$ : radius grows as $n^2$ (rapidly, and orbits spread out), while speed falls as $1/n$ (electrons in outer orbits are slower) — these trends run in opposite directions and are the single most common mix-up in exam answers.
Forgetting the $Z$ -dependence for hydrogen-like ions: using hydrogen's $r_n = n^2a_0$ or $v_n \propto 1/n$ formulas directly for He⁺, Li²⁺, or any other multi-proton one-electron ion, instead of $r_n = n^2a_0/Z$ and $v_n \propto Z/n$ .
Treating $E_n = -13.6/n^2$ eV as a positive, increasing quantity — the negative sign matters: as $n$ increases, $E_n$ becomes less negative, meaning the electron is less tightly bound, not more energetic in a loose everyday sense.

Common trapA stationary orbit is not a motionless electron and it is not "allowed because classical physics permits it"

The single biggest misconception is imagining the stationary orbits as ordinary classical orbits that just happen to be picked out. They are not classically permitted at all — an accelerating charge radiating nothing is a flat contradiction of Maxwell's electromagnetism, and Bohr's first postulate is an explicit, deliberate override of that law, justified only by the fact that it works. The electron in a stationary orbit is moving fast, accelerating constantly, and radiating nothing — all three at once — purely because quantum behaviour, not classical behaviour, governs it. Do not try to justify the postulate using classical arguments; it is a rule imposed on top of classical mechanics, not a consequence of it.

You should be able to…

State Bohr's two postulates in your own words: stationary (non-radiating) orbits, and quantised angular momentum $mvr = nh/2\pi$ .
Derive, or at least follow the derivation of, $r_n = n^2a_0/Z$ by eliminating $v$ between the quantisation condition and the Coulomb force-balance equation.
Compute orbit radius, orbital speed, and orbit energy for a given $n$ using $r_n = n^2a_0/Z$ , $v_n \propto Z/n$ , and $E_n = -13.6Z^2/n^2$ eV.
Correctly extend hydrogen's formulas to a hydrogen-like ion (He⁺, Li²⁺, …) by including the atomic number $Z$ .
Explain qualitatively why quantisation restricts the electron to a discrete ladder of orbits instead of the continuous range classical mechanics would allow.
Describe, in words, how radius and orbital speed each change as the orbit number $n$ increases.

Ask the tutor: want a walk-through of how the same quantisation-plus-Coulomb-balance method is later reused to derive the energy formula $E_n = -13.6Z^2/n^2$ eV from scratch — ask the AI tutor to show that algebra next.

Why doesn't the electron just fall into the nucleus if it's constantly accelerating?

In classical electromagnetism it would have to — any accelerating charge is supposed to radiate energy continuously and spiral inward within a tiny fraction of a second, which is exactly the puzzle Bohr set out to fix. His first postulate simply declares that, in the special stationary orbits fixed by the quantisation rule, this classical radiation loss does not happen; the electron keeps a constant radius, speed, and energy indefinitely. It is not that acceleration stops — it is that the radiation consequence of acceleration is switched off while the electron occupies an allowed orbit. This is a postulate, not something derivable from older physics, and that is precisely why it counted as a revolutionary idea in 1913.

Are there really only whole-number orbits, or is that just an approximation?

Within the Bohr model, it is exact: the angular momentum quantisation condition $mvr = nh/2\pi$ only has solutions for whole-number $n$ , so only a countable, discrete set of orbits exists — no orbit with, say, $n = 2.5$ is permitted. The Bohr model itself is a simplified picture that a later, fuller quantum-mechanical treatment refines (with concepts beyond this syllabus), but the basic result that atomic orbits and energies are discrete rather than continuous survives into that fuller theory too. For CBSE Class 12 purposes, treat $n$ strictly as a positive integer $1, 2, 3, \dots$

Why does the orbit radius grow so fast — four times, then nine times, then sixteen times?

It falls straight out of the algebra: the quantisation condition ties $v$ to $n/r$ , while the Coulomb force-balance ties $v^2$ to $1/r$ ; eliminating $v$ between these two relations leaves $r$ multiplied by $n^2$ , not by $n$ . Physically, going to a higher orbit means slowing the electron down (lower $v$ ) at the same time as needing a weaker restoring force (since the Coulomb pull weakens with distance) — both effects push the equilibrium radius outward, and together they compound into the quadratic growth. It is a good habit to remember this comes from squaring $n$ inside the derivation, not from any separate rule.

Do I need to memorise the full derivation of r_n for my board exam?

No — the syllabus explicitly treats the derivation as optional background, not the examinable spine of the topic. What you are expected to know and use confidently are the final results: the quantisation condition $mvr = nh/2\pi$ , and the quoted formulas $r_n = n^2a_0/Z$ , $v_n \propto Z/n$ , and $E_n = -13.6Z^2/n^2$ eV, applied correctly to compute radius, speed, or energy for a given $n$ and $Z$ . Seeing the one-time derivation (as shown above) helps you trust and remember the formulas, but reproducing it is not typically demanded unless a question explicitly asks you to derive $r_n$ .