Bohr model of hydrogen

Rutherford's nuclear atom told students where the mass and the positive charge of an atom live — packed into a tiny central nucleus, with electrons somewhere outside it. What it could not tell anyone was why the atom holds together at all. A classical electron circling a classical nucleus is an accelerating charge, and an accelerating charge must radiate electromagnetic energy continuously. Do the arithmetic and a classical hydrogen atom should collapse — the electron spiralling into the nucleus — in a fraction of a nanosecond. Real hydrogen atoms plainly do not do this; they sit there, stable, for as long as the universe has existed, and when disturbed they emit light only at a few sharp, reproducible colours rather than a continuous rainbow. In 1913 Niels Bohr cut through the contradiction not by abandoning classical mechanics but by grafting two new rules onto it — rules borrowed from the still-new idea of quantisation that Planck and Einstein had used for radiation. Bohr proposed that the electron is only permitted to occupy certain special orbits, and that it can jump between them by emitting or absorbing a photon of exactly the right energy. Nothing about the electron's motion inside an allowed orbit needed to change; the trick was in restricting which orbits exist. This article rebuilds that reasoning step by step: the three postulates, the resulting formulas for the size, speed and energy of each orbit, and how those numbers explain both atomic stability and the discrete colours of the hydrogen spectrum. Every formula here is the one your CBSE Class 12 board exam will test — quoted and applied with confidence, not re-derived from scratch line by line.

Key takeaways

Bohr rescued Rutherford's nuclear atom from classical collapse by proposing that electrons occupy only certain stationary orbits in which they do not radiate, even while accelerating.
Angular momentum in a stationary orbit is quantised: $L = mvr = \dfrac{nh}{2\pi}$ , with $n = 1, 2, 3, \dots$ the principal quantum number.
The radius of the $n$ th orbit is $r_n = n^2 a_0$ , where $a_0 = 0.529$ Å is the Bohr radius (the $n=1$ orbit of hydrogen).
Orbital speed falls off as $v_n \propto 1/n$ — outer electrons move more slowly, not faster.
Orbital energy is $E_n = -\dfrac{13.6}{n^2}$ eV; it is always negative (bound), and its magnitude shrinks as $n$ grows, reaching zero (a free electron) only as $n \to \infty$ .
A photon is emitted or absorbed only when an electron jumps between two allowed orbits, with $h\nu = E_i - E_f$ — this is why hydrogen's spectrum is a set of sharp lines, not a smear of colour.
Key idea: Bohr quantised the electron's angular momentum in circular orbits ( $L = nh/2\pi$ ), which fixes a discrete ladder of radii ( $r_n = n^2a_0$ ) and energies ( $E_n = -13.6/n^2$ eV), and photons appear only when the electron jumps between rungs of that ladder.

The core insight

Quantising the electron's angular momentum as $L = nh/2\pi$ forces hydrogen's orbits, speeds and energies onto a discrete ladder ( $r_n = n^2a_0$ , $E_n = -13.6/n^2$ eV), and a photon is emitted or absorbed only when the electron jumps between two rungs of that ladder, $h\nu = E_i - E_f$ .

Where it fits in the syllabus

The Bohr model is the centrepiece of the CBSE Class 12 Physics chapter "Atoms" (Unit IX, Atoms and Nuclei), and it is one of the highest-yield derivation-plus-numerical topics in the whole board paper. The syllabus expects you to state all three postulates, and to use — not re-derive from scratch — the standard results $r_n = n^2h^2\varepsilon_0/(\pi m e^2) = n^2a_0$ , $v_n \propto 1/n$ , and $E_n = -13.6/n^2$ eV for numerical problems on hydrogen and hydrogen-like ions. Hydrogen's line spectra (Lyman, Balmer and the rest) are explicitly a qualitative topic on this syllabus — you should know that such series exist and roughly why, but the full Rydberg-formula numerics for series-line wavelengths sit outside the 2025-26 prescription and are not built into this article's spine.

The physical picture

Bohr's model rests on three postulates, laid one on top of another. Take them in order, because each is a patch on a specific classical failure.

Postulate 1 — the nuclear atom (inherited from Rutherford). The atom has a tiny, massive, positively charged nucleus at its centre, with an electron held in a circular orbit around it by the ordinary electrostatic (Coulomb) force of attraction — nothing new here, this is Rutherford's scattering result carried over unchanged.

Postulate 2 — the stationary-orbit postulate. Of all the circular orbits a classical electron could in principle occupy, only a special discrete set is actually allowed — the stationary orbits — and while the electron sits in one of these, it does not radiate energy, even though it is accelerating. What singles out the allowed orbits is a quantisation condition on angular momentum:

L = mvr = \frac{nh}{2\pi}, \qquad n = 1, 2, 3, \dots

where: $m$ is the electron mass, $v$ its orbital speed, $r$ the orbit radius, $h$ is Planck's constant, and $n$ is a positive integer called the principal quantum number, labelling which orbit the electron occupies. The physical reading is blunt but effective: angular momentum, which classically could take any value, is here forced onto an evenly spaced ladder of allowed values ( $h/2\pi$ , $2h/2\pi$ , $3h/2\pi$ , …). Only orbits whose angular momentum lands exactly on a rung of that ladder are permitted to exist; everything in between is simply not available to the electron.

Postulate 3 — the frequency postulate. An atom emits electromagnetic radiation only when an electron makes a transition (a jump) from a stationary orbit of higher energy $E_i$ down to one of lower energy $E_f$ , and the emitted photon's frequency $\nu$ satisfies

h\nu = E_i - E_f

where: $h$ is Planck's constant and $E_i$ , $E_f$ are the electron's total energies in the initial and final orbits. Read physically, this says the atom cannot leak energy continuously — it can only pay out energy in fixed parcels, exactly equal to the gap between two rungs of its allowed-orbit ladder. Running the same equation in reverse describes absorption: an electron jumps from a lower orbit to a higher one only if it absorbs a photon whose energy exactly matches that gap; a photon of the wrong energy is simply not absorbed by that transition. Combining postulates 1 and 2 — balancing the Coulomb force against the centripetal requirement of circular motion, then substituting the quantised angular momentum — fixes the radius, speed and energy of every allowed orbit in hydrogen ( $Z = 1$ ) in terms of $n$ alone:

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

v_n \propto \frac{1}{n}

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

where: $\varepsilon_0$ is the permittivity of free space, $e$ is the electron's charge, $a_0$ is the Bohr radius — the radius of the smallest ( $n=1$ ) orbit, $0.529$ Å — and $E_n$ is the total (kinetic plus potential) energy of the electron in the $n$ th orbit. Three physical readings matter here. First, $r_n \propto n^2$ : orbits get rapidly larger as $n$ climbs — the $n=2$ orbit is four times the area of $n=1$ , not merely double. Second, $v_n \propto 1/n$ : contrary to a common intuition, electrons in outer orbits move slower, because a weaker net inward pull is needed to hold a wider, more slowly turning orbit in balance. Third, and most important for numerical work, $E_n$ carries a minus sign — the electron is bound to the nucleus, so its energy is lower than that of a free electron at rest (taken as the zero of energy). The magnitude of $E_n$ shrinks as $n$ grows, and $E_n \to 0$ only in the limit $n \to \infty$ , corresponding to a fully freed, ionised electron. The $n=1$ state is the ground state, the atom's lowest and most tightly bound configuration, with $E_1 = -13.6$ eV for hydrogen; every $n>1$ state is an excited state. The energy required to remove the electron completely from the ground state — to carry it from $n=1$ out to $n=\infty$ — is the ionisation energy, exactly $13.6$ eV for hydrogen. For a hydrogen-like ion — a single electron orbiting a nucleus of charge $+Ze$ , such as He⁺ ( $Z=2$ ) or Li²⁺ ( $Z=3$ ) — the stronger nuclear pull scales the same formulas: $r_n \propto n^2/Z$ (orbits shrink for a more highly charged nucleus) and $E_n = -\dfrac{13.6\,Z^2}{n^2}$ eV (orbits bind more tightly). These scaled formulas are fair game for board and competitive numericals alike.

Figure

A ladder of horizontal energy levels for hydrogen labelled n equals 1 through n equals infinity with values minus 13.6, minus 3.4, minus 1.5 and 0 electron volts, and a downward arrow between two levels marked emitted photon

Derivation AFrom Coulomb force and quantised angular momentum to the hydrogen energy formula

The goal of this derivation is to show, using only the balance of forces in circular motion and the single quantisation rule $L = nh/2\pi$ , how Bohr's two postulates squeeze out a definite formula for the energy of the $n$ th orbit, $E_n = -13.6/n^2$ eV — turning an abstract quantisation rule into a concrete, testable number. CBSE does not require reproducing every algebraic step in an exam answer, but seeing the chain once makes the final formula far easier to trust and to apply correctly.

Step 1. Set up the force balance. For the electron to move in a stable circular orbit of radius $r$ around a nucleus of charge $+e$ , the Coulomb attraction must supply exactly the centripetal force needed for that circular path: $\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2} = \dfrac{mv^2}{r}$ . This is pure classical mechanics — nothing quantised yet, just "the electrostatic pull equals the centripetal requirement."

Step 2. Bring in the quantisation postulate. Bohr's second postulate fixes the angular momentum as $mvr = \dfrac{nh}{2\pi}$ , so $v = \dfrac{nh}{2\pi mr}$ . Substituting this expression for $v$ into the force-balance equation from Step 1 and solving for $r$ eliminates the unknown speed entirely, leaving the orbit radius written purely in terms of $n$ and fundamental constants: $r_n = \dfrac{n^2h^2\varepsilon_0}{\pi m e^2}$ . This is exactly the radius formula quoted above, and it shows explicitly why $r_n$ grows as $n^2$ — the quantisation condition, squared through the algebra, is what produces that power.

Step 3. Assemble the total energy. The electron's total energy is kinetic plus electrostatic potential energy, $E = \tfrac{1}{2}mv^2 - \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r}$ . Using the force-balance relation from Step 1, the kinetic term simplifies to $\tfrac{1}{2}mv^2 = \dfrac{1}{8\pi\varepsilon_0}\dfrac{e^2}{r}$ — exactly half the magnitude of the potential term and opposite in its effective sign, a standard feature of any inverse-square orbit. Adding the two gives $E = -\dfrac{1}{8\pi\varepsilon_0}\dfrac{e^2}{r}$ : the total energy is negative, and equal in magnitude to half the potential energy.

Step 4. Substitute the quantised radius. Plugging $r_n$ from Step 2 into the energy expression from Step 3 and collecting the constants ( $m$ , $e$ , $h$ , $\varepsilon_0$ ) into a single number gives $E_n = -\dfrac{me^4}{8\varepsilon_0^2h^2}\dfrac{1}{n^2}$ . Evaluating that constant numerically for hydrogen turns it into the familiar $E_n = -13.6/n^2$ eV — the same formula quoted and used throughout this article, now traced back to its two starting postulates rather than simply asserted.

Worked examples

Board level — energy of the third excited state and the emitted photon

Question: Calculate the energy of an electron in the third excited state of a hydrogen atom, and then find the energy of the photon emitted when this electron falls to the first excited state.

Step 1. Translate the words into a value of $n$ . The ground state is $n=1$ ; counting up, the first excited state is $n=2$ , the second excited state is $n=3$ , and the third excited state is $n=4$ — a common place to lose a mark is reading "third excited state" as $n=3$ instead of $n=4$ .

Step 2. Compute the energy of $n=4$ using $E_n = -13.6/n^2$ eV: $E_4 = -13.6/16 = -0.85$ eV.

Step 3. Identify the final state for the transition. "First excited state" is $n=2$ , with $E_2 = -13.6/4 = -3.40$ eV.

Step 4. The photon energy equals the energy the electron loses, higher level minus lower level: $E_{\text{photon}} = E_4 - E_2 = (-0.85) - (-3.40) = 2.55$ eV.

Answer: The third excited state has energy $E_4 = \mathbf{-0.85\ eV}$ , and the electron's fall from $n=4$ to $n=2$ emits a photon of energy $\mathbf{2.55\ eV}$ — this transition belongs to the Balmer series, since it ends on $n=2$ .

Competitive level — comparing hydrogen with a hydrogen-like ion

Question: An electron in a hydrogen atom falls from $n=3$ to $n=2$ , emitting a photon of energy $1.89$ eV. Find the energy of the corresponding $n=3$ to $n=2$ transition in a doubly ionised lithium ion, Li²⁺ ( $Z=3$ ), and compare the two photon energies.

Step 1. Recall the hydrogen result (Pass-1 numbers, unchanged): $E_3 = -13.6/9 = -1.51$ eV and $E_2 = -13.6/4 = -3.40$ eV, so $E_{\text{photon}} = E_3 - E_2 = 1.89$ eV for hydrogen — matches the value quoted in the question, confirming the setup.

Step 2. Scale to the hydrogen-like ion. For a nucleus of charge $Z$ , $E_n = -13.6\,Z^2/n^2$ eV, so every energy level for Li²⁺ ( $Z=3$ , hence $Z^2=9$ ) is simply $9$ times the corresponding hydrogen level: $E_3(\text{Li}^{2+}) = 9 \times (-1.51) = -13.6$ eV and $E_2(\text{Li}^{2+}) = 9 \times (-3.40) = -30.6$ eV.

Step 3. Find the Li²⁺ photon energy: $E_{\text{photon}} = E_3 - E_2 = (-13.6) - (-30.6) = 17.0$ eV.

Step 4. Compare: since every energy level scales by $Z^2 = 9$ , the energy gap between any fixed pair of levels also scales by $Z^2$ , so $E_{\text{photon}}(\text{Li}^{2+}) = 9 \times 1.89 = 17.0$ eV — consistent with Step 3, and a useful shortcut for future problems: same- $n$ -pair transitions in a hydrogen-like ion of charge $Z$ are always $Z^2$ times the hydrogen photon energy.

Answer: The Li²⁺ ( $Z=3$ ) $n=3\to n=2$ transition emits a photon of $\mathbf{17.0\ eV}$ — nine times the $1.89$ eV hydrogen photon, exactly the $Z^2$ scaling. Timing note: in a competitive-exam setting, do not recompute $E_3$ and $E_2$ for Li²⁺ from scratch — recognising that every hydrogen-like energy level (and hence every same-transition photon energy) scales by $Z^2$ against hydrogen saves a full round of arithmetic and is exactly the kind of shortcut that separates a 90-second solve from a 30-second one.

Common traps

Reading " $k$ th excited state" as $n=k$ instead of $n=k+1$ — the ground state already uses up $n=1$ , so the first excited state is $n=2$ , the second is $n=3$ , and so on; this single misreading silently shifts every energy value in a problem.
Writing or leaving $E_n$ as a positive number — a bound electron's total energy is always negative relative to a free electron at rest (the chosen zero); only the magnitude of $E_n$ shrinks as $n$ increases, it never becomes a large positive quantity.
Confusing which direction of transition emits versus absorbs a photon — falling from a higher $n$ to a lower $n$ (more tightly bound) releases a photon; climbing from a lower $n$ to a higher $n$ requires absorbing a photon of exactly matching energy, not any photon that happens to arrive.
Assuming outer orbits mean faster electrons — $v_n \propto 1/n$ means orbital speed decreases as $n$ increases; it is the radius, not the speed, that grows with $n$ .
Applying the hydrogen formulas unmodified to a multi-electron atom — the Bohr formulas above are derived for a single electron orbiting a bare nuclear charge and hold exactly only for hydrogen and hydrogen-like single-electron ions (He⁺, Li²⁺, …), not for atoms with more than one electron, where electron-electron repulsion breaks the simple picture.
Forgetting the $Z^2$ scaling (or the $1/Z$ scaling on radius) when a question switches from hydrogen to a hydrogen-like ion — every energy level, and hence every transition energy, must be multiplied by $Z^2$ ; the radius, in contrast, is divided by $Z$ .

Common trapthe electron does not radiate while orbiting, and it does not spiral inward — this is a postulate, not a classical result

The single deepest misconception to avoid is thinking that Bohr somehow proved, from classical electromagnetism, that an orbiting electron is stable. He did not, and could not — classical electromagnetism flatly predicts that any accelerating charge radiates continuously and spirals into the nucleus almost instantly. Bohr's stationary-orbit postulate is an assumption, imposed by hand on top of classical mechanics precisely because it contradicts a piece of classical physics (Maxwell's radiation law) that Bohr chose to override for orbits satisfying the angular-momentum quantisation condition. The model works — its numbers for $r_n$ , $v_n$ and $E_n$ match hydrogen's observed spectrum extremely well — but it works by postulating away the classical instability, not by resolving it from first principles. A fully first-principles explanation of why bound electrons do not radiate away their energy needed the later, more complete framework of quantum mechanics, which lies beyond this syllabus; here, treat "no radiation in a stationary orbit" as a rule you are given, not a conclusion you derive.

You should be able to…

State Bohr's three postulates in your own words: the nuclear atom, the quantised non-radiating stationary orbits ( $L = nh/2\pi$ ), and the frequency condition for emission and absorption ( $h\nu = E_i - E_f$ ).
Compute $r_n$ , $v_n$ , and $E_n$ for hydrogen at any given $n$ , and correctly identify the ground state, an excited state, and the ionisation energy from the energy formula.
Compute the energy (and, in principle, the frequency via $h\nu$ ) of a photon emitted or absorbed for a stated transition between two values of $n$ .
Extend every hydrogen formula to a hydrogen-like ion of atomic number $Z$ , including the $Z^2$ scaling of energy and the $1/Z$ scaling of radius.
Explain, qualitatively, why hydrogen's emission spectrum is a set of sharp discrete lines rather than a continuous smear of colour, and why this was impossible to explain before Bohr's quantisation postulate.
Explain why the Bohr model is a postulated patch on classical physics rather than a first-principles proof of atomic stability, and why it applies exactly only to one-electron systems.

Ask the tutor: want to practise computing $r_n$ , $v_n$ , and $E_n$ together for a few different values of $n$ , or work through more $Z^2$ -scaling problems comparing hydrogen with He⁺ or Li²⁺?

If the electron is accelerating in its circular orbit, why doesn't classical physics make it lose energy and crash into the nucleus, the way Bohr's model claims it doesn't?

Classical electromagnetism really does predict exactly that collapse for an ordinary accelerating charge, and that prediction is correct for things like electrons oscillating in an antenna. Bohr's stationary-orbit postulate is a deliberate, explicit override of that classical prediction, applied only to orbits whose angular momentum satisfies $L = nh/2\pi$ . It is not something Bohr derived from Maxwell's equations — it is an extra rule bolted on, justified only because the atoms and spectra it predicts match what we actually observe. The real, from-first-principles reason bound electrons in real atoms do not radiate away their energy comes from full quantum mechanics, which describes the electron as a standing wave rather than a tiny orbiting ball — a topic beyond this syllabus.

Why is the energy negative? What does a negative energy even mean physically?

The zero of energy is a choice, and Bohr's model (like most bound-system physics) chooses it to be the energy of an electron completely removed from the atom and left at rest, infinitely far from the nucleus, with zero kinetic energy. Any electron that is actually bound to the nucleus has less energy than that free, at-rest state, because energy had to be released (as a photon, historically, when the atom first formed) to pull the electron down into orbit. "Less than zero" is therefore just "less than the energy of a free electron at rest" — a negative number is the natural way to record that the electron is in a lower-energy, bound configuration. The more negative $E_n$ is, the more tightly the electron is held, and the more energy you would have to supply to free it.

Does the Bohr model work for atoms other than hydrogen, like helium or carbon?

Only in a very limited sense — for hydrogen-like ions with exactly one electron orbiting a bare nucleus of charge $+Ze$ , such as He⁺ or Li²⁺, the same formulas apply with $Z$ inserted, and they work well. For any atom with two or more electrons — neutral helium onward — the model as taught here breaks down, because the formulas were derived assuming only one electron feels only the nucleus's pull. In a real multi-electron atom, each electron also repels every other electron, and that extra interaction is not captured by Bohr's simple circular-orbit picture at all. Multi-electron atoms need the far more complete (and mathematically heavier) framework of quantum mechanics, which is out of reach for this syllabus.

The syllabus says hydrogen's line spectra are only qualitative — so do I need to memorise the Lyman and Balmer series formulas and wavelength numbers?

No, not for this topic as scoped here. You are expected to know, qualitatively, that hydrogen emits light only at specific discrete wavelengths (not a continuous spectrum), that this happens because electrons jump between the discrete energy levels the Bohr model predicts, and that these jumps group into named series (such as the Balmer series lying partly in visible light, as in the first worked example above). The full quantitative Rydberg formula $1/\lambda = R(1/n_1^2 - 1/n_2^2)$ with numerical series-line wavelengths sits outside the 2025-26 CBSE prescription for this chapter — you should recognise the idea, not compute wavelengths from that formula for a board answer.