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E.1

Structure of the atom

Theme E · Nuclear and quantum physics · SL and HL, with additional HL material marked

This replaces most of the old Unit 7. Be aware that the particle physics part of that unit — quarks, leptons, exchange particles, Feynman diagrams — has been very substantially reduced, so old notes will cover a good deal that is no longer examined.

🎯What you need to be able to do

  • Describe the Geiger–Marsden–Rutherford experiment and what it established.
  • Use emission and absorption spectra as evidence for discrete atomic energy levels.
  • Relate photon energy to transitions between levels using \( E = hf \).
  • Use nuclide notation, and describe isotopes.
  • Use the nuclear radius relation \( R = R_0 A^{1/3} \).
  • HLThe hydrogen energy level equation, and distance of closest approach.

📚The physics

The Geiger–Marsden experiment fired alpha particles at thin gold foil. Almost all passed through with little deflection, a small fraction deflected substantially, and about one in eight thousand bounced back. Each observation carries a conclusion, and the exam wants the pairing: mostly empty space; a concentrated positive charge; and a nucleus that is both tiny and massive. Rutherford’s remark that it was as surprising as a shell bouncing off tissue paper is worth quoting because it captures why the plum-pudding model died.

Spectra are the evidence for energy levels. A hot gas emits only certain wavelengths — a line emission spectrum; light passed through a cool gas loses exactly those same wavelengths — a line absorption spectrum. If electrons could have any energy the spectrum would be continuous. That it is not proves the levels are discrete. Because the pattern is unique to each element, spectra also identify what distant stars are made of.

Photons carry the difference. When an electron falls from a higher level to a lower one it emits a photon of energy exactly equal to the gap:

\[ hf = E_2 - E_1 \]

Combined with \( c = f\lambda \) this gives the wavelength. Note the direction of the relationship: a larger energy gap gives a shorter wavelength.

Nuclide notation. In \( ^{A}_{Z}\mathrm{X} \), \(Z\) is the proton number (which fixes the element) and \(A\) the nucleon number. Neutron number is \( A - Z \). Isotopes share \(Z\) but differ in \(A\) — chemically near-identical, nuclearly quite different, which is why \( ^{235}\mathrm{U} \) and \( ^{238}\mathrm{U} \) behave so differently in a reactor.

Nuclear size.

\[ R = R_0 A^{1/3} \qquad R_0 \approx 1.2\ \text{fm} \]

The cube root is the interesting part: it says volume is proportional to \(A\), so nucleons pack at essentially constant density regardless of the nucleus. Doubling the nucleon number does not double the radius; it multiplies it by only 1.26.

HLHydrogen levels

\( E_n = -\dfrac{13.6}{n^{2}}\ \text{eV} \). The energies are negative because the electron is bound, and they crowd together as \(n\) rises, converging on zero at ionisation. The ionisation energy from the ground state is therefore 13.6 eV.

HLDistance of closest approach

An alpha particle fired straight at a nucleus stops when all its kinetic energy has become electric potential energy: \( E_k = kQ_1Q_2/d \). Solving for \(d\) gives an upper bound on the nuclear radius, and it was how the nucleus was first sized.

✏️Worked example

(a) An electron in hydrogen falls from \( n = 3 \) to \( n = 2 \). Find the wavelength emitted. Take \( h = 6.63 \times 10^{-34} \) J s and 1 eV \( = 1.60 \times 10^{-19} \) J. (b) Estimate the radius of a gold nucleus, \( A = 197 \).

(a) \( E_3 = -13.6/9 = -1.51 \) eV and \( E_2 = -13.6/4 = -3.40 \) eV. The gap is \( 3.40 - 1.51 = 1.89 \) eV \( = 1.89 \times 1.60 \times 10^{-19} = 3.02 \times 10^{-19} \) J. Then

\[ \lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{3.02 \times 10^{-19}} = 6.6 \times 10^{-7}\ \text{m} \]

or 660 nm — red light. This is the H-alpha line, the red glow of hydrogen nebulae, and getting a visible answer is a good sign the arithmetic worked.

(b) \( R = 1.2 \times 10^{-15} \times 197^{1/3} = 1.2 \times 10^{-15} \times 5.82 = 7.0 \times 10^{-15} \) m.

Put that in perspective. A gold atom is about \( 1.4 \times 10^{-10} \) m across, so the nucleus is roughly twenty thousand times smaller than the atom. If the atom were a sports stadium, the nucleus would be a grain of rice at the centre spot. That ratio is what Geiger and Marsden actually measured, and it is why almost every alpha particle sailed straight through.

🔭See it happen

PhET, Rutherford Scattering. Switch between the plum-pudding and nuclear models and fire alpha particles at each. The plum pudding never produces a backscatter; the nuclear model does, rarely. Watching the rarity is the point — a small target is exactly what one-in-eight-thousand implies.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. An electron transition releases a photon of energy 10.2 eV. Find its wavelength. Take \( h = 6.63 \times 10^{-34} \) J s and 1 eV \( = 1.60 \times 10^{-19} \) J.
Convert first: \( E = 10.2 \times 1.60 \times 10^{-19} = 1.63 \times 10^{-18} \) J. Then \( \lambda = \dfrac{hc}{E} = \dfrac{1.99 \times 10^{-25}}{1.63 \times 10^{-18}} = 1.2 \times 10^{-7} \) m, about 122 nm — ultraviolet.
2. HLFind the energy of the \( n = 4 \) level of hydrogen.
\( E_4 = -13.6/4^{2} = -13.6/16 = -0.85 \) eV. Negative because the electron is bound.
3. HLHow much energy is needed to ionise a hydrogen atom already excited to \( n = 2 \)?
Ionisation means reaching \( E = 0 \). Since \( E_2 = -13.6/4 = -3.40 \) eV, the energy needed is 3.40 eV — much less than the 13.6 eV required from the ground state.
4. Estimate the radius of an aluminium nucleus, \( A = 27 \), taking \( R_0 = 1.2 \) fm.
\( R = R_0A^{1/3} = 1.2 \times 10^{-15} \times 27^{1/3} = 1.2 \times 10^{-15} \times 3 = 3.6 \times 10^{-15} \) m. The cube root of 27 is exactly 3, which is why this value is often chosen.
5. State the number of protons and neutrons in \( ^{56}_{26}\mathrm{Fe} \).
\( Z = 26 \) protons. Neutrons \( = A - Z = 56 - 26 = 30 \).
6. Explain why the existence of line spectra, rather than continuous ones, is evidence for discrete energy levels in atoms.
A photon is emitted with energy exactly equal to the gap between two levels, \( hf = E_2 - E_1 \). If an electron could take any energy, the gaps would form a continuum and the spectrum would be continuous. Observing only certain sharp wavelengths means only certain energy differences exist, so the levels themselves must be discrete. Because the pattern is unique to each element, spectra also reveal what distant stars are made of.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • HyperPhysics — the Rutherford experiment, hydrogen spectrum and nuclear size
  • The Physics Hypertextbook — atomic structure and spectra