ChemistryISC Class 11 · Structure of Atom

Bohr Model and the Hydrogen Spectrum: Energy Levels and Spectral Lines

Updated 17 October 2026 · Fact-checked

In the Bohr model the electron moves in fixed orbits with angular momentum nh/2π. The radius is rₙ = 0.529 n²/Z Å and the energy is Eₙ = −13.6 Z²/n² eV. A transition gives a line with 1/λ = R Z²(1/n₁² − 1/n₂²), R = 1.097 × 10⁷ m⁻¹, and lines fall into the Lyman, Balmer and Paschen series.

Understand Bohr Model and the Hydrogen Spectrum

Postulates. (1) The electron in a hydrogen-like atom revolves in certain stationary circular orbits without radiating energy. (2) Only those orbits are allowed in which the angular momentum is an integral multiple of h/2π: mvr = nh/2π, n = 1, 2, 3... (3) Energy is absorbed or emitted only when the electron jumps between orbits: ΔE = E₂ − E₁ = hν.

Results. For an atom or ion of atomic number Z with one electron: rₙ = 0.529 n²/Z Å, vₙ = 2.18 × 10⁶ (Z/n) m s⁻¹, and Eₙ = −13.6 Z²/n² eV (= −2.18 × 10⁻¹⁸ Z²/n² J). The negative sign means the electron is bound; the ground state (n = 1) is the most stable, and E = 0 at n = ∞.

Spectral lines. When an electron falls from n₂ to n₁, 1/λ = R Z²(1/n₁² − 1/n₂²), and the photon energy is ΔE = 13.6 Z²(1/n₁² − 1/n₂²) eV. Since hc = 1240 eV nm (derived from h = 6.626 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m s⁻¹), λ(nm) = 1240/ΔE(eV). Series of hydrogen: Lyman (n₁ = 1, ultraviolet), Balmer (n₁ = 2, visible and near UV), Paschen (n₁ = 3), Brackett (n₁ = 4) and Pfund (n₁ = 5), all infrared. The number of different lines from level n to the ground state is n(n − 1)/2.

Limitations. The model fails for atoms with more than one electron, cannot explain the fine structure of lines or the intensity of lines, ignores the wave nature of the electron (de Broglie) and violates the uncertainty principle by fixing both radius and velocity.

De Broglie link. The allowed orbit has a circumference equal to a whole number of electron wavelengths: 2πr = nλ.

Bohr Model and the Hydrogen Spectrum: formulas and key facts

Orbit radius and speed: $r_{n} = 0.529 n^{2} / Z \overset{˚}{A} ∣ v_{n} = 2.18 \times 1 0^{6} (Z / n) m s^{- 1}$; Z = atomic number.
Energy: $E_{n} = - 13.6 Z^{2} / n^{2} e V$; Eₙ = −2.18 × 10⁻¹⁸ Z²/n² J.
Spectral lines: $1/ λ = R Z^{2} (1/ n_{1}^{2} - 1/ n_{2}^{2})$; R = 1.097 × 10⁷ m⁻¹.
Photon wavelength: $λ (nm) = 1240/Δ E (e V)$; From E = hc/λ.

Bohr Model and the Hydrogen Spectrum in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

The red line of a hydrogen lamp

Suppose a hydrogen discharge lamp glows with a visible red line from the electron falling from n = 3 to n = 2 (R = 1.097 × 10⁷ m⁻¹).

Find the wavelength and photon energy.

Try it first, then show the solutionSolution

1/λ = R(1/2² − 1/3²) = 1.097 × 10⁷ × 5/36 = 1.524 × 10⁶ m⁻¹.
λ = 6.563 × 10⁻⁷ m = 656 nm (red).
Energy = 13.6 × 5/36 = 1.89 eV; check 1240/656 = 1.89 eV.

Answer: 656 nm; 1.89 eV.

Memory hook: Balmer lines end on n = 2.

Tricks to remember Bohr Model and the Hydrogen Spectrum

Short memory aids. Say each one aloud once, then use it on a question the same day.

n² for radius, 1/n² for energy
Radius grows as n² while the energy rises towards zero as 1/n².
Use it when: Quick ratios between orbits.
Lyman, Balmer, Paschen = 1, 2, 3
The series name is set by the lower level n₁.
Use it when: Identifying a series.

Bohr Model and the Hydrogen Spectrum: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Energy of an orbit · 1 mark · MCQBoard level

The energy of an electron in the second orbit of a hydrogen atom is:
(a) −13.6 eV
(b) −6.8 eV
(c) −3.4 eV
(d) −1.51 eV

Try it first, then show the solutionSolution

E₂ = −13.6/2² = −3.4 eV.

Answer: (c) −3.4 eV

Example 2: Radius and energy of an orbit · 3 marks · Short answerBoard level

Calculate the radius and the energy of the electron in the third orbit of the He⁺ ion (Z = 2), and the energy needed to remove it from there.

Try it first, then show the solutionSolution

r₃ = 0.529 × 3²/2 = 2.38 Å. (1 mark)
E₃ = −13.6 × 4/9 = −6.04 eV. (1 mark)
Ionisation from n = 3 needs +6.04 eV. (1 mark)

Answer: r₃ = 2.38 Å; E₃ = −6.04 eV; 6.04 eV needed to ionise it.

Example 3: Wavelength of a Balmer line · 3 marks · Short answerHarder board level

Find the wavelength of the first line of the Balmer series (n = 3 → n = 2) and the energy of the photon. (R = 1.097 × 10⁷ m⁻¹)

Try it first, then show the solutionSolution

1/λ = R(1/4 − 1/9) = 1.097 × 10⁷ × 0.1389. (1 mark)
= 1.524 × 10⁶ m⁻¹, so λ = 6.56 × 10⁻⁷ m = 656 nm. (1 mark)
E = 13.6 × 0.1389 = 1.89 eV. (1 mark)

Answer: λ = 656 nm; E = 1.89 eV.

Example 4: Excitation by absorption · 3 marks · Short answerHarder board level

Hydrogen atoms in the ground state absorb radiation that raises the electron to n = 3. Find the energy absorbed and the wavelength of the radiation. (hc = 1240 eV nm)

Try it first, then show the solutionSolution

ΔE = 13.6(1 − 1/9) = 13.6 × 0.8889. (1 mark)
= 12.09 eV. (1 mark)
λ = 1240/12.09 = 102.6 nm (ultraviolet, the first Lyman line is 121.6 nm; this is the second Lyman line). (1 mark)

Answer: ΔE = 12.09 eV; λ = 102.6 nm.

Example 5: Lines from n = 4 · 5 marks · Long answerHOTS / competency

Hydrogen atoms excited to n = 4 return to lower levels. Find the number of spectral lines, the shortest and longest wavelengths and the series to which they belong. (R = 1.097 × 10⁷ m⁻¹)

Try it first, then show the solutionSolution

Number of lines = 4 × 3/2 = 6 (4→3, 4→2, 4→1, 3→2, 3→1, 2→1). (1 mark)
Shortest wavelength is for 4 → 1: 1/λ = R(1 − 1/16) = 1.0284 × 10⁷ m⁻¹. (1 mark)
λ = 97.2 nm, Lyman series (ultraviolet). (1 mark)
Longest is for 4 → 3: 1/λ = R(1/9 − 1/16) = 1.097 × 10⁷ × 0.04861 = 5.33 × 10⁵ m⁻¹. (1 mark)
λ = 1.875 × 10⁻⁶ m = 1875 nm, Paschen series (infrared). (1 mark)

Answer: 6 lines; 97.2 nm (Lyman) and 1875 nm (Paschen).

Where students lose marks in Bohr Model and the Hydrogen Spectrum

The slips examiners see most often, and what to write instead.

Mistake: Wrong sign for energy
Quoting E₂ = +3.4 eV.
Do this: Bound-state energies are negative: Eₙ = −13.6/n² eV.
Usually costs: 1 mark.
Mistake: Mixing n₁ and n₂ in the Rydberg formula
Getting a negative 1/λ.
Do this: n₁ is the lower level and n₂ the higher: (1/n₁² − 1/n₂²) > 0.
Usually costs: 1 to 2 marks.
Mistake: Forgetting Z² for ions
Using hydrogen energies for He⁺ or Li²⁺.
Do this: Multiply the energy by Z² and divide the radius by Z.
Usually costs: 2 marks.

Exam tips

State the postulates in the order: stationary orbits, quantised angular momentum, energy of the transition.
Name the series by the lower level n₁.
Use 1240 eV nm to convert between energy and wavelength quickly.
Mention the limitations of the model when a question asks for them.

Practice questions on Bohr Model and the Hydrogen Spectrum

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The Lyman series of hydrogen lies in the:
(a) infrared region
(b) visible region
(c) ultraviolet region
(d) microwave region
Hint
Lyman lines end on n = 1.
Answer
(c) ultraviolet region
Q2 · 2 marks · Short answerBoard level
Find the radius and the energy of the electron in the fourth orbit of a hydrogen atom.
Hint
rₙ = 0.529 n² Å; Eₙ = −13.6/n² eV.
Answer
r₄ = 0.529 × 16 = 8.46 Å; E₄ = −13.6/16 = −0.85 eV.
Q3 · 3 marks · Short answerHarder board level
Find the wavelength of the line emitted when an electron falls from n = 5 to n = 2 in hydrogen. (R = 1.097 × 10⁷ m⁻¹)
Hint
1/λ = R(1/4 − 1/25).
Answer
1/λ = 1.097 × 10⁷ × 0.21 = 2.304 × 10⁶ m⁻¹, so λ = 434 nm.
Q4 · 3 marks · Short answerHarder board level
Find the speed of the electron in the second orbit of hydrogen and the de Broglie wavelength of the electron in the first orbit. (m = 9.11 × 10⁻³¹ kg, h = 6.626 × 10⁻³⁴ J s)
Hint
v₁ = 2.18 × 10⁶ m s⁻¹; vₙ = v₁/n for hydrogen.
Answer
v₂ = 1.09 × 10⁶ m s⁻¹. λ₁ = h/mv = 6.626 × 10⁻³⁴/(9.11 × 10⁻³¹ × 2.18 × 10⁶) = 3.34 × 10⁻¹⁰ m, which equals 2πr₁ = 2π × 0.529 Å = 3.32 Å.
Q5 · 3 marks · Short answerHarder board level
Find the ionisation energy of the one-electron ion Be³⁺ (Z = 4) from its ground state and the radius of its first orbit.
Hint
Scale by Z² and 1/Z.
Answer
E₁ = −13.6 × 16 = −217.6 eV, so the ionisation energy is 217.6 eV. r₁ = 0.529/4 = 0.132 Å.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a hydrogen atom in its ground state absorbs a photon of energy 12.75 eV.
(i) To which level is the electron raised? (1 mark)
(ii) Find the radius of that orbit. (1 mark)
(iii) Find the speed of the electron there. (1 mark)
(iv) Find its angular momentum in J s (h/2π = 1.055 × 10⁻³⁴ J s). (1 mark)
Hint
Final energy = −13.6 + 12.75 = −0.85 eV.
Answer
(i) −13.6/n² = −0.85 → n = 4. (ii) r = 0.529 × 16 = 8.46 Å. (iii) v = 2.18 × 10⁶/4 = 5.45 × 10⁵ m s⁻¹. (iv) L = 4 × 1.055 × 10⁻³⁴ = 4.22 × 10⁻³⁴ J s.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Bohr's quantisation condition.Show answer
mvr = nh/2π.
2.Energy of the nth level of hydrogen?Show answer
−13.6/n² eV.
3.Which series is in the visible region?Show answer
Balmer.
4.Number of lines from level n?Show answer
n(n − 1)/2.

Bohr Model and the Hydrogen Spectrum: frequently asked questions

Why is the energy of the electron negative?

The zero is the electron at rest infinitely far away. A bound electron has lower energy, so its energy is negative.

Does the Bohr model work for helium?

Not for neutral helium, which has two electrons, but it works for He⁺ and other one-electron ions with the factor Z².

Why are the hydrogen lines discrete?

Because the electron's energy takes only certain values, only particular photon energies are emitted or absorbed.