PhysicsISC Class 12 · Atoms

Energy Levels of Hydrogen and the Hydrogen Spectrum: Series, Rydberg Formula and Transitions

Updated 15 October 2026 · Fact-checked

The electron's energy in the nth orbit of hydrogen is $E_n = -\frac{13.6}{n^2}$ eV. A jump from $n_i$ to a lower $n_f$ emits a photon with $h\nu = E_i - E_f$, giving $\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ with $R = 1.097 \times 10^7$ m⁻¹. Lines ending on $n_f$ = 1, 2, 3, 4, 5 form the Lyman to Pfund series.

Levels crowd together towards 0 eV; jumps down to n = 1 give Lyman (UV) lines and jumps to n = 2 give Balmer (visible) lines.

Illustration drawn for QuestionBuddy.

Understand Energy Levels and the Hydrogen Spectrum

Energy of the electron. From the force balance, the kinetic energy is $K = \frac{1}{2} m v^{2} = \frac{e ^{2}}{8 π ε _{0} r}$ and the potential energy is $U = - \frac{e ^{2}}{4 π ε _{0} r}$ . The total energy is
$E = K + U = - \frac{e ^{2}}{8 π ε _{0} r} .$
Substituting $r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m e ^{2}}$ :
$E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} n ^{2} h ^{2}} = - \frac{13.6}{n ^{2}} eV .$
The negative sign means the electron is bound; zero energy corresponds to the electron at rest far away (ionised).

Energy level diagram. n = 1: −13.6 eV (ground state); n = 2: −3.4 eV; n = 3: −1.51 eV; n = 4: −0.85 eV; n = 5: −0.544 eV; ... n = ∞: 0. Levels crowd together towards zero.

Ionisation energy (from the ground state): 13.6 eV.
Excitation energy to level n: $E_{n} - E_{1}$ , e.g. 10.2 eV to n = 2, 12.09 eV to n = 3.

Emission and the Rydberg formula. When the electron falls from $n_{i}$ to $n_{f}$ , a photon of energy $h ν = E_{n_{i}} - E_{n_{f}}$ is emitted. With $λ = \frac{h c}{Δ E}$ :
$\frac{1}{λ} = R (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}), R = \frac{m e ^{4}}{8 ε _{0}^{2} h ^{3} c} = 1.097 \times 1 0^{7} m^{- 1} .$
The calculated wavelengths agree closely with the measured lines, which was the great success of Bohr's theory. A quick route: $λ (nm) = \frac{1240}{Δ E ( eV )}$ .

Spectral series.

Series	$n_{f}$	$n_{i}$	Region
Lyman	1	2, 3, ...	Ultraviolet
Balmer	2	3, 4, ...	Visible (and near UV)
Paschen	3	4, 5, ...	Infrared
Brackett	4	5, 6, ...	Infrared
Pfund	5	6, 7, ...	Infrared

The first line of a series ( $n_{i} = n_{f} + 1$ ) has the longest wavelength; the series limit ( $n_{i} \to \infty$ ) has the shortest, $λ_{l imi t} = \frac{n _{f}^{2}}{R}$ .

Number of lines. Atoms excited to level n can emit $\frac{n ( n - 1 )}{2}$ different lines as they cascade down.

Emission and absorption spectra. Hot, excited hydrogen gives bright lines on a dark background (emission line spectrum; arrows point down on the level diagram). Cool hydrogen in front of a white source removes the same wavelengths, giving dark lines (absorption spectrum; arrows point up). At room temperature nearly all atoms are in n = 1, so cool hydrogen absorbs mainly Lyman lines. A photon is absorbed only if its energy exactly matches a level difference (or exceeds 13.6 eV, which ionises).

Energy Levels and the Hydrogen Spectrum: formulas and key facts

Total energy: $E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} n ^{2} h ^{2}} = - \frac{13.6}{n ^{2}} eV$; Z = 1.
K and U: $K = - E_{n}, U = 2 E_{n}$; Signs matter.
Photon emitted: $h ν = E_{n_{i}} - E_{n_{f}}$; Bohr's frequency condition.
Rydberg formula: $\frac{1}{λ} = R (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})$; R = 1.097 × 10⁷ m⁻¹.
Quick wavelength: $λ (nm) = \frac{1240}{Δ E ( eV )}$; hc ≈ 1240 eV nm.
Series limit: $λ_{l imi t} = \frac{n _{f}^{2}}{R}$; Shortest line.
Number of lines: $N = \frac{n ( n - 1 )}{2}$; From level n.

How to solve Energy Levels and the Hydrogen Spectrum questions

Decide the two levels, then use either energies or the Rydberg formula; check one with the other.

1Write the levels: $E_{n} = - \frac{13.6}{n ^{2}}$ eV.
2ΔE = upper level − lower level (a positive number for emission).
3λ = $\frac{1240}{Δ E}$ nm, or $\frac{1}{λ} = R (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})$ .
4Name the series from $n_{f}$ and the region from λ.
5Longest line: $n_{i} = n_{f} + 1$ ; shortest (limit): $n_{i} = \infty$ .
6Absorption: the photon energy must equal $E_{n} - E_{1}$ exactly for some n.

Energy Levels 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.

Reading a star's absorption line

Suppose an astronomer sees a dark hydrogen line at 486 nm in a star's spectrum.

Which transition absorbed this light, and what does it say about the absorbing atoms?

Try it first, then show the solutionSolution

Photon energy $= \frac{1240}{486} \approx 2.55$ eV $= E_{4} - E_{2} = - 0.85 - (- 3.40)$ .
The absorbing atoms must start in n = 2, so the star's outer gas is hot enough for many atoms to be already excited.

Answer: n = 2 → n = 4 (Balmer absorption); the gas is hot.

Memory hook: Dark Balmer lines mean warm atoms sitting in n = 2.

Electron bombardment

Imagine a discharge tube in which electrons of energy 12.5 eV collide with hydrogen atoms in the ground state.

Up to which level can the atoms be excited, and which lines can they then emit?

Try it first, then show the solutionSolution

Excitation to n = 3 needs 12.09 eV (possible); to n = 4 needs 12.75 eV (not possible).
From n = 3 the atom can emit 3 → 2 (≈ 656 nm), 3 → 1 (≈ 102.6 nm) and 2 → 1 (≈ 121.6 nm): $\frac{3 \times 2}{2} = 3$ lines.

Answer: Up to n = 3; three lines (one visible red, two ultraviolet).

Memory hook: Energy in decides the highest step; steps down decide the lines.

Tricks to remember Energy Levels and the Hydrogen Spectrum

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

Thirteen-six over n squared
−13.6, −3.4, −1.51, −0.85, −0.544 eV.
Use it when: Energy levels.
Little Boys Prefer Bright Pens
Lyman, Balmer, Paschen, Brackett, Pfund: $n_{f}$ = 1 to 5.
Use it when: Series order.
UV, visible, then IR
Lyman UV; Balmer visible; the rest infrared.
Use it when: Regions.
Twelve-forty again
λ(nm) = 1240/ΔE(eV).
Use it when: Fast wavelengths.

Energy Levels and the Hydrogen Spectrum: solved examples

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

Example 1: Ionisation energy · 1 mark · MCQBoard level

The energy needed to ionise a hydrogen atom in its ground state is:
(a) 3.4 eV
(b) 10.2 eV
(c) 13.6 eV
(d) 27.2 eV

Try it first, then show the solutionSolution

From −13.6 eV to 0: 13.6 eV.

Answer: (c).

Example 2: Levels and excitation · 2 marks · Short answerBoard level

Find the energy of the electron in the n = 5 level of hydrogen, and the energy needed to excite it from the ground state to n = 3.

Try it first, then show the solutionSolution

$E_{5} = - \frac{13.6}{25} = - 0.544$ eV. (1)
$E_{3} - E_{1} = - 1.51 - (- 13.6) \approx 12.09$ eV. (1)

Answer: −0.544 eV; ≈ 12.09 eV.

Example 3: Derivation of the energy levels · 3 marks · Long answerHarder board level

Using Bohr's theory, obtain the total energy of the electron in the nth orbit of hydrogen and show that it is $- \frac{13.6}{n ^{2}}$ eV.

Try it first, then show the solutionSolution

From $\frac{m v ^{2}}{r} = \frac{e ^{2}}{4 π ε _{0} r ^{2}}$ : $K = \frac{e ^{2}}{8 π ε _{0} r}$ ; $U = - \frac{e ^{2}}{4 π ε _{0} r}$ . (1)
$E = K + U = - \frac{e ^{2}}{8 π ε _{0} r}$ ; with $r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m e ^{2}}$ , $E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} n ^{2} h ^{2}}$ . (1)
Substituting constants, $\frac{m e ^{4}}{8 ε _{0}^{2} h ^{2}} \approx 2.18 \times 1 0^{- 18}$ J = 13.6 eV, so $E_{n} = - \frac{13.6}{n ^{2}}$ eV. (1)

Answer: $E_{n} = - \frac{13.6}{n ^{2}}$ eV.

Example 4: The blue-green Balmer line · 3 marks · Short answerHarder board level

Find the wavelength of the photon emitted when the electron in hydrogen falls from n = 4 to n = 2, (a) from the energy levels and (b) from the Rydberg formula. Name its colour.

Try it first, then show the solutionSolution

(a) ΔE = −0.85 − (−3.40) = 2.55 eV; λ = $\frac{1240}{2.55}$ ≈ 486 nm. (1)
(b) $\frac{1}{λ} = R (\frac{1}{4} - \frac{1}{16}) = \frac{3 R}{16}$ ; λ = $\frac{16}{3 \times 1.097 \times 1 0 ^{7}}$ ≈ 486.2 nm. (1)
Balmer series, blue-green (visible). (1)

Answer: ≈ 486 nm, blue-green, Balmer.

Example 5: Identifying an infrared line · 3 marks · Short answerHOTS / competency

A hydrogen line is observed at 1282 nm. Identify the transition and the series.

Try it first, then show the solutionSolution

$\frac{1}{λ R} = \frac{1}{1282 \times 1 0 ^{- 9} \times 1.097 \times 1 0 ^{7}} \approx 0.0711$ . (1)
Try $n_{f} = 3$ : $\frac{1}{9} - \frac{1}{n _{i}^{2}} = 0.0711$ gives $\frac{1}{n _{i}^{2}} = 0.0400$ , $n_{i} = 5$ . (1)
Transition 5 → 3: Paschen series, infrared. (1)

Answer: n = 5 → n = 3, Paschen (infrared).

Example 6: Reading an energy-level diagram · 4 marks · Case-basedHOTS / competency

Suppose a diagram of hydrogen shows downward arrows from n = 5, 4 and 3, all ending on n = 2.
(i) What type of spectrum do these transitions give? (1 mark)
(ii) Name the series. (1 mark)
(iii) Find the wavelength for 5 → 2. (1 mark)
(iv) What spectrum would upward arrows between the same levels represent? (1 mark)

Try it first, then show the solutionSolution

(i) An emission (line) spectrum: downward jumps release photons.
(ii) Balmer series ( $n_{f}$ = 2).
(iii) $\frac{1}{λ} = R (\frac{1}{4} - \frac{1}{25}) = 0.21 R$ ; λ ≈ 434 nm (violet).
(iv) An absorption spectrum: dark lines at the same wavelengths.

Answer: (i) Emission line spectrum; (ii) Balmer; (iii) ≈ 434 nm; (iv) absorption.

Where students lose marks in Energy Levels and the Hydrogen Spectrum

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

Mistake: Calling it just "hydrogen spectrum"
The ISC 2024 examiners (Q7(i)) found many unaware of emission versus absorption, writing "hydrogen spectrum" or "line spectrum" only.
Do this: Downward arrows: emission (line) spectrum; upward: absorption.
Usually costs: 1 mark.
Mistake: Naming the wrong series
Also 2024: some named Lyman for transitions ending on n = 2.
Do this: Series is fixed by the lower level $n_{f}$ : 1 Lyman, 2 Balmer, 3 Paschen.
Usually costs: 1 mark.
Mistake: Subtracting energies the wrong way
Negative ΔE leads to a negative wavelength.
Wrong working
ΔE = −3.40 − (−0.85) = −2.55 eV.
Right working
ΔE = E(upper) − E(lower) = 2.55 eV.
Do this: Take the magnitude of the difference.
Usually costs: 1 mark.
Mistake: Using Å
The syllabus asks for nm.
Do this: Quote wavelengths in nm.
Usually costs: ½ mark.

Exam tips

Draw the level diagram with energies and arrows when explaining series.
Show both routes (energy and Rydberg) when a question allows; one checks the other.
State the region (UV, visible, IR) with any wavelength you calculate.
Explain the negative sign of energy in one line: bound electron.

Practice questions on Energy Levels 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 Paschen series of hydrogen lies in the:
(a) ultraviolet
(b) visible
(c) infrared
(d) X-ray region
Hint
$n_{f}$ = 3.
Answer
(c).
Q2 · 1 mark · Short answerBoard level
How many different spectral lines can hydrogen atoms excited to n = 5 emit?
Hint
$\frac{n ( n - 1 )}{2}$ .
Answer
10.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The hydrogen emission spectrum consists of sharp lines.
Reason (R): The electron can have only certain discrete energies, so only certain photon energies are emitted.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Discrete levels.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
Find the wavelength of the second line of the Lyman series (3 → 1).
Hint
ΔE ≈ 12.09 eV.
Answer
≈ 102.6 nm (ultraviolet).
Q5 · 2 marks · Short answerHarder board level
A ground-state hydrogen atom absorbs a photon of 12.75 eV. To which level is it raised, and how many lines can it emit afterwards?
Hint
−13.6 + 12.75.
Answer
n = 4 (−0.85 eV); up to 6 lines.
Q6 · 2 marks · Short answerHOTS / competency
Will a ground-state hydrogen atom absorb a photon of 10.5 eV? Explain.
Hint
Look for a level at −3.1 eV.
Answer
No. It would need a level at −13.6 + 10.5 = −3.1 eV; none exists (n = 2 is −3.4 eV), and 10.5 eV is too little to ionise, so the photon passes through.
Q7 · 1 mark · Short answerHOTS / competency
Find the series limit of the Brackett series.
Hint
$λ = \frac{16}{R}$ .
Answer
≈ 1459 nm (infrared).
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student views a hydrogen lamp through a spectroscope and sees visible lines at 656, 486, 434 and 410 nm.
(i) Name the series. (1 mark)
(ii) Identify the transition that gives 410 nm. (1 mark)
(iii) Find the photon energy at 410 nm. (1 mark)
(iv) Why are no Lyman lines seen? (1 mark)
Hint
$\frac{1}{4} - \frac{1}{n ^{2}}$ .
Answer
(i) Balmer. (ii) $\frac{1}{λ} = R (\frac{1}{4} - \frac{1}{36})$ gives 410 nm: 6 → 2. (iii) ≈ 3.02 eV. (iv) Lyman lines are ultraviolet, invisible to the eye and absorbed by glass.

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.Energy of n = 2 level?Show answer
−3.4 eV.
2.Rydberg constant?Show answer
$1.097 \times 1 0^{7}$ m⁻¹.
3.Series in the visible region?Show answer
Balmer.
4.Ionisation energy from n = 2?Show answer
3.4 eV.
5.Lines from n = 4?Show answer
6.

Energy Levels and the Hydrogen Spectrum: frequently asked questions

Why do the energy levels get closer at higher n?

Because E_n \propto -\frac{1}{n^2}: the differences between successive values shrink as n grows, approaching zero at the ionisation limit.

Why is the first line of a series the longest in wavelength?

It comes from the smallest energy gap (n_f + 1 \to n_f), and λ is inversely proportional to the energy difference.

Why does cool hydrogen absorb only Lyman lines?

Almost all its atoms are in the ground state, so absorption starts from n = 1.

What does the ISC 2027 syllabus need here?

Bohr's energy expressions (Z = 1), the energy level diagram, ΔE, frequency and wavelength of emission lines, the formulae for the Lyman, Balmer, Paschen, Brackett and Pfund series, the Rydberg constant, and agreement with observed values; λ in nm.

How was this asked recently?

ISC 2024 Q7(i) (2 marks) gave an energy-level diagram with transitions to n = 2 and asked the type of spectrum and the series.