Bohr Model — Radius and Energy of the nth Orbit

By Aniket Bhardwaj · 23 September 2026 · Maths & Physics

The Bohr model is wrong. Every chemistry teacher says so, and they are right — it fails the moment an atom has two electrons. Yet it stays in the Class 11 syllabus, in NEET, in JEE and in IIT-JAM, for one very good reason: for hydrogen and for any ion with a single electron, it gives exactly the right energies. Those energies explain the hydrogen spectrum line by line. So the sensible approach is to learn its two formulas properly, learn where they apply, and learn where they collapse.

The two formulas that matter

r_n = 0.529 × (n² / Z) Å = 52.9 × (n² / Z) pm
E_n = −13.6 × (Z² / n²) eV
v_n = 2.18 × 10⁶ × (Z / n) m/s
m v r = n h / 2π (Bohr's quantisation condition)

Symbol	Meaning	Notes
n	Principal quantum number of the orbit	n = 1, 2, 3, … — a whole number, never zero, never a fraction
Z	Atomic number (nuclear charge)	H: Z = 1 · He⁺: Z = 2 · Li²⁺: Z = 3
r_n	Radius of the nth orbit	0.529 Å is the Bohr radius, a₀
E_n	Total energy of the electron in that orbit	Always negative — the electron is bound

The two n-dependences run in opposite directions, and that pairing is the whole story of the model. Radius grows as n²: the fourth orbit is sixteen times wider than the first. Energy rises as −1/n²: the levels crowd closer and closer together as they climb towards zero. Far from the nucleus, the electron is barely held at all.

Why the energy is negative

Zero energy is defined as the electron sitting infinitely far from the nucleus, at rest and completely free. Anything bound to the nucleus has less energy than that, so its energy is below zero. A more negative number therefore means a more tightly held electron. E₁ = −13.6 eV for hydrogen is the most negative level there is, which is why n = 1 is the ground state.

Worked example 1 — hydrogen, third orbit

Question: For a hydrogen atom in the n = 3 orbit, find the radius, the energy and the electron's speed.

Hydrogen, so Z = 1, and n = 3, so n² = 9.

Radius: r₃ = 0.529 × 9 ÷ 1 = 4.761 Å = 476.1 pm

Energy: E₃ = −13.6 × 1 ÷ 9 = −1.5111 ≈ −1.51 eV

Speed: v₃ = 2.18 × 10⁶ × 1 ÷ 3 = 7.27 × 10⁵ m/s

Check with the quantisation condition. The angular momentum should equal 3h/2π. Using m_e = 9.11 × 10⁻³¹ kg, r = 4.761 × 10⁻¹⁰ m:
m v r = (9.11 × 10⁻³¹)(7.27 × 10⁵)(4.761 × 10⁻¹⁰) = 3.153 × 10⁻³⁴ J·s
3h / 2π = 3 × (6.626 × 10⁻³⁴) ÷ 6.2832 = 3.164 × 10⁻³⁴ J·s. ✔ (agreeing to the rounding in the three-figure constants)

Worked example 2 — He⁺, and a neat coincidence

Question: Find the radius and energy of the n = 2 orbit of He⁺ and compare them with hydrogen's ground state.

He⁺ has one electron and Z = 2.

Radius: r₂ = 0.529 × (2²) ÷ 2 = 0.529 × 4 ÷ 2 = 1.058 Å, which is exactly twice hydrogen's 0.529 Å.

Energy: E₂ = −13.6 × (2²) ÷ (2²) = −13.6 × 4 ÷ 4 = −13.6 eV, which is exactly hydrogen's ground-state energy.

This is not a fluke. Whenever n/Z is the same, the energy is the same, because E depends on (Z/n)². He⁺ at n = 2 and H at n = 1 both have Z/n = 1. Examiners like this pairing precisely because it looks like a coincidence and is not.

Warning: these formulas apply to He⁺ but not to neutral helium. The moment there are two electrons, they repel each other, the model has no way to include that, and every number it produces is wrong.

Worked example 3 — the red hydrogen line

Question: An electron in a hydrogen atom falls from n = 3 to n = 2. Find the energy released and the wavelength of the emitted light.

Energy of each level:
E₃ = −13.6 ÷ 9 = −1.5111 eV
E₂ = −13.6 ÷ 4 = −3.4000 eV

Energy released = E₃ − E₂ = −1.5111 − (−3.4000) = 1.8889 eV

Wavelength, using E(eV) = 1240 / λ(nm):
λ = 1240 ÷ 1.8889 = 656.4 nm — deep red, and this is the Hα line every hydrogen discharge tube shows.

Check with the Rydberg formula, which is an entirely separate route:
1/λ = R(1/n₁² − 1/n₂²) with R = 1.097 × 10⁷ m⁻¹
1/λ = 1.097 × 10⁷ × (1/4 − 1/9)
1/4 = 0.25000, 1/9 = 0.11111, difference = 0.13889
1/λ = 1.097 × 10⁷ × 0.13889 = 1.5236 × 10⁶ m⁻¹
λ = 1 ÷ (1.5236 × 10⁶) = 6.563 × 10⁻⁷ m = 656.3 nm. ✔

Two independent formulas, agreeing to one part in a thousand. That agreement is the strongest evidence a Class 11 student ever sees that the model, for hydrogen at least, is genuinely describing something real.

Worked example 4 — ionisation energy, in eV and in kJ/mol

Question: Find the ionisation energy of a hydrogen atom in its ground state, in eV and in kJ/mol. Then do the same for He⁺.

Ionising means moving the electron from n = 1 out to n = ∞, where E = 0.

Hydrogen: IE = E_∞ − E₁ = 0 − (−13.6) = 13.6 eV

Per mole: 1 eV per particle = (1.602 × 10⁻¹⁹ J) × (6.022 × 10²³ mol⁻¹) = 96,470 J/mol = 96.47 kJ/mol.
IE = 13.6 × 96.47 = 1312 kJ/mol

That figure, 1312 kJ/mol, is the value printed in every data book for hydrogen's first ionisation enthalpy — a measured quantity that the model reproduces from nothing but n = 1 and Z = 1.

He⁺: E₁ = −13.6 × (2²) ÷ 1 = −54.4 eV, so IE = 54.4 eV = 54.4 × 96.47 = 5248 kJ/mol.
Four times hydrogen's, because Z² = 4. Pulling an electron off a doubly charged nucleus is much harder, which is exactly what you would expect.

Where the model breaks — state this in the theory answer

Only one-electron species. H, He⁺, Li²⁺, Be³⁺. Nothing else. Electron–electron repulsion is not in the model.
No fine structure. Spectral lines that are really several close lines appear as one.
No explanation of intensities — it says which lines exist, not how bright they are.
It contradicts the uncertainty principle. A fixed orbit means knowing position and momentum together, which quantum mechanics forbids. Orbits were replaced by orbitals: regions of probability, not paths.
No account of chemical bonding, the Zeeman effect or the Stark effect.

None of this makes the arithmetic above wrong for hydrogen. It makes it limited, which is a different thing, and saying so precisely earns marks in the theory question.

Common mistakes that cost marks

Dropping the minus sign on E_n. A positive energy means a free electron. Writing E₂ = 3.4 eV instead of −3.4 eV reverses the physics of the answer.
Putting Z in the wrong power. Radius has Z in the denominator to the first power; energy has Z² in the numerator. Mixing them up is the most frequent error in He⁺ and Li²⁺ questions.
Using these formulas for multi-electron atoms. Applying −13.6 Z²/n² to sodium or to neutral helium gives a number that means nothing.
Sign confusion in transitions. Energy is emitted when the electron falls to a lower n and absorbed when it climbs. Take the magnitude for the photon, then say in words which way it went.
Forgetting the Å to metre conversion. 1 Å = 10⁻¹⁰ m = 100 pm. Feeding Ångströms into an SI formula is a factor-of-10¹⁰ error.
Treating n as continuous. There is no n = 2.5 orbit. That whole-number restriction is the quantisation.

Where this appears in exams

Exam	Typical use
CBSE/ICSE Class 11 Chemistry	Structure of Atom — radius, energy, transitions, spectral series, limitations of the model
CBSE Class 12 Physics	Atoms — Bohr's postulates, hydrogen spectrum, derivation of r_n and E_n
NEET	Single-step numericals on radius ratios, energy ratios and ionisation energy of hydrogen-like ions
JEE Main & Advanced	Multi-step transitions, photon momentum, recoil, and questions comparing two different Z values
IIT-JAM / CSIR-NET	As the starting point before the true quantum-mechanical treatment of the hydrogen atom

Check your working in seconds. The Bohr Model tool takes n and Z and returns the orbit radius, the energy level and the electron velocity, so you can confirm a transition energy before converting it to a wavelength.

Open the Bohr Model Calculator →

Structure of Atom is the chapter where Class 11 chemistry suddenly starts demanding physics, and it is where many students first fall behind. ABC Chemistry runs Class 11–12 coaching at the Gurugram centre and online classes across India — details at abcchemistry.in.