Rutherford's nuclear model of the atom
Rutherford pictured the atom as a tiny positive nucleus with electrons orbiting it like planets round the Sun. The model could not explain atomic stability or line spectra.
The nuclear model
Rutherford's model says that all the positive charge and almost all the mass of an atom are packed into a tiny nucleus at its centre. The electrons revolve round the nucleus in orbits, as planets go round the Sun. The electrostatic attraction between the nucleus and each electron provides the centripetal force for its circular motion. A nucleus of charge +Ze has Z electrons round it, so the atom is neutral overall. Most of the atom is empty space.
Thomson's model and Rutherford's model
| Feature | Thomson's model | Rutherford's model |
|---|---|---|
| Positive charge | Spread evenly through the whole atom | Packed into a tiny central nucleus |
| Electrons | Embedded in the positive sphere, like seeds in a watermelon (plum pudding model) | Revolve round the nucleus |
| Large-angle alpha scattering | Cannot explain it | Explains it: a close approach to the nucleus |
Energy of the electron in a hydrogen orbit
Assume the electron (mass m, charge −e) moves round the proton (+e) in a circle of radius r with speed v.
- 1Centripetal force = electrostatic attraction: mv²/r = (1/4πε₀)(e²/r²).
- 2Multiply both sides by r/2: (1/2)mv² = e²/(8πε₀r). So kinetic energy K = e²/(8πε₀r).
- 3Potential energy of charge −e at distance r from charge +e: U = −e²/(4πε₀r).
- 4Total energy E = K + U = e²/(8πε₀r) − e²/(4πε₀r).
- 5★So, E = −e²/(8πε₀r). The minus sign shows the electron is bound; energy must be supplied to pull it away.
★ key step
Two failures of the Rutherford model
Bohr's model removed both problems for hydrogen.
Worked example: orbit radius from the energy
The total energy of the electron in a hydrogen atom is −13.6 eV. Using Rutherford's model, find the radius of its orbit and its potential energy. (1/4πε₀ = 9 × 10⁹ N m² C⁻², e = 1.6 × 10⁻¹⁹ C)
- Given: E = −13.6 eV = −13.6 × 1.6 × 10⁻¹⁹ J = −2.176 × 10⁻¹⁸ J.
- Formula: E = −e²/(8πε₀r), so r = (1/4πε₀) × e²/(2|E|).
- r = 9 × 10⁹ × (1.6 × 10⁻¹⁹)² ÷ (2 × 2.176 × 10⁻¹⁸).
- r = 2.304 × 10⁻²⁸ ÷ 4.352 × 10⁻¹⁸ = 5.3 × 10⁻¹¹ m.
- Potential energy U = 2E = 2 × (−13.6 eV) = −27.2 eV.
r ≈ 5.3 × 10⁻¹¹ m (0.53 Å); U = −27.2 eV
If an atom were as big as a cricket stadium, its nucleus would be about the size of a mustard seed (rai) at the centre of the pitch. Almost everything around you is mostly empty space.
2 marks: “State two limitations of Rutherford's model of the atom.” An orbiting electron accelerates, so it should radiate energy and spiral into the nucleus: the atom would be unstable. It would give a continuous spectrum, but atoms give line spectra.
3 marks: “Show that the total energy of the electron in a hydrogen atom in Rutherford's model is −e²/(8πε₀r). What does the negative sign mean?” mv²/r = e²/(4πε₀r²) gives K = e²/(8πε₀r). U = −e²/(4πε₀r). E = K + U = −e²/(8πε₀r). Negative: the electron is bound to the nucleus.
1 marks: “In Rutherford's model, what provides the centripetal force on an orbiting electron?” The electrostatic attraction between the electron and the nucleus.
Nucleus small, atom huge; electron orbits but should crash: Rutherford got the nucleus right and the stability wrong.