§ 2.4NCERT Class 11 · Chemistry · Chapter 2

Bohr’s Model for Hydrogen Atom: Postulates, Formulas & Numericals

In 1913 Niels Bohr did the unthinkable: he forced Planck’s quanta onto Rutherford’s planet-like atom — and the hydrogen spectrum surrendered every one of its lines. Postulates, the radius–velocity–energy trio, the Rydberg machine, and a live solver for any transition.

QCC Notes · Class 11 Chemistry NCERT & NTA aligned UPDATED 24 SEP 2026 13 MIN READ

−0.0eV

Ground-state energy of hydrogen
The zero sits at n = ∞ (a free electron)
Ionization energy = +13.6 eV — the sign flips, the number stays

What is Bohr’s Model? — Complete Theory

Rutherford’s nuclear atom (§ 2.2) had a fatal flaw: a charged electron circling a nucleus must radiate energy continuously and spiral into the nucleus in about 10⁻¹¹ s. Atoms obviously survive. Bohr’s rescue was to keep Rutherford’s picture but quantize it — borrowing Planck’s idea that energy comes in packets. The model works exactly for hydrogen and H-like ions (He⁺, Li²⁺, Be³⁺ — one electron, any nucleus), and its three postulates are the whole machine:

Postulate 1 · Stationary orbits

Electrons revolve without radiating

The electron moves in certain permitted circular orbits — stationary states — with fixed energy, and while in one, it does not radiate. This single break with classical physics saves the atom from collapse.

Postulate 2 · Quantized angular momentum

Only special orbits are allowed

An orbit exists only if its angular momentum is an integral multiple of h/2π: mvr = nh/2π, n = 1, 2, 3… Quantum numbers enter physics right here — n is the first quantum number.

Postulate 3 · The jump condition

Energy changes only in jumps

Moving between orbits absorbs or emits a photon whose frequency obeys hν = E₂ − E₁. Jump down → emission (a spectral line); jump up → absorption. No in-between orbits, no continuous spectrum.

From these three lines of algebra fall the four formulas that run every numerical in this section. For an electron in the nth orbit of a one-electron species with atomic number Z:

Radiusrₙ = 0.529 n²/Z Å Velocityvₙ = 2.18×10⁶ Z/n m/s EnergyEₙ = −13.6 Z²/n² eV Jumpν̄ = R_HZ²(1/n₁²−1/n₂²)

Table 1 — The hydrogen ladder, level by level (memorise the eV column)
Level	Radius rₙ	Velocity vₙ	Energy Eₙ (J)	Energy (eV)
n = 1	0.529 Å	2.18 × 10⁶ m/s	−2.18 × 10⁻¹⁸	−13.6
n = 2	2.12 Å	1.09 × 10⁶ m/s	−5.45 × 10⁻¹⁹	−3.40
n = 3	4.76 Å	7.27 × 10⁵ m/s	−2.42 × 10⁻¹⁹	−1.51
n = 4	8.47 Å	5.45 × 10⁵ m/s	−1.36 × 10⁻¹⁹	−0.85
n = 5	13.2 Å	4.36 × 10⁵ m/s	−8.72 × 10⁻²⁰	−0.54
n = ∞	∞	0	0	0 — ionization limit

Now the jump condition in its most used costume — the Rydberg equation. For a downward transition from n₂ to n₁ in a H-like species:

Wave numberν̄ = R_H × Z² × (1/n₁² − 1/n₂²) Photon energyE = hν = hc/λ Line countn(n−1)/2

with R_H = 109677 cm⁻¹. Each choice of the lower level n₁ defines a series — the five named families of the hydrogen spectrum. The higher members of each series (n₂ → ∞) crowd together at the “series limit”.

Table 2 — The five spectral series of hydrogen
Series	Lower level n₁	Upper level n₂	Region	First line (n₂ = n₁+1)
Lyman	1	2, 3, 4 …	Ultraviolet	121.6 nm (2→1)
Balmer	2	3, 4, 5 …	Visible	656.3 nm (3→2)
Paschen	3	4, 5, 6 …	Infrared	1875 nm (4→3)
Brackett	4	5, 6, 7 …	Infrared	4051 nm (5→4)
Pfund	5	6, 7, 8 …	Infrared	7458 nm (6→5)

Two counting facts close the theory. When an electron cascades from level n down to the ground state, the maximum number of spectral lines it can produce is n(n−1)/2 — every pair of levels contributes one line (from n = 5: ten lines). And angular momentum in orbit n is n·h/2π: in the second orbit it is h/π, a favourite one-liner. The model’s limits — no fine structure, no Zeeman splitting, silent for multi-electron atoms, and clashing with de Broglie and Heisenberg — are exactly where § 2.5 picks up the story.

Visualising Energy Levels & the Hydrogen Spectrum

Ek seedhi seedhi si seedhi — levels ka ladder, paanch series, aur ek live solver jo har transition ka λ nikaal deta hai.

FIG. 1 — The ladder: each rung is a stationary state with energy −13.6/n² eV, and the rungs crowd together as n grows (that compression is the −1/n² scaling made visible). Each downward arrow is one emitted photon — one spectral line. Arrows landing on the same level (n₁) belong to the same series.

Try it live

Bohr Transition Solver

Species (Z)

From n₂

To n₁

Constants: R_H = 109677 cm⁻¹ · h = 6.626 × 10⁻³⁴ J s · c = 3.0 × 10⁸ m/s · 1 eV = 1.602 × 10⁻¹⁹ J. Diagram energy labels show hydrogen values — multiply by Z² for H-like ions.

Solved Examples (Step-by-Step)

Formula → substitute → unit-check. Jo steps yahan dikhte hain, wahi rough sheet par likhne se full marks milte hain.

EXAMPLE 01Foundation · NEET

The n = 2 trio: radius, velocity, energy

For the electron in the second orbit of hydrogen, calculate the radius, velocity and energy.

Radiusr₂ = 0.529 × 2²/1 = 0.529 × 4 = 2.12 Å — four times the Bohr radius (n² scaling).
Velocityv₂ = 2.18×10⁶ × 1/2 = 1.09 × 10⁶ m/s — half of n = 1 (1/n scaling).
EnergyE₂ = −13.6 × 1/2² = −13.6/4 = −3.40 eV = −5.45 × 10⁻¹⁹ J (1/n² scaling).
VerifyThree formulas, three different scalings — cross-check against Table 1 ✓

2.12 Å · 1.09 × 10⁶ m/s · −3.40 eV

EXAMPLE 02JEE Main · Rydberg

The 4 → 2 line: wavelength and region

Calculate the wavelength of light emitted when an electron in hydrogen falls from n = 4 to n = 2. Name the series and the spectral region. (R_H = 109677 cm⁻¹)

Formulaν̄ = R_H(1/n₁² − 1/n₂²) = 109677 × (1/4 − 1/16)
Substitute= 109677 × 3/16 = 20564 cm⁻¹
Invertλ = 1/20564 = 4.863 × 10⁻⁵ cm = 486.3 nm
ClassifyLands on n₁ = 2 → Balmer series, visible region (blue-green) — verify on the solver above.

λ = 486.3 nm · Balmer · visible

EXAMPLE 03JEE Main · Z² trap

Ionization energies of the H-like family

Calculate the ionization energy of He⁺ and Li²⁺ (both from the ground state), and compare with hydrogen’s 13.6 eV.

FormulaIE = 13.6 × Z²/n² eV — ground state n = 1, so IE = 13.6 Z².
He⁺IE = 13.6 × 2² = 54.4 eV — four times hydrogen.
Li²⁺IE = 13.6 × 3² = 122.4 eV — nine times hydrogen.
WhyNuclear charge binds the lone electron harder — the same Z² that shrinks orbit radii deepens the energy well. E scale ×Z², λ scale ÷Z².

He⁺: 54.4 eV · Li²⁺: 122.4 eV — Z² at work

Key Formulas & Takeaways

Formula card

Eight lines that solve this topic

mvr = nh/2πThe quantization postulate — only orbits obeying this exist.

rₙ = 0.529 × n²/Z ÅBohr radius 0.529 Å for H, n = 1; grows as n², shrinks with Z.

vₙ = 2.18 × 10⁶ × Z/n m/sFastest in n = 1; falls as 1/n — never rises with orbit.

Eₙ = −13.6 × Z²/n² eV= −2.18 × 10⁻¹⁸ Z²/n² J; zero at n = ∞, negative when bound.

hν = E₂ − E₁The jump condition — emission down, absorption up.

ν̄ = R_HZ²(1/n₁² − 1/n₂²)Rydberg machine; R_H = 109677 cm⁻¹; n₁ names the series.

IE from level n = 13.6 Z²/n² eVIonization from excited states — 3.4 eV from n = 2 of H.

Lines from level n = n(n−1)/2Maximum lines in a cascade to the ground state.

Constants: R_H = 109677 cm⁻¹ · a₀ = 0.529 Å · v₁ = 2.18 × 10⁶ m/s · E₁(H) = −13.6 eV = −2.18 × 10⁻¹⁸ J · h = 6.626 × 10⁻³⁴ J s · c = 3.0 × 10⁸ m/s

Three postulates, four formulas — quantized orbits, mvr = nh/2π, and the jump condition generate every numerical in this section.
Three scalings, never mixed — r ∝ n²/Z, v ∝ Z/n, E ∝ −Z²/n²; each quantity obeys only its own trend.
The Rydberg equation carries a Z² — all H-like spectra are hydrogen’s shifted by Z²; series names come from the lower level n₁.
Bohr’s wins are also its walls — perfect for H-like atoms, silent on multi-electron atoms, fine structure and Zeeman shifts; those walls open § 2.5.

FAQs

What are the postulates of Bohr’s model of the hydrogen atom?

Bohr proposed that (1) the electron moves only in certain circular, stationary orbits without radiating energy; (2) only those orbits are allowed whose angular momentum is an integral multiple of h/2π (mvr = nh/2π); and (3) energy is emitted or absorbed only when the electron jumps between orbits, as a photon of energy hν = E₂ − E₁.

What is the Bohr radius?

The Bohr radius is the radius of the first orbit of the hydrogen atom, 0.529 Å (52.9 pm). In general, the radius of the nth orbit is rₙ = 0.529 × n²/Z Å, where Z is the atomic number — so orbits grow as n² and shrink as Z increases.

How do you calculate the wavelength of a spectral line using the Rydberg equation?

Use 1/λ (wave number) = R_H × Z² × (1/n₁² − 1/n₂²), with R_H = 109677 cm⁻¹ and n₁ the lower level. For the hydrogen 4 → 2 transition: 109677 × (1/4 − 1/16) = 20564 cm⁻¹, giving λ ≈ 486 nm — the second line of the Balmer series in the visible region.

Why is the energy of an electron in an atom negative?

The zero of energy is assigned to a free electron at infinite separation (n = ∞). A bound electron has less energy than a free one, so its energy is negative: Eₙ = −13.6 Z²/n² eV. The ionization energy is the positive amount needed to raise the electron from its level to zero — 13.6 eV for hydrogen’s ground state.

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QCC Notes — Class 11 Chemistry

Strictly NCERT-aligned notes for JEE Main & NEET, prepared by QCC Notes (Padho Likho JEE). Content follows the latest NCERT edition and current NTA exam pattern.

Last updated
24 Sep 2026