ChemistryCBSE Class 11 · Structure of Atom

Bohr Model and the Hydrogen Spectrum Class 11

Updated 17 October 2026 · Fact-checked

In Bohr's model an electron in the nth orbit of a hydrogen-like species (nuclear charge Z) has radius r_n = 52.9 n²/Z pm and energy E_n = −2.18 × 10⁻¹⁸ Z²/n² J. The line emitted in a jump from n₂ to n₁ has wavenumber 1/λ = R_H Z²(1/n₁² − 1/n₂²), with R_H = 1.097 × 10⁷ m⁻¹. Lyman lines lie in the ultraviolet, Balmer in the visible, and Paschen, Brackett and Pfund in the infrared.

Notice that each series ends on one level: n=1 (Lyman), n=2 (Balmer), n=3 (Paschen). Lyman jumps are largest, and levels crowd together near E = 0.

Illustration drawn for QuestionBuddy.

Understand Bohr Model and the Hydrogen Spectrum

Postulates. (i) An electron revolves in certain fixed circular orbits (stationary states) without radiating. (ii) Allowed orbits have angular momentum $m v r = n \frac{h}{2 π}$ , $n = 1, 2, 3, \dots$ (equivalently $2 π r = nλ$ for the electron's de Broglie wavelength). (iii) A transition between orbits emits or absorbs a photon with $Δ E = E_{hi g h} - E_{l o w} = h ν$ .

Radius and energy for a one-electron species of nuclear charge $Z$ (H, He⁺, Li²⁺, …):
$ $r_{n} = 52.9 pm \times \frac{n ^{2}}{Z}, E_{n} = - 2.18 \times 1 0^{- 18} J \times \frac{Z ^{2}}{n ^{2}} .$ $
Energy is negative (bound), the ground state ( $n = 1$ ) is lowest, and the ionisation energy of hydrogen from $n = 1$ is $2.18 \times 1 0^{- 18}$ J per atom ( $1312$ kJ mol⁻¹). Orbits get farther apart in energy near $n = 1$ and crowd together at larger $n$ .

Spectral lines.
$ $\frac{1}{λ} = \overset{ν}{ˉ} = R_{H} Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}), Δ E = h c \overset{ν}{ˉ}, R_{H} = 1.097 \times 1 0^{7} m^{- 1} .$ $

Lyman: $n_{1} = 1$ , $n_{2} = 2, 3, \dots$ (ultraviolet)
Balmer: $n_{1} = 2$ , $n_{2} = 3, 4, \dots$ (visible)
Paschen: $n_{1} = 3$ , $n_{2} = 4, 5, \dots$ (infrared)
Brackett: $n_{1} = 4$ ; Pfund: $n_{1} = 5$ (infrared)

The series limit ( $n_{2} = \infty$ ) is the shortest wavelength of a series. From a level $n$ , the number of different lines possible is $n (n - 1) /2$ .

Emission versus absorption. Emission lines appear when excited electrons fall to lower levels; absorption lines appear when ground-state atoms absorb exactly the energies needed to move up.

Limitations. The model works for one-electron species only; it cannot explain the splitting of lines in a magnetic field (Zeeman) or electric field (Stark), the intensities of lines or the spectra of multi-electron atoms, and it is inconsistent with the uncertainty principle and with the wave nature of the electron.

Bohr Model and the Hydrogen Spectrum: formulas and key facts

Radius: $r_{n} = 52.9 \frac{n ^{2}}{Z} pm$; Hydrogen-like species.
Energy: $E_{n} = - 2.18 \times 1 0^{- 18} \frac{Z ^{2}}{n ^{2}} J$; Per atom.
Rydberg formula: $\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$; $R_{H} = 1.097 \times 1 0^{7} m^{- 1}$ .
Photon energy: $Δ E = h ν = \frac{h c}{λ}$; $h = 6.626 \times 1 0^{- 34}$ J s.
Quantisation: $m v r = \frac{nh}{2 π}, 2 π r = nλ$; Bohr and de Broglie.
Number of lines: $\frac{n ( n - 1 )}{2}$; From level $n$ .

How to solve Bohr Model and the Hydrogen Spectrum questions

Pick the species (Z) and the two levels, find the energy difference, and convert with E = hc/λ.

1Identify Z (1 for H, 2 for He⁺, 3 for Li²⁺).
2Write n₁ (lower) and n₂ (higher) for the transition.
3Use the Rydberg formula for 1/λ, or E_n for the energy difference.
4Convert: λ = 1/ν̄ and E = hc/λ (λ in metres).
5Identify the series and region from n₁ and the wavelength.
6For lines from level n, count n(n − 1)/2.

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.

A neon-sign-like glow of hydrogen

Suppose a hydrogen discharge tube shows a red line from the jump n = 3 to n = 2.

Find its wavelength.

Try it first, then show the solutionSolution

1/λ = 1.097 × 10⁷(1/4 − 1/9) = 1.097 × 10⁷ × 5/36 = 1.524 × 10⁶ m⁻¹.
λ = 6.56 × 10⁻⁷ m = 656 nm, a red line.

Answer: 656 nm (red).

Memory hook: The first Balmer line is the red one.

An astronomer's measurement

Suppose a star's spectrum shows the Lyman series limit.

Find the wavelength of the series limit.

Try it first, then show the solutionSolution

n₁ = 1, n₂ = ∞: 1/λ = R_H = 1.097 × 10⁷ m⁻¹.
λ = 9.12 × 10⁻⁸ m = 91.2 nm (ultraviolet).

Answer: 91.2 nm.

Memory hook: The series limit is the shortest wavelength.

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.

Z² and n²
E ∝ Z²/n² and r ∝ n²/Z.
Use it when: Hydrogen-like ions.
Series by n₁
Lyman 1, Balmer 2, Paschen 3.
Use it when: Series identification.
Limit means ∞
n₂ = ∞ gives the shortest wavelength.
Use it when: Series limit.
n(n−1)/2
Number of lines from a level.
Use it when: Counting lines.

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 and radius of the n = 2 orbit · 2 marks · Very short answerBoard level

Find the energy and radius of the electron in the n = 2 orbit of a hydrogen atom.

Try it first, then show the solutionSolution

E₂ = −2.18 × 10⁻¹⁸/4 = −5.45 × 10⁻¹⁹ J. (1)
r₂ = 52.9 × 4 = 211.6 pm. (1)

Answer: −5.45 × 10⁻¹⁹ J; 212 pm.

Example 2: A Balmer line · 3 marks · Short answerBoard level

Calculate the wavelength and the photon energy for the transition n = 4 to n = 2 in hydrogen (R_H = 1.097 × 10⁷ m⁻¹, h = 6.626 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m/s). Name the series and the region.

Try it first, then show the solutionSolution

1/λ = 1.097 × 10⁷(1/4 − 1/16) = 1.097 × 10⁷ × 3/16 = 2.057 × 10⁶ m⁻¹, λ = 486 nm. (1)
E = hc/λ = 6.626 × 10⁻³⁴ × 3.0 × 10⁸/4.86 × 10⁻⁷ = 4.09 × 10⁻¹⁹ J. (1)
Balmer series, visible region. (1)

Answer: 486 nm; 4.09 × 10⁻¹⁹ J; Balmer, visible.

Example 3: A hydrogen-like ion · 3 marks · Short answerHarder board level

Find the ionisation energy of Li²⁺ from its ground state in J per ion and in kJ per mole, and the radius of its first orbit.

Try it first, then show the solutionSolution

E₁ = −2.18 × 10⁻¹⁸ × 9 = −1.96 × 10⁻¹⁷ J, so the ionisation energy is 1.96 × 10⁻¹⁷ J per ion. (1)
Per mole: 1.962 × 10⁻¹⁷ × 6.022 × 10²³ = 1.18 × 10⁷ J = 1.18 × 10⁴ kJ mol⁻¹. (1)
r₁ = 52.9/3 = 17.6 pm. (1)

Answer: 1.96 × 10⁻¹⁷ J; 1.18 × 10⁴ kJ/mol; 17.6 pm.

Example 4: Lyman and Paschen lines · 4 marks · Long answerHarder board level

Find (a) the wavelength of the first line of the Lyman series, (b) the series limit of the Lyman series, and (c) the wavelength of the first line of the Paschen series in hydrogen. State the region of each.

Try it first, then show the solutionSolution

(a) 1/λ = R_H(1 − 1/4) = 0.75 × 1.097 × 10⁷, λ = 121.5 nm, ultraviolet. (1)
(b) 1/λ = R_H, λ = 91.2 nm, ultraviolet. (1)
(c) n = 4 → 3: 1/λ = R_H(1/9 − 1/16) = R_H × 7/144, λ = 1.875 × 10⁻⁶ m = 1875 nm. (1)
This lies in the infrared region. (1)

Answer: (a) 121.5 nm; (b) 91.2 nm; (c) 1875 nm (infrared).

Example 5: Excitation and emission · 4 marks · Long answerHOTS / competency

Hydrogen atoms in the ground state absorb photons and are excited to n = 3. (a) Find the wavelength of the photon absorbed. (b) List all the lines that can be emitted as the atoms return to n = 1, with their wavelengths and series. (c) Which line is in the visible region?

Try it first, then show the solutionSolution

(a) ΔE = 2.18 × 10⁻¹⁸ (1 − 1/9) = 1.94 × 10⁻¹⁸ J; λ = hc/ΔE = 1.026 × 10⁻⁷ m = 102.6 nm. (1)
(b) Three lines (3 × 2/2): 3 → 1 (102.6 nm, Lyman), 2 → 1 (121.5 nm, Lyman), 3 → 2 (656 nm, Balmer). (1)
1/λ(3→2) = R_H × 5/36 gives 656 nm; 1/λ(2→1) = R_H × 3/4 gives 121.5 nm. (1)
(c) Only 3 → 2 (656 nm) is visible; the other two are ultraviolet. (1)

Answer: (a) 102.6 nm; (b) 102.6, 121.5, 656 nm; (c) 656 nm.

Where students lose marks in Bohr Model and the Hydrogen Spectrum

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

Mistake: Putting n₂ below n₁ in the Rydberg formula.
Reversing the levels gives a negative wavelength.
Do this: n₂ is the higher level; 1/n₁² − 1/n₂² > 0.
Usually costs: The answer mark.
Mistake: Forgetting Z² for ions.
Using the hydrogen formula for He⁺.
Do this: Multiply by Z².
Usually costs: The answer mark.
Mistake: Using nm in hc/λ without converting.
Unit slip.
Do this: Convert λ to metres.
Usually costs: Numerical mark.
Mistake: Calling the Balmer series ultraviolet.
Memory slip.
Do this: Balmer: visible; Lyman: ultraviolet; Paschen: infrared.
Usually costs: 1 mark.

Exam tips

State the Rydberg formula with R_H and Z before substituting.
Name the series from n₁ and give the region.
For ions, write Z² explicitly.
Show unit conversions (nm to m) in a separate line.
Give energy in J per atom and per mole when asked.

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 Balmer series of hydrogen lies in the
(a) ultraviolet region (b) visible region (c) infrared region (d) X-ray region
Hint
n₁ = 2.
Answer
(b) visible region.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Bohr's model cannot explain the spectra of atoms with more than one electron.
Reason (R): The model ignores the repulsion between electrons and the wave nature of the electron.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
One-electron theory.
Answer
(a).
Q3 · 3 marks · Short answerHarder board level
Show that Bohr's condition mvr = nh/2π is equivalent to 2πr = nλ for the de Broglie wavelength of the electron, and find λ for the electron in the n = 4 orbit of hydrogen (r₄ = 846.4 pm).
Hint
λ = h/mv.
Answer
With λ = h/mv: mvr = nh/2π gives 2πr = nλ. For n = 4, λ = 2π × 846.4/4 = 1330 pm = 1.33 nm.
Q4 · 3 marks · Short answerHarder board level
Find the wavelength of the line emitted when the electron in Li²⁺ falls from n = 3 to n = 2. In which region does it lie?
Hint
Z² = 9.
Answer
1/λ = 1.097 × 10⁷ × 9 × (1/4 − 1/9) = 1.097 × 10⁷ × 9 × 5/36 = 1.371 × 10⁷ m⁻¹, λ = 72.9 nm, ultraviolet.
Q5 · 4 marks · Long answerHOTS / competency
The first line of the Lyman series (n = 2 → 1) of a hydrogen-like ion has wavelength 30.4 nm. (a) Identify the ion. (b) Find its ionisation energy from the ground state. (c) Find the radius of its n = 2 orbit.
Hint
1/λ = R_H Z² × 3/4.
Answer
(a) Z² = 1/(1.097 × 10⁷ × 0.75 × 30.4 × 10⁻⁹) = 4.0, so Z = 2: He⁺. (b) IE = 2.18 × 10⁻¹⁸ × 4 = 8.72 × 10⁻¹⁸ J. (c) r₂ = 52.9 × 4/2 = 105.8 pm.

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.Radius of nth orbit?Show answer
52.9 n²/Z pm.
2.Energy of nth orbit?Show answer
−2.18 × 10⁻¹⁸ Z²/n² J.
3.Rydberg constant?Show answer
1.097 × 10⁷ m⁻¹.
4.Balmer series n₁?Show answer
2.
5.Number of lines from n?Show answer
n(n−1)/2.
6.Lyman region?Show answer
Ultraviolet.

Bohr Model and the Hydrogen Spectrum: frequently asked questions

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

It is bound to the nucleus; zero energy corresponds to the electron at rest at infinite distance.

What is the series limit?

The shortest wavelength of a series, for transitions from n = ∞.

What are stationary states?

Orbits in which the electron does not radiate energy.

Why does Bohr's model fail for multi-electron atoms?

It cannot account for electron-electron repulsion or the wave nature of the electron.

How is an absorption line different from an emission line?

Absorption removes specific wavelengths from a continuous spectrum; emission produces bright lines at those wavelengths.