Year 12 Physics Module 8 ⏱ ~20 min 5 MC · 2 Short Answer Lesson 10 of 17

Bohr's Model of the Hydrogen Atom

You'll be able to use Rydberg formula.

In 1913, Niels Bohr at Copenhagen applied Planck's quantum hypothesis to Rutherford's nuclear atom, proposing that electrons orbit only in states with angular momentum L = nh/2π. He calculated hydrogen energy levels as Eₙ = −13.6/n² eV and predicted Balmer series wavelengths to within 0.02% of measured values. The Nobel Prize was awarded to Bohr in 1922. His model also successfully predicted the Lyman series (UV) and Paschen series (IR), spectral series that had been measured before any theoretical explanation existed.

Today's hook: In 1913, Niels Bohr at Copenhagen calculated that hydrogen electrons can only orbit at specific energies Eₙ = −13.6/n² eV, where n = 1, 2, 3... When an electron jumps from n = 3 to n = 2, it emits a photon with exactly 1.89 eV of energy, the red H-alpha line at 656 nm that gives hydrogen nebulae their distinctive colour. Bohr's formula matched every known hydrogen spectral line to within 0.02%. Why can the hydrogen atom only emit these specific wavelengths, and never the colours in between?

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Orient and predict

Set up the vocabulary for quantised energy levels, then predict why a hydrogen atom emits only certain wavelengths.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions

Before you read, predict

The hydrogen emission spectrum shows light at only specific wavelengths, not a continuous rainbow.

Before reading on, answer:

Why might a hydrogen atom emit light at only certain wavelengths?
What does this tell us about the energy levels of electrons in hydrogen?
How could we calculate the wavelength of light emitted when an electron jumps between levels?

Warm-up: Rutherford's nuclear model failed because classical physics predicted that orbiting electrons would:

Learning Intentions

goals

Know, Bohr's Postulates

Quantised orbits
Angular momentum quantisation
Photon emission on transition

Understand, Energy Levels

$E_n = -13.6/n^2$ eV
Ground state, excited states
Ionisation energy

Can Do, Calculate Spectral Lines

Use Rydberg formula
Calculate photon energy/wavelength
Identify spectral series

Scan these before reading

vocab

Ground state ($n = 1$)Lowest energy state; $E_1 = -13.6$ eV for hydrogen. Electron is closest to nucleus.

Excited stateHigher energy level ($n > 1$); electron occupies orbit farther from nucleus.

QuantisationPhysical quantities restricted to discrete values; $L = n\hbar$ for Bohr orbits.

Rydberg constant$R_H = 1.097 \times 10^7$ m⁻¹; appears in the formula that predicts hydrogen spectral wavelengths.

Ionisation energyEnergy required to remove an electron completely from the atom; 13.6 eV for hydrogen in the ground state ($n \rightarrow \infty$).

Cross-lesson links: L11 established nuclear structure from scattering. L12 imposes quantisation on the orbits, Bohr's 1913 model explains why hydrogen emits only specific wavelengths (from M7 L09) and predicts them quantitatively. The Bohr model is the bridge between classical atomic physics and quantum mechanics.

Core Content

State Bohr's three postulates

Read the three postulates off the energy level diagram, then pin down $E_n = -13.6/n^2$ eV and the Bohr radius.

Bohr's Postulates

+5 XP

Three radical assumptions that explained hydrogen

Pass an electric current through a glass tube filled with hydrogen gas at low pressure: it glows with a distinctive pink-red light. Examine that light through a diffraction grating and instead of a rainbow you see four sharp coloured lines, red at 656 nm, blue-green at 486 nm, violet at 434 nm, and deep violet at 410 nm. These four lines are the same every time, in every hydrogen sample, anywhere on Earth or in a distant star. In 1913, Niels Bohr at Copenhagen explained this pattern by proposing three postulates:

Postulate 1, Quantised orbits: Electrons orbit the nucleus only in certain allowed circular orbits. In these orbits, the electron does not radiate energy (contrary to classical electromagnetism).

Postulate 2, Angular momentum quantisation: The angular momentum of the electron is quantised in units of $\hbar = h/2\pi$:

$$L = m_e v r = n\hbar \quad\text{where}\quad n = 1, 2, 3, \ldots$$

Postulate 3, Photon emission/absorption: Electrons emit or absorb photons only when transitioning between orbits. The photon energy equals the energy difference:

$$E_\text{photon} = |E_i - E_f| = hf = \dfrac{hc}{\lambda}$$

Combining these postulates with classical Coulomb force and centripetal motion gives the allowed energy levels:

$$E_n = -\dfrac{13.6}{n^2} \text{ eV}$$

and the orbital radii:

$$r_n = n^2 a_0 \quad\text{where}\quad a_0 = 0.529 \times 10^{-10} \text{ m (Bohr radius)}$$

The negative energy indicates bound states. The ground state ($n = 1$) has $E_1 = -13.6$ eV. The ionisation energy is 13.6 eV, the energy needed to remove the electron entirely ($n \rightarrow \infty$).

Figure 1, Hydrogen energy levels showing transitions for Balmer series (Hα, Hβ) and Lyman series (Lyman-α, Lyman-β). Each arrow represents a photon emitted as an electron drops to a lower level.

Text description

Hydrogen’s energy levels, drawn to scale: n = 1 at −13.6 eV, n = 2 at −3.40 eV, n = 3 at −1.51 eV, n = 4 at −0.85 eV, and n = ∞ at E = 0 (ionisation).
Balmer series (ending on n = 2): Hα is n = 3 → 2 and Hβ is n = 4 → 2.
Lyman series (ending on n = 1): Lyman-α is n = 2 → 1 and Lyman-β is n = 3 → 1. These are much larger drops, so much more energetic photons.
Each arrow is a photon emitted as the electron drops to a lower level.

Stop and check

Calculate the wavelength of light emitted when an electron transitions from $n = 3$ to $n = 2$ in hydrogen. In which spectral series does this line appear?

Bohr's three postulates: (1) electrons in allowed orbits do not radiate; (2) L = nℏ (angular momentum quantised); (3) photon emitted/absorbed on transition, E = hf. Energy levels: E_n = −13.6/n² eV; ground state E₁ = −13.6 eV; ionisation energy = 13.6 eV. Orbital radii: r_n = n²a₀, a₀ = 0.529 Å.

Write the three postulates and the energy level formula, they are the core of every Bohr question.

According to Bohr's model, the energy of hydrogen in the ground state ($n = 1$) is:

Predict wavelengths with the Rydberg formula

Turn any transition $n_i \rightarrow n_f$ into a wavelength, and sort the named series by their final level.

The Hydrogen Spectrum and the Rydberg Formula

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From theory to precise prediction

We just saw that Bohr's postulates give hydrogen energy levels E_n = −13.6/n² eV and quantised orbits. That raises a question: how do these energy levels connect to the actual wavelengths of light we observe? This card answers it → the Rydberg formula links any transition (n_i → n_f) directly to a measurable wavelength, and the five named series group lines by their final level.

The energy levels predict the wavelengths of hydrogen's spectral lines. The Rydberg formula gives the wavelength of any transition:

$$\dfrac{1}{\lambda} = R_H \left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$$

where $R_H = 1.097 \times 10^7$ m⁻¹ is the Rydberg constant, $n_f$ is the final (lower) level, and $n_i > n_f$ is the initial (higher) level.

The spectral lines group into series named after their discoverers:

Lyman series: $n_f = 1$ (ultraviolet), transitions to ground state
Balmer series: $n_f = 2$ (visible), transitions to first excited state
Paschen series: $n_f = 3$ (infrared), transitions to second excited state
Brackett series: $n_f = 4$ (infrared)
Pfund series: $n_f = 5$ (infrared)

The Balmer series is historically significant because its visible lines were observed and measured before Bohr's theory. Bohr's model not only explained why these lines exist but predicted their wavelengths to remarkable accuracy.

However, Bohr's model has important limitations:

It only works accurately for hydrogen (single-electron systems).
It cannot explain fine structure (small splitting of lines).
It cannot predict the relative intensities of spectral lines.
It does not explain why angular momentum is quantised, this requires quantum mechanics (the Schrödinger equation).

Key Formulae, Bohr Model

$E_n = -13.6/n^2$ eV, hydrogen energy levels

$r_n = n^2 a_0$, Bohr orbit radii ($a_0 = 0.529$ Å)

$1/\lambda = R_H(1/n_f^2 - 1/n_i^2)$, Rydberg formula

$E_\text{photon} = hf = hc/\lambda$, photon energy

$E_\text{ionisation} = 13.6$ eV; $R_H = 1.097\times10^7$ m⁻¹; $h = 6.626\times10^{-34}$ J·s; 1 eV $= 1.6\times10^{-19}$ J

Figure 2, Hydrogen spectral series arranged by region of the electromagnetic spectrum. Only the Balmer series falls in the visible range ($n_f = 2$). The Lyman series (UV) requires the largest energy transitions; far-IR series involve smaller gaps.

Text description

Each series groups the transitions that end on the same final level n_f. In order of increasing wavelength:
Lyman, n_f = 1: ultraviolet. Balmer, n_f = 2: visible. Paschen, n_f = 3: near infrared. Brackett and beyond, n_f ≥ 4: far infrared.
Shorter wavelength means a larger energy transition.

Stop and check

Calculate the shortest wavelength in the Balmer series. What electron transition produces the shortest wavelength in a series?

Rydberg formula: 1/λ = R_H(1/n_f² − 1/n_i²), R_H = 1.097×10⁷ m⁻¹. Lyman: n_f=1 (UV); Balmer: n_f=2 (visible); Paschen: n_f=3 (IR). Shortest λ in any series = n_i=∞→n_f (largest gap). Bohr's model only works for single-electron atoms, it cannot explain fine structure, intensities, or why L is quantised.

Record the Rydberg formula and series names with their n_f values, the Balmer = visible link is a classic MC trap.

The shortest wavelength in the Lyman series corresponds to the transition $n=2 \rightarrow n=1$.

The Balmer series falls in the visible part of the electromagnetic spectrum because $n_f = 2$.

Bohr's model accurately predicts the spectrum of helium because it has only two protons.

Avoid the standard Bohr traps

Check the direction of a transition, the shortest-wavelength limit in a series, and where the model stops working.

Misconceptions, Final Check

HSC Tip, Bohr Model Calculations

When using the Rydberg formula, $n_i > n_f$ for emission (electron drops down) and $n_i < n_f$ for absorption (electron jumps up). The result $1/\lambda$ must be positive for emission. The shortest wavelength in a series corresponds to the largest energy transition: $n_i = \infty \rightarrow n_f$. The longest wavelength corresponds to the smallest gap: $n_i = n_f + 1 \rightarrow n_f$. Memorise $E_1 = -13.6$ eV and $E_n = E_1/n^2$. Converting: 1 eV $= 1.6\times10^{-19}$ J. The Bohr model is only valid for single-electron systems.

Wrong: "Shorter wavelength means the electron fell from a higher level to a lower level (e.g., $n=3\rightarrow n=1$ gives longer $\lambda$ than $n=2\rightarrow n=1$)."

Right: Shorter wavelength = higher energy photon = larger $\Delta E$. The transition $n=3\rightarrow n=1$ has a larger energy gap than $n=2\rightarrow n=1$ and therefore produces a shorter wavelength.

Wrong: "Bohr's model works for all atoms, it just needs more orbits for heavier elements."

Right: Bohr's model only works for single-electron systems (hydrogen-like ions such as He⁺). Multi-electron atoms require quantum mechanics because electron–electron interactions are not accounted for in Bohr's postulates.

Three of these statements about Bohr's model are correct. Pick the odd one out.

Calculate spectral lines

Run four Rydberg calculations, naming the series and the region of the spectrum each time.

Activities

Activity 1, Spectral Line Calculations

ApplyBand 4

Practice using the Rydberg formula and energy levels

Use the Rydberg formula to calculate the wavelength of the H$\alpha$ line ($n=3 \rightarrow n=2$). State which spectral series it belongs to and whether it is visible.
Calculate the wavelength of the Lyman-$\alpha$ line ($n=2 \rightarrow n=1$). Is it UV, visible, or IR?
Calculate the shortest wavelength in the Balmer series. What transition produces it?
An electron in hydrogen drops from $n=5$ to $n=2$. Calculate the energy of the emitted photon in eV, then convert to joules. Find the wavelength and identify the series.

Evaluate Bohr's model

Set each postulate against what classical physics predicted, then judge why the model still mattered.

Activity 2, Evaluating Bohr's Model

AnalyseBand 5

Compare Bohr's model with classical physics and later quantum mechanics

State Bohr's three postulates. For each, explain what classical physics would have predicted instead.
Calculate the radius of the $n=3$ orbit in hydrogen. How does this compare to the ground-state Bohr radius $a_0$?
A student claims: "Because the Balmer series is in the visible range, the Lyman series must also be visible." Identify the flaw and calculate the Lyman-$\alpha$ wavelength to support your answer.
Explain why Bohr's model was a significant achievement even though it was later replaced by quantum mechanics.

Assess7 marks

Assess the limitations of Bohr's model of the atom. (7 marks)

Credit what the model achieved first

You cannot assess limitations without establishing what is being limited. A model with no successes has no interesting failures.

Bohr's model succeeded where Rutherford's could not: by postulating that electrons occupy only certain allowed orbits with quantised energies, and radiate only when moving between them, it explained why atoms are stable and predicted the wavelengths of the hydrogen line spectrum, including the Balmer series, with remarkable accuracy.

Naming the specific success — the hydrogen spectrum — is what makes the later failures significant rather than expected.

Analyse6 marks

Analyse how the quantised energy levels of the hydrogen atom account for the discrete lines of its emission spectrum, and for the fact that the lines in the Balmer series crowd together towards a short-wavelength limit. (6 marks)

State the structural claim the whole answer rests on

Discrete lines require discrete levels. Say so before any equation.

An electron in a hydrogen atom can occupy only certain allowed energy levels, so the energy it can lose in a transition is restricted to the differences between those levels rather than taking any value.

"Differences between levels" is the key phrase. A spectrum is built from gaps, not from the levels themselves.

Independent practice

Show what you can do without prompts

Complete a shuffled question-bank set, then write full Bohr model responses in HSC style.

Quick recall, Bohr's model and hydrogen spectrum

+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 8 marks

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ApplyBand 4(3 marks) 1. An electron in hydrogen transitions from $n=4$ to $n=2$. (a) State which spectral series this line belongs to. (b) Calculate the wavelength of the emitted photon using the Rydberg formula. (c) State whether this wavelength is visible, UV, or IR.

1 mark: correct series · 1 mark: correct wavelength calculation · 1 mark: correct region identification

AnalyseBand 6(5 marks) 2. (a) State Bohr's three postulates for the hydrogen atom. (b) Explain why Bohr postulated that electrons in allowed orbits do not radiate, what problem was he solving? (c) Calculate the ionisation energy of hydrogen in joules. (d) Calculate the wavelength of the shortest line in the Lyman series. (e) Explain one limitation of Bohr's model that required a more complete theory to resolve.

1 mark per part (a–e)

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (3 marks): (a) Balmer series ($n_f = 2$) (1 mark). (b) $1/\lambda = R_H(1/n_f^2 - 1/n_i^2) = 1.097\times10^7(1/4 - 1/16) = 1.097\times10^7 \times (4/16 - 1/16) = 1.097\times10^7 \times 3/16 = 2.057\times10^6$ m⁻¹; $\lambda = 1/2.057\times10^6 \approx 486$ nm (1 mark). (c) Visible, this is the H$\beta$ line (blue-green) in the Balmer series (1 mark).

Q2 (5 marks): (a) Postulate 1: electrons orbit in allowed circular orbits without radiating energy; Postulate 2: angular momentum is quantised, $L = n\hbar$; Postulate 3: photons are emitted or absorbed only when electrons transition between levels, $E_\text{photon} = hf$ (1 mark). (b) Classical electromagnetism predicts that an accelerating charge radiates energy. An orbiting electron continuously accelerates (changes direction), so it would continuously radiate, lose energy, and spiral into the nucleus in ~$10^{-10}$ s, Bohr's first postulate simply forbids this (1 mark). (c) Ionisation energy = 13.6 eV $= 13.6 \times 1.6\times10^{-19} = 2.18\times10^{-18}$ J (1 mark). (d) Shortest Lyman: $n_i = \infty \rightarrow n_f = 1$; $1/\lambda = R_H(1/1 - 0) = 1.097\times10^7$ m⁻¹; $\lambda = 91.2$ nm (UV) (1 mark). (e) Bohr's model cannot explain the spectra of multi-electron atoms (e.g., helium) because it ignores electron–electron repulsion; it also cannot explain the fine structure of spectral lines or why angular momentum is quantised, quantum mechanics (Schrödinger equation) resolves these (1 mark).

Final step

Retrieve, reflect and finish

Check what stuck, revisit your opening prediction about hydrogen's four visible lines and record the method you will reuse.

Check what actually stuck

How did your thinking change?

At the start you were asked about Niels Bohr's 1913 discovery: hydrogen emits only four visible wavelengths (656 nm, 486 nm, 434 nm, 410 nm), which Bohr's formula Eₙ = −13.6/n² eV predicted to within 0.02%. Review your predictions:

Did you predict hydrogen emits only specific wavelengths because electrons can only jump between discrete quantised energy levels? Correct, Bohr's 1913 postulate that angular momentum is restricted to L = nh/2π directly produces the four Balmer lines.
Did you predict discrete energy levels rather than a continuum? Correct, Bohr's quantised orbits imply Eₙ = −13.6/n² eV; only these specific energy differences correspond to observable photon wavelengths.
Did you predict using $E = hc/\lambda$ to relate energy difference to wavelength? Correct, this is the fundamental relationship Bohr used to connect his energy level formula to the measured spectral lines.

Extend: A hydrogen-like ion He⁺ has nuclear charge $Z = 2$. The energy levels are $E_n = -Z^2 \times 13.6/n^2$ eV. (a) Calculate $E_1$ for He⁺. (b) Calculate the wavelength of the transition $n=3 \rightarrow n=1$ for He⁺. (c) Compare this wavelength to the equivalent Lyman transition in hydrogen, explain why the He⁺ photon has shorter wavelength.

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