IB DP Physics: Atomic structure: Rutherford scattering, spectra and the Bohr model (DP Physics)

This guide covers sub-topic E.1 (Structure of the atom) from the IB Physics guide, first assessment 2025. The core ideas are SL and HL. The additional higher level (AHL) understandings are HL only: nuclear radius and density, deviations from Rutherford scattering, distance of closest approach and the Bohr model. For quick recall use the revision notes, then test yourself with the practice questions. Find your weak spots first with the free 10-minute diagnostics.

Related pages: the course hub and the printable checklist.

Values used in the worked examples: c = 3.00 × 10⁸ m s⁻¹, h = 6.63 × 10⁻³⁴ J s, e = 1.60 × 10⁻¹⁹ C, electron mass 9.11 × 10⁻³¹ kg, atomic mass unit u = 1.66 × 10⁻²⁷ kg and Coulomb constant k = 8.99 × 10⁹ N m² C⁻².

What this unit covers

Part	What you must be able to do	Tier
E.1	Describe the Geiger-Marsden-Rutherford experiment and how it led to the nucleus (qualitative at SL)	SL and HL
E.1	Use nuclear notation with nucleon number A, proton number Z and chemical symbol X	SL and HL
E.1	Explain how emission and absorption spectra are evidence for discrete energy levels	SL and HL
E.1	Describe photon emission and absorption in atomic transitions and use E = hf	SL and HL
E.1	Use spectra to identify chemical composition	SL and HL
E.1	Use R = R₀A^(1/3) and discuss nuclear density	HL only
E.1	Explain deviations from Rutherford scattering at high energy	HL only
E.1	Find the distance of closest approach in a head-on collision	HL only
E.1	Use the Bohr levels E = −13.6/n² eV and angular momentum mvr = nh/2π	HL only

You do not need to recall chemical symbols. If a symbol matters, the question will give it.

The Geiger-Marsden-Rutherford experiment

Before the experiment, a common picture of the atom was a diffuse positive sphere with electrons spread through it. In the experiment, a narrow beam of alpha particles was aimed at a very thin gold foil in an evacuated chamber. A detector, a screen that flashes where a particle lands, was moved around the foil to count how many particles arrived at each angle.

Observations and what each one shows:

• Almost all alpha particles passed straight through or were deflected by tiny angles. The atom is largely empty space.
• A small fraction were deflected through large angles. Something in the atom carries a large positive charge and exerts a strong repulsive electric force on the positive alpha particle.
• A very small number came almost straight back. The charge must be concentrated in a region far smaller than the atom, and that region must be massive. A light, spread-out charge could not reverse a fast alpha particle.

The conclusion is the nuclear model: a tiny, dense, positively charged nucleus holds nearly all the atom’s mass, with electrons occupying the space around it. Only a qualitative account is needed at SL.

The foil is thin so most alpha particles meet at most one nucleus, and the chamber is evacuated so air does not scatter them.

Nuclear notation

A nuclide is written as ᴬ_Z X, with the nucleon number A at the top left, the proton number Z at the bottom left and the chemical symbol X.

• A = number of protons + number of neutrons (the nucleon number).
• Z = number of protons, which also fixes the element.
• Neutron number N = A − Z.

Worked example 1. A nuclide of strontium is written ⁸⁸₃₈Sr. State its numbers of protons, neutrons and (if neutral) electrons.

Protons = Z = 38. Neutrons = A − Z = 88 − 38 = 50. A neutral atom has as many electrons as protons, so 38 electrons.

Spectra and discrete energy levels

A hot, dense solid gives a continuous spectrum: every wavelength in a band. A low-pressure gas that has been excited, for example by an electric discharge, gives an emission spectrum: bright lines at particular wavelengths on a dark background. White light passed through a cool gas gives an absorption spectrum: dark lines on a continuous background, at the same wavelengths that the gas emits.

Line spectra are evidence for discrete energy levels. If an atom could have any energy, it could emit any photon energy and the spectrum would be continuous. Sharp lines mean the atom can only exist with certain energies, so the possible energy differences, and therefore photon energies, are fixed.

Photons in transitions

• Emission: an electron falls from a higher level to a lower level and the atom emits one photon. The photon energy equals the difference between the levels.
• Absorption: an atom takes in a photon only if the photon energy equals the gap between its current level and a higher level. Photons with other energies pass through.

The relationship is E = hf, so for levels E_upper and E_lower:

hf = E_upper − E_lower, and with c = fλ, λ = hc/(E_upper − E_lower).

Widen the gap and the photon frequency rises while its wavelength shrinks. Energies on level diagrams are usually in electronvolts (eV). Convert with 1 eV = 1.60 × 10⁻¹⁹ J before using h in J s.

Worked example 2. An atom has levels at −5.4 eV, −3.1 eV, −1.2 eV and 0 (ionised). Find the frequency and wavelength of the photon emitted when an electron falls from −1.2 eV to −5.4 eV, and the number of different emission lines possible from four levels.

Energy gap = −1.2 − (−5.4) = 4.2 eV = 4.2 × 1.60 × 10⁻¹⁹ = 6.72 × 10⁻¹⁹ J.

f = E/h = 6.72 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 1.01 × 10¹⁵ Hz.

λ = c/f = 3.00 × 10⁸ / 1.01 × 10¹⁵ = 2.96 × 10⁻⁷ m (296 nm, ultraviolet).

With four levels, every pair can give a line: 4 × 3 / 2 = 6 lines.

Composition from spectra

Each element has its own set of energy levels, so its set of line wavelengths is like a fingerprint. Matching the lines in a spectrum to laboratory spectra of known elements shows which elements are present. In a star, light from the hot interior passes through cooler outer gas. Atoms in that gas absorb their own characteristic wavelengths, leaving dark lines in the star’s spectrum. The pattern of dark lines identifies the elements in the star’s atmosphere.

Nuclear size and density (HL only)

Scattering and other experiments show that nuclear radius grows with the nucleon number:

R = R₀ A^(1/3)

R₀ is a constant, taken here as 1.2 fm (1 fm = 10⁻¹⁵ m). Volume goes as R³, so nuclear volume is proportional to A: every nucleon claims roughly the same share of space, as if the nucleons were packed together.

Worked example 3. Find the radius of a nucleus with A = 512, and the ratio of the radii of nuclei with A = 216 and A = 27.

R = 1.2 × 512^(1/3) = 1.2 × 8 = 9.6 fm.

Ratio = (216/27)^(1/3) = 8^(1/3) = 2. A nucleus with eight times the nucleons has only twice the radius.

Nuclear density. Take the mass of a nucleus as A × u. Then:

ρ = A u / (4/3 π R³) = A u / (4/3 π R₀³ A) = 3u / (4π R₀³)

The A cancels, so the density is the same for all nuclei:

ρ = 3 × 1.66 × 10⁻²⁷ / (4π × (1.2 × 10⁻¹⁵)³) = 2.3 × 10¹⁷ kg m⁻³.

The implication to state: nuclear matter is closely packed, and the density is far greater than that of ordinary solids.

Closest approach and high-energy scattering (HL only)

At the energies used here only the electric force acts, so energy conservation is enough.

In a head-on collision the alpha particle slows as it climbs the electric potential energy hill of the nucleus. It stops for an instant at the closest point, where its initial kinetic energy has all been stored as electric potential energy. Treat the nucleus as fixed and the particles as point charges:

E_k = k (2e)(Ze) / d, so d = 2kZe² / E_k

Worked example 4. A 6.0 MeV alpha particle is fired head-on at a nickel nucleus (Z = 28, A = 58). Find the distance of closest approach and compare it with the nuclear radius.

E_k = 6.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 9.6 × 10⁻¹³ J.

d = 2 × 8.99 × 10⁹ × 28 × (1.60 × 10⁻¹⁹)² / 9.6 × 10⁻¹³ = 1.34 × 10⁻¹⁴ m (13.4 fm).

R = 1.2 × 58^(1/3) = 4.6 fm. Since d is almost three times R, the alpha particle never reaches the nucleus, and only the electric force acts. This is why Rutherford’s prediction held.

Deviations at high energy. If the alpha energy is increased, d falls. When d approaches the nuclear radius, the alpha particle comes within range of the strong nuclear force, which is attractive and acts between nucleons. The measured scattering then departs from the prediction that assumes only electric repulsion. Setting d equal to R gives a rough energy at which the departure starts:

E_k = 2kZe² / R, which for nickel is 2 × 28 × 1.44 MeV fm / 4.6 fm ≈ 17 MeV

(using ke² = 1.44 MeV fm, which follows from the constants above).

The Bohr model for hydrogen (HL only)

Bohr proposed that the electron in hydrogen moves in circular orbits, but only certain orbits are allowed. The condition comes from quantised angular momentum:

mvr = nh/2π, with n = 1, 2, 3 …

Only orbits where the electron’s angular momentum is a whole-number multiple of h/2π exist. Each allowed orbit has a fixed energy:

E = −13.6 / n² eV

The energy is negative because the electron is bound; zero is the ionised state. n = 1 is the ground state, with energy −13.6 eV, and the levels crowd together as n increases. Light is emitted or absorbed when the electron changes level, with hf equal to the energy difference.

Worked example 5. Find the wavelength of the photon emitted in the transition n = 4 to n = 2 in hydrogen.

E₄ = −13.6/16 = −0.85 eV. E₂ = −13.6/4 = −3.40 eV.

ΔE = −0.85 − (−3.40) = 2.55 eV = 2.55 × 1.60 × 10⁻¹⁹ = 4.08 × 10⁻¹⁹ J.

λ = hc/ΔE = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 4.08 × 10⁻¹⁹ = 4.88 × 10⁻⁷ m (488 nm, in the visible range).

Worked example 6. In the ground state the orbit radius is 5.29 × 10⁻¹¹ m. Find the electron speed.

mvr = h/2π, so v = h / (2π m r) = 6.63 × 10⁻³⁴ / (2π × 9.11 × 10⁻³¹ × 5.29 × 10⁻¹¹) = 2.19 × 10⁶ m s⁻¹.

Absorption works in reverse: a 10.2 eV photon is absorbed by a ground-state hydrogen atom (n = 1 to n = 2), but an 11.0 eV photon is not absorbed, because 11.0 eV does not match any gap. The model suits hydrogen, not atoms with more electrons.

Common errors

• Writing the photon energy as the energy of one level rather than the difference between two levels.
• Using eV in E = hf without converting to joules.
• Saying the experiment showed “electrons orbit the nucleus”. It showed a small, dense, positive nucleus.
• Saying an absorbed photon can have any energy. It must match a gap exactly (otherwise it passes through).
• Treating the distance of closest approach as the nuclear radius. It is only an upper limit on the radius when the alpha particle has not reached the nucleus.
• Giving the Bohr energy as positive. Bound levels are negative.

Official syllabus

IB Physics guide, first assessment 2025, published by the International Baccalaureate Organization: sub-topic E.1 Structure of the atom.