Bohr's postulates

Rutherford's nuclear atom had a fatal flaw. Classical electrodynamics insists that any accelerating charge radiates energy continuously, so an electron circling a nucleus should lose energy every instant, spiral inward, and crash into it in a fraction of a second — atoms should not exist, and the light they emit should form a smooth rainbow, not the sharp discrete lines actually seen in a spectrometer. In 1913 Niels Bohr rescued the model with three bold postulates that simply forbid the classical breakdown for a special set of orbits. This article states those three postulates precisely, derives the standard formulas for the radius, velocity and energy of the nn-th orbit of hydrogen, and works through the two calculations you are expected to reproduce in an ISC exam: an orbit's size and energy, and the wavelength of a photon emitted in a transition.

The core insight

Bohr postulated that an electron can occupy only those orbits where its angular momentum is an integer multiple of h/2πh/2\pi, and it neither radiates while in such a stationary orbit nor loses energy, emitting a photon of energy hν=Ei−Efh\nu = E_i - E_f only when it jumps between two such orbits.

The core idea

Bohr kept Rutherford's picture of a small, heavy, positively charged nucleus with electrons orbiting it under electrostatic attraction, but overruled classical physics with three assumptions, applied to an electron of charge −e-e orbiting a nucleus of charge +Ze+Ze (for hydrogen, Z=1Z = 1).

Postulate 1 — stationary orbits. An electron can revolve only in certain permitted orbits, called stationary orbits, in which it does not radiate energy even though it is undergoing centripetal acceleration. In such an orbit the Coulomb force of attraction supplies exactly the centripetal force needed to keep the electron circling:

14πε0Ze2r2=mv2r\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r^2} = \dfrac{mv^2}{r}

where: ε0\varepsilon_0 is the permittivity of free space, ee is the electronic charge, ZZ is the atomic number of the nucleus, rr is the orbit radius, mm is the electron mass, and vv is its orbital speed. This equation alone is just Newtonian mechanics; it says nothing about why only certain radii are allowed.

Postulate 2 — quantisation of angular momentum. That missing restriction is Bohr's genuinely new physics: the angular momentum of the electron in a stationary orbit can take only the discrete values

mvr=nh2π,n=1,2,3,…mvr = \dfrac{nh}{2\pi}, \qquad n = 1, 2, 3, \dots

where: hh is Planck's constant and nn, the principal quantum number, labels which orbit the electron sits in. There is no classical justification for this rule — it is imposed to match experiment, and its deeper origin (de Broglie's matter waves fitting a whole number of wavelengths around the orbit) came only a decade later.

Postulate 3 — the frequency condition. An atom radiates energy only when an electron jumps from a higher-energy stationary orbit to a lower one, and the photon carries away exactly the energy difference:

hν=Ei−Efh\nu = E_i - E_f

where: ν\nu is the frequency of the emitted photon and EiE_i, EfE_f are the total energies of the initial (higher) and final (lower) orbits. The reverse jump, from a lower to a higher orbit, requires absorbing a photon of the same energy. While the electron sits in any one stationary orbit, postulate 1 keeps it stable and no radiation occurs at all — this is what fixes Rutherford's collapse problem, and postulate 3 is what produces sharp spectral lines instead of a continuous glow, since only photons of energy exactly Ei−EfE_i - E_f for some pair of allowed orbits can ever be emitted.

Figure
Bohr atom diagram showing electron orbits n equals 1 2 3 around a nucleus with an arrow marking a downward jump from orbit 3 to orbit 1 emitting a photon
Stationary orbits and a quantum jump n=3 to n=1 nucleus +Ze n=1 n=2 n=3 photon emitted hv = E3 - E1 No radiation while orbiting; a photon is emitted only on a jump between orbits.

Combining postulates 1 and 2 by eliminating vv gives the derived orbit quantities for hydrogen-like atoms, shown in full in the derivation below:

rn=n2h2ε0πmZe2,vn=Ze22ε0nh,En=−14πε0Ze22rnr_n = \dfrac{n^2h^2\varepsilon_0}{\pi mZe^2}, \qquad v_n = \dfrac{Ze^2}{2\varepsilon_0 nh}, \qquad E_n = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{2r_n}

For hydrogen (Z=1Z = 1), substituting the constants collapses this to the compact and exam-famous result

En=−13.6n2 eVE_n = -\dfrac{13.6}{n^2}\ \text{eV}

The total energy is negative because the electron is bound; as nn grows, rn∝n2r_n \propto n^2 increases and En→0E_n \to 0, the limit of a free (ionised) electron.

Derivation AOrbit radius and energy of the n-th Bohr orbit

The goal is to combine the centripetal-force condition of postulate 1 with the angular-momentum quantisation of postulate 2 to eliminate the unmeasurable speed vv and obtain the orbit radius rnr_n purely in terms of the quantum number nn and fundamental constants, then use that radius to get the total energy EnE_n — the two formulas every worked problem in this topic is built on.

Step 1. Start from the two postulate equations for an electron of charge −e-e orbiting a nucleus of charge +Ze+Ze: the force balance 14πε0Ze2r2=mv2r\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r^2} = \dfrac{mv^2}{r} from postulate 1, and the quantisation condition mvr=nh2πmvr = \dfrac{nh}{2\pi} from postulate 2. The first has two unknowns, vv and rr; the second lets us eliminate one of them.

Step 2. Solve the quantisation condition for vv: v=nh2πmrv = \dfrac{nh}{2\pi mr}. Substitute this into the force balance in place of vv. Squaring vv gives v2=n2h24π2m2r2v^2 = \dfrac{n^2h^2}{4\pi^2m^2r^2}, so the force equation becomes 14πε0Ze2r2=mr⋅n2h24π2m2r2\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r^2} = \dfrac{m}{r}\cdot\dfrac{n^2h^2}{4\pi^2m^2r^2}.

Step 3. Multiply both sides by r2r^2 and simplify. The left side becomes Ze24πε0\dfrac{Ze^2}{4\pi\varepsilon_0}, and the right side becomes n2h24π2mr\dfrac{n^2h^2}{4\pi^2mr}. Cross-multiplying to isolate rr gives r=rn=n2h2ε0πmZe2r = r_n = \dfrac{n^2h^2\varepsilon_0}{\pi mZe^2}. Every rr has been replaced by the single allowed radius rnr_n, showing that only orbits with r∝n2r \propto n^2 are permitted — the orbits are quantised, exactly as postulate 2 demands.

Step 4. Find the energy. Kinetic energy is KEn=12mvn2KE_n = \tfrac12mv_n^2, which from the force-balance equation equals 14πε0Ze22rn\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{2r_n}. Potential energy of the electron-nucleus pair is PEn=−14πε0Ze2rnPE_n = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r_n}, which is exactly −2KEn-2KE_n — the potential energy is negative and twice the magnitude of the kinetic energy. Adding them, En=KEn+PEn=14πε0Ze22rn−14πε0Ze2rn=−14πε0Ze22rnE_n = KE_n + PE_n = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{2r_n} - \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r_n} = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{2r_n}, a negative total energy confirming the electron is bound to the nucleus. Substituting rnr_n from Step 3 and the values of the fundamental constants for hydrogen (Z=1Z=1) reduces this to the standard En=−13.6/n2E_n = -13.6/n^2 eV used in every worked problem below.

Worked example

An electron in a hydrogen atom drops from the n=3n = 3 level to the n=2n = 2 level, emitting a photon. Given E1=−13.6E_1 = -13.6 eV, find (a) the radius of the n=2n = 2 orbit using r1=0.529×10−10r_1 = 0.529\times10^{-10} m, and (b) the wavelength of the emitted photon.

Step 1. Find E2E_2 and E3E_3 using En=E1/n2E_n = E_1/n^2: E2=−13.6/4=−3.4E_2 = -13.6/4 = -3.4 eV, and E3=−13.6/9=−1.51E_3 = -13.6/9 = -1.51 eV.

Step 2. Find the n=2n = 2 orbit radius using rn=r1n2r_n = r_1 n^2 (from postulates 1 and 2 combined, derivation above): r2=0.529×10−10×4=2.12×10−10r_2 = 0.529\times10^{-10}\times4 = 2.12\times10^{-10} m.

Step 3. Apply postulate 3 to get the photon energy: ΔE=E3−E1=−1.51−(−3.4)=1.89\Delta E = E_3 - E_1 = -1.51-(-3.4) = 1.89 eV. Converting to joules, ΔE=1.89×1.6×10−19=3.02×10−19\Delta E = 1.89\times1.6\times10^{-19} = 3.02\times10^{-19} J.

Step 4. Use hν=ΔEh\nu = \Delta E and c=νλc = \nu\lambda to get the wavelength directly: λ=hcΔE=(6.63×10−34)(3×108)3.02×10−19=6.59×10−7 m=659 nm\lambda = \dfrac{hc}{\Delta E} = \dfrac{(6.63\times10^{-34})(3\times10^{8})}{3.02\times10^{-19}} = 6.59\times10^{-7}\ \text{m} = 659\ \text{nm}.

The n=2n=2 orbit has radius 2.12×10−102.12\times10^{-10} m, and the 3→23 \to 2 transition emits a photon of wavelength ≈659\approx 659 nm, in the visible red region — this is HαH_\alpha, the first line of the Balmer series.

Competitive stretch

For a hydrogen-like ion with Z=2Z = 2 (singly ionised helium, He+^+), every energy level scales as En=−13.6Z2/n2=−54.4/n2E_n = -13.6Z^2/n^2 = -54.4/n^2 eV and every radius scales as rn=r1(hydrogen) n2/Zr_n = r_1(\text{hydrogen})\, n^2/Z. So the He+^+ ground state (n=1n=1) has E1=−54.4E_1 = -54.4 eV and r1=0.529×10−10/2=0.265×10−10r_1 = 0.529\times10^{-10}/2 = 0.265\times10^{-10} m — a more tightly bound, smaller orbit than hydrogen because the doubled nuclear charge pulls harder.

Common traps

  • Writing the angular momentum condition as mvr=nhmvr = nh instead of mvr=nh/2πmvr = nh/2\pi — the 2π2\pi is the single most commonly dropped factor in this topic.
  • Forgetting the sign convention: PEnPE_n is negative and twice the magnitude of KEnKE_n, so En=KEn+PEnE_n = KE_n + PE_n comes out negative, not the sum of two positive quantities.
  • Confusing "energy released on de-excitation" with "energy absorbed on excitation" — always take ΔE\Delta E as (energy of the higher level) minus (energy of the lower level), and check whether the transition is upward (absorption) or downward (emission) before assigning a sign to the photon.
  • Using En=−13.6/n2E_n = -13.6/n^2 eV for a non-hydrogen ion without the Z2Z^2 scaling factor — that compact form is valid only for Z=1Z = 1; a hydrogen-like ion needs En=−13.6Z2/n2E_n = -13.6Z^2/n^2 eV.

Ask the tutor: Want to practice deriving rnr_n from scratch, or work through another energy-level transition to check your formula-scaling instincts?

Why does postulate 1 not violate classical electromagnetism if the electron is still accelerating?

It does violate the classical prediction, and Bohr knew it — that is precisely the point of stating it as a postulate rather than deriving it. Classical electrodynamics genuinely says any accelerating charge must radiate, but atoms are observably stable, so Bohr simply asserted that this classical rule fails for a special discrete set of orbits fixed by the angular-momentum condition in postulate 2. The justification is that the resulting formulas correctly predict the hydrogen spectrum; a proper theoretical reason (the electron's wave nature) only came later with de Broglie and full quantum mechanics.

Why is potential energy negative and exactly twice the magnitude of kinetic energy?

Potential energy is negative because it is defined as zero at infinite separation and the electron is bound closer in, in an attractive Coulomb field, so pulling it away from the nucleus requires adding energy — meaning its current state has less than zero. The exact factor of two, PEn=−2KEnPE_n = -2KE_n, is not a coincidence; it falls straight out of the force-balance equation of postulate 1, which fixes KEn=12mvn2KE_n = \tfrac12mv_n^2 to equal exactly half the magnitude of PEnPE_n for any circular Coulomb orbit. Adding them gives En=−KEnE_n = -KE_n, a compact way to remember the total energy once you know the kinetic energy alone.

Do Bohr's postulates work for atoms with more than one electron?

No, not accurately. The postulates were built and verified for a single electron orbiting a nucleus — hydrogen and hydrogen-like ions such as He+^+ or Li2+^{2+}, where En=−13.6Z2/n2E_n = -13.6Z^2/n^2 eV holds well. In multi-electron atoms the electrons repel each other as well as being attracted to the nucleus, an interaction Bohr's simple model has no way to include, so the formulas from this article do not extend to atoms like helium or lithium in their neutral form. That limitation is one of the reasons Bohr's model was eventually superseded by full quantum mechanics, though it remains an excellent and syllabus-correct approximation for the single-electron case.

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