Is the Rydberg equation on the HSC data sheet?

Yes. The 2019 data sheet gives 1/λ = R(1/nf^2 − 1/ni^2) with R = 1.097 × 10^7 m^−1, plus E = hf, v = fλ and λ = h ÷ mv. The energy levels En = −13.6 ÷ n^2 eV are not on the sheet; questions that need them usually give the values.

What is the difference between an emission spectrum and an absorption spectrum?

An emission spectrum comes from a hot, low-pressure gas and shows bright lines only. An absorption spectrum is a continuous spectrum with dark lines, made when light passes through a cooler gas that absorbs certain photons. For the same element the lines are at the same wavelengths.

Why are the Balmer lines visible?

Because they all end on n = 2. The energy gaps from higher levels down to n = 2 run from 1.89 eV up to 3.40 eV, which corresponds to wavelengths from 656 nm down to 365 nm. The first four lines (656, 486, 434 and 410 nm) are visible; the rest crowd together just past the violet end, in the near ultraviolet. Drops to n = 1 are much bigger (ultraviolet).

What are the limitations of the Bohr model?

It only works for hydrogen-like (one-electron) atoms, can't explain the different brightness of lines or the splitting of lines, and adds quantum rules to classical orbits without a reason for them. de Broglie's matter waves later explained the orbits, and Schrödinger's model replaced fixed orbits with orbitals.

How did de Broglie explain Bohr's orbits?

He treated the electron as a wave with λ = h ÷ mv. A stable orbit is one where a whole number of these wavelengths fits around the circumference (2πr = nλ), forming a standing wave. That condition is exactly Bohr's quantised angular momentum.

White light passes through cool hydrogen gas and then a spectroscope. What is seen?

A continuous spectrum with dark lines at hydrogen's emission wavelengths.

Bright lines only, at hydrogen's emission wavelengths.

A continuous spectrum with dark lines at wavelengths different from hydrogen's emission lines.

A continuous spectrum with no lines, because the gas is cool.

How did de Broglie's idea explain why only certain electron orbits are allowed?

Only orbits that fit a whole number of electron wavelengths make a stable standing wave.

Electrons in other orbits are pulled into the nucleus by the strong force.

Only orbits with a whole number of photons in them can exist.

Electrons radiate energy until they reach an orbit where they stop.

HSC Physics · Module 8 · From the Universe to the Atom · 13 min read

The Bohr model and hydrogen spectra

Pass electricity through hydrogen gas and it glows, but not in a smooth rainbow. Through a spectroscope you see a few sharp coloured lines: red, blue-green and violet. Bohr's model of the atom explains exactly where those lines are, and it is the centre of a set of HSC questions on atomic models, spectra and the Rydberg equation.

This guide covers why Rutherford's atom needed fixing, Bohr's energy levels, how a jump between levels makes one photon, the Balmer series, emission and absorption spectra, and how de Broglie's matter waves explain Bohr's orbits. Then come the common mistakes, a quick quiz and a full worked example.

What you need to know

Account for how emission and absorption spectra each arise, and compare each with the continuous spectrum of a black body.
Assess where the atomic models of Rutherford and Bohr fall short.
Investigate hydrogen's line emission spectrum to examine its Balmer series.
Relate hydrogen's line emission spectrum, in words and in numbers, to its quantised energy levels and to conservation of energy, using E = hf, E = hc ÷ λ and 1/λ = R(1/n_f² − 1/n_i²).
Investigate the matter waves de Broglie proposed (λ = h ÷ mv), and the experiments that back them up.

The equations (from the 2019 HSC data sheet unless marked):

1/λ = R(1/n_f² − 1/n_i²)Rydberg equation for hydrogen, R = 1.097 × 10⁷ m⁻¹. n_f is the lower (final) level, n_i the higher (initial) level.

E = hfThe photon's energy equals the energy gap: hf = E_i − E_f.

E = hc ÷ λ not on the data sheetE = hf with c = fλ (v = fλ on the sheet). Turns a line's wavelength into a photon energy.

E_n = −13.6 ÷ n² eV not on the data sheetEnergy levels of hydrogen: −13.6 eV for n = 1, −3.40 eV for n = 2, and so on up to 0 at n = ∞.

λ = h ÷ mvde Broglie wavelength of a moving particle, such as an electron.

Dot points: PH-M8-04, PH-M8-10, PH-M8-11, PH-M8-12, PH-M8-13 (2017 Physics syllabus, HSC 2026 and 2027), in our own words.

Mini lesson

Rutherford's atom and its problem

Geiger and Marsden's gold-foil results led Rutherford (1911) to a nuclear atom: a tiny, dense, positive nucleus with electrons orbiting around it, and mostly empty space.

It had two big gaps. First, an orbiting electron is always accelerating (its direction keeps changing), and classical electromagnetism says an accelerating charge radiates energy. The electron should spiral into the nucleus in a tiny fraction of a second, so the atom should not be stable. Second, a spiralling electron would give out a smooth spread of frequencies, not the sharp lines that hydrogen actually shows.

Bohr's postulates and hydrogen's energy levels

Bohr (1913) kept Rutherford's nucleus but added rules that classical physics could not justify:

Stationary states. The electron can only be in certain orbits, each with a fixed energy, and while it stays in one it does not radiate.
Photons from jumps. When the electron drops from a higher level to a lower one, it emits one photon whose energy equals the gap: hf = E_i − E_f. Absorbing a photon of exactly that energy lifts it back up. This is conservation of energy.
Quantised angular momentum. The allowed orbits are those where mvr = nh ÷ 2π, with n = 1, 2, 3 …
Result for hydrogen. These rules give energy levels E_n = −13.6 ÷ n² eV. The energies are negative because the electron is bound: 13.6 eV is needed to remove it from the ground state (n = 1). The levels crowd together as n increases, towards 0 at n = ∞.

Hydrogen's levels (not to scale). Drops that end on n = 2 are the Balmer series: 656, 486 and 434 nm, all visible. The drop to n = 1 is much bigger, so it is ultraviolet (the Lyman series).

The Balmer series and the Rydberg equation

Every hydrogen line comes from one jump, so the wavelength follows from the two levels. The Rydberg equation does this directly: 1/λ = R(1/n_f² − 1/n_i²), where n_f is the level the electron lands on and n_i the one it started from. Balmer found the pattern for the visible lines in 1885 by fitting the data; Bohr's model explained why it works.

The visible lines all end on n_f = 2 (the Balmer series). For n_i = 3: 1/λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.524 × 10⁶ m⁻¹, so λ = 656 nm, the red line. Drops to n = 1 (Lyman) are ultraviolet and drops to n = 3 (Paschen) are infrared.

To get the photon energy, use E = hc ÷ λ, or subtract the two levels: E₃ − E₂ = −1.51 − (−3.40) = 1.89 eV for the red line. The two methods agree, which is a good check.

Emission and absorption spectra

A hot, dense object such as a filament or a star's surface gives a continuous spectrum: every wavelength, in a black body curve. A hot, low-pressure gas gives an emission spectrum: bright lines at only the wavelengths its electrons emit as they drop between levels.

Put a cooler gas in front of a continuous source and you get an absorption spectrum: the full rainbow with dark lines. Atoms in the gas absorb only photons whose energy exactly matches a gap, lifting electrons up. They re-emit that energy, but in all directions, so much less of it reaches you along the original line of sight. The dark lines sit at the same wavelengths as that gas's emission lines, which is how astronomers identify elements in stars.

Same gas, same wavelengths. Emission: bright lines (peaks) on nothing. Absorption: dark lines (dips) in a continuous spectrum, where photons matching an energy gap were absorbed and re-emitted in other directions.

de Broglie: why the orbits are stable

Bohr's quantised orbits were a rule without a reason. In 1924 de Broglie proposed that moving particles have a wavelength, λ = h ÷ mv. An electron in an orbit is then a wave wrapped around the nucleus. It only lasts if a whole number of wavelengths fits the circumference, 2πr = nλ, making a standing wave. Any other orbit would interfere with itself and cancel out. Putting λ = h ÷ mv into 2πr = nλ gives Bohr's rule mvr = nh ÷ 2π.

The numbers fit. In the ground state the electron moves at about 2.19 × 10⁶ m s⁻¹, so λ = 6.626 × 10⁻³⁴ ÷ (9.109 × 10⁻³¹ × 2.19 × 10⁶) = 3.32 × 10⁻¹⁰ m. That equals the circumference of the n = 1 orbit (radius 5.29 × 10⁻¹¹ m): exactly one wavelength. Davisson and Germer's electron diffraction experiment (1927) confirmed that electrons really do behave as waves.

A standing electron wave for n = 4: four whole wavelengths fit the orbit's circumference. Only orbits where 2πr = nλ survive, which is Bohr's quantisation explained.

Where Bohr's model falls short

Bohr's model was a big step: it explained hydrogen's lines and the Rydberg constant from basic constants. But it has clear limits, which is why Schrödinger's wave model replaced it.

It only works for hydrogen and other one-electron ions. It fails for helium and every larger atom.
It cannot explain why some lines are brighter than others, or why some lines split into several close lines (fine structure, and splitting in a magnetic field, the Zeeman effect).
It mixes ideas: classical orbits, with quantum rules simply added on. It gives no reason for the stationary states not radiating, or for quantised angular momentum (de Broglie later supplied one).
It pictures electrons as particles in fixed, flat orbits; the modern model describes them as probability clouds (orbitals).

Common mistakes

Swapping n_f and n_i in the Rydberg equation, which gives a negative wavelength.n_f is the lower level the electron ends on. For emission n_i > n_f, so the bracket is positive. Balmer lines always have n_f = 2.
Saying the Balmer series ends on n = 1, or that every hydrogen line is visible.Balmer lines end on n = 2; the first four (656, 486, 434 and 410 nm) are visible. Drops to n = 1 (Lyman) are ultraviolet; drops to n = 3 (Paschen) are infrared.
Describing absorption lines as wavelengths the gas 'blocks' completely, with no mechanism.Electrons absorb only photons with energy exactly equal to a gap, then re-emit in all directions. Less light continues forward at that wavelength, so the line is dark compared with its surroundings.
Crediting de Broglie with the energy levels, or Bohr with matter waves.Bohr postulated the quantised orbits (1913). de Broglie explained them a decade later: an electron's matter wave must fit the orbit as a whole number of wavelengths.
Writing 'the Rutherford–Bohr model' as if it were one model, or saying Bohr fixed every problem.Treat them separately: Rutherford's electrons should spiral in and radiate a continuous spectrum; Bohr's stationary states fix that for hydrogen only.
Using the proton mass, or forgetting to convert eV to J, in λ = h ÷ mv for an electron.Use m_e = 9.109 × 10⁻³¹ kg and v in m s⁻¹. If you are given kinetic energy in eV, convert to joules before finding v.

Worked example

A student looks at a hydrogen discharge tube through a spectroscope and measures a violet line at 434 nm. All the visible lines of hydrogen end on the n = 2 level.

Use the Rydberg equation to identify the energy level the electron started from, and calculate the energy of the emitted photon in electronvolts.

(4 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. The red line, 656 nm, is the drop from n = 3, and the blue-green line, 486 nm, is from n = 4. 434 nm is shorter still, so the photon has more energy and the drop is bigger: I predict n_i = 5. The red photon is about 1.9 eV and the most any Balmer photon can have is 3.4 eV (from n = ∞), so I expect a little under 3 eV.

K · Knowns

I sketch the two levels with the drop and the wavelength on it, then write the values in SI units. The nanometres go to metres straight away.

λ = 434 nm = 434 × 10⁻⁹ m, n_f = 2
R = 1.097 × 10⁷ m⁻¹, h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹
1 eV = 1.602 × 10⁻¹⁹ J

Knowns on the sketch: λ = 434 nm and n_f = 2. Unknowns: the starting level n_i and the photon energy E.

N · Not known

Two things to find: a whole number for the starting level, then an energy in eV.

n_i = ? (a whole number), E = ? eV

O · Organise

I write the Rydberg equation and rearrange it for 1/n_i² before putting numbers in. For the energy I use E = hc ÷ λ, then convert joules to eV.

1/λ = R(1/n_f² − 1/n_i²) → 1/n_i² = 1/n_f² − 1/(Rλ)
E = hc ÷ λ, then E (eV) = E (J) ÷ 1.602 × 10⁻¹⁹

W · Work it out

I keep four figures in the middle so the whole number shows up clearly at the end. n_i must come out very close to a whole number; if it didn't, I'd suspect I had swapped n_f and n_i.

1/(Rλ) = 1 ÷ (1.097 × 10⁷ × 434 × 10⁻⁹) = 0.21004
1/n_i² = 1/2² − 0.2100 = 0.25000 − 0.21004 = 0.03996
n_i² = 1 ÷ 0.03996 = 25.03, so n_i = 5.00 → n_i = 5
E = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (434 × 10⁻⁹) = 4.580 × 10⁻¹⁹ J
E = (4.580 × 10⁻¹⁹) ÷ (1.602 × 10⁻¹⁹) = 2.859 eV

U · Units

The level number has no unit. The energy was asked for in eV, and the wavelength was given to three significant figures, so I give three.

n_i = 5, E = 2.86 eV

D · Direction (sense check)

Energy is a scalar, so instead of a direction I check it against the energy levels, a completely separate route. E₅ − E₂ should match.

E₅ = −13.6 ÷ 25 = −0.544 eV, E₂ = −13.6 ÷ 4 = −3.40 eV
E₅ − E₂ = −0.544 − (−3.40) = 2.856 eV ≈ 2.86 eV ✓

The electron started in level n = 5 and dropped to n = 2, emitting a photon of energy 2.86 eV (4.58 × 10⁻¹⁹ J).

Check against the prediction. I predicted n_i = 5 and a little under 3 eV. The working gives n_i = 5 and 2.86 eV, between the red line's 1.9 eV and the 3.4 eV limit, and the energy-level route gives the same answer.

HSC advice

What markers keep saying

Know Bohr's postulates well enough to state them: fixed-energy stationary states, no radiation while in one, and a photon of energy hf = E_i − E_f when the electron changes level. Link the photon's energy to conservation of energy.NESA marking feedback, 2020, summarised
In Rydberg calculations, show every substitution, and remember the equation gives 1/λ: invert it to get λ before using c = fλ to find a frequency. Put the lower level in as n_f.NESA marking feedback, 2024, summarised
To explain an absorption line, say which photons are absorbed (only those whose energy matches a gap between levels) and why the line looks dark (the energy is re-emitted in all directions, so less reaches the observer).NESA marking feedback, 2024, summarised
Keep Bohr and de Broglie apart: Bohr proposed quantised orbits, and de Broglie's standing electron waves explained why only those orbits are stable. When you describe the electron as a wave, include its wavelength, λ = h ÷ mv.NESA marking feedback, 2019, summarised
When comparing atomic models, describe Rutherford's and Bohr's separately and focus on what each says about the electrons, rather than on the nucleus.NESA marking feedback, 2025, summarised

Summaries in our own words. Read NESA's own feedback with the exam packs: 2019 · 2020 · 2024 · 2025.

How often it is examined

The Bohr model and hydrogen spectra questions were worth 32 marks across the seven HSC Physics exams from 2019 to 2025 (about 4.6 marks a paper), and appeared in 6 of 7 papers.

Marks per paper (Module 8, Quantum Mechanical Nature of the Atom). Source: CortexHSC's own tagging of the NESA HSC Physics exams 2019–2025, each question counted under its main topic, so treat the numbers as close, not exact.

FAQ

Is the Rydberg equation on the HSC data sheet?: Yes. The 2019 data sheet gives 1/λ = R(1/n_f² − 1/n_i²) with R = 1.097 × 10⁷ m⁻¹, plus E = hf, v = fλ and λ = h ÷ mv. The energy levels E_n = −13.6 ÷ n² eV are not on the sheet; questions that need them usually give the values.
What is the difference between an emission spectrum and an absorption spectrum?: An emission spectrum comes from a hot, low-pressure gas and shows bright lines only. An absorption spectrum is a continuous spectrum with dark lines, made when light passes through a cooler gas that absorbs certain photons. For the same element the lines are at the same wavelengths.
Why are the Balmer lines visible?: Because they all end on n = 2. The energy gaps from higher levels down to n = 2 run from 1.89 eV up to 3.40 eV, which corresponds to wavelengths from 656 nm down to 365 nm. The first four lines (656, 486, 434 and 410 nm) are visible; the rest crowd together just past the violet end, in the near ultraviolet. Drops to n = 1 are much bigger (ultraviolet).
What are the limitations of the Bohr model?: It only works for hydrogen-like (one-electron) atoms, can't explain the different brightness of lines or the splitting of lines, and adds quantum rules to classical orbits without a reason for them. de Broglie's matter waves later explained the orbits, and Schrödinger's model replaced fixed orbits with orbitals.
How did de Broglie explain Bohr's orbits?: He treated the electron as a wave with λ = h ÷ mv. A stable orbit is one where a whole number of these wavelengths fits around the circumference (2πr = nλ), forming a standing wave. That condition is exactly Bohr's quantised angular momentum.

Related guides

Explore more of CortexHSC

CortexHSC is a free HSC Physics study app built by a NSW Physics teacher. Pick up where this guide leaves off:

Free to start, no card needed. Some features need a free account; a few are part of a paid pass.