Problem statement
Using Thomson's model, find the radius of a hydrogen atom and the wavelength of the light it gives out. The ionization energy is E = 13.6 eV.
The big idea in plain words
Thomson pictured the atom as a soft ball of positive charge +e, like a pudding. One electron (charge −e) sits inside, like a plum. Push the plum a little and the ball pulls it back, like a spring. A spring makes the plum swing to and fro. A swinging charge gives out light of the same frequency. So the ionization energy gives the ball's size. The swing rate then gives the colour of the light.
Constants used: e = 1.602×10⁻¹⁹ C (size of the electron's charge), ε₀ = 8.854×10⁻¹² F/m (the constant that sets the strength of electric force), m = 9.109×10⁻³¹ kg (electron mass), c = 2.998×10⁸ m/s (speed of light), and e²/(4πε₀) = 1.44 eV·nm.
Solution in each frame
1. The model. We assume the charge +e is spread evenly through a ball of radius R. This means the charge in any part of the ball is in proportion to its volume. We also assume the ball stays still, so only the electron moves. The electron is a point charge −e.
2. The electric field of the ball. We want the force on the electron, so we first need the field at distance r from the centre. The idea is Gauss's law: the electric flux through a closed surface equals the enclosed charge divided by ε₀. We draw a sphere of radius r.
Inside (r < R): the sphere holds the fraction r³/R³ of the charge, so \(Q_{\text{in}}=e\,r^3/R^3\). The field is the same all over the sphere, so \(E\cdot4\pi r^2=\dfrac{e\,r^3}{\varepsilon_0R^3}\). Therefore \(E_{in}=\dfrac{e\,r}{4\pi\varepsilon_0R^3}\).
Outside (r > R): the sphere holds all of e, so \(E_{out}=\dfrac{e}{4\pi\varepsilon_0r^2}\).
3. The potential at the centre. The electron starts at rest at the centre, so we need the potential φ₀ there. Potential is the work per unit charge to bring a test charge in from far away. We take the potential at infinity as zero. So we add up the field along a path from the centre to infinity:
\(=\dfrac{e}{4\pi\varepsilon_0R^3}\cdot\dfrac{R^2}{2}+\dfrac{e}{4\pi\varepsilon_0R}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{e}{2R}+\dfrac{1}{4\pi\varepsilon_0}\dfrac{e}{R}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{3e}{2R}\)
The first part (e/2R) is the climb inside the ball. The second part (e/R) is the climb outside it. Together they give the factor 3/2.
4. The radius from the ionization energy. We want R. Ionizing means pulling the electron from rest at the centre to rest far away. Its energy at the centre is potential energy −eφ₀ plus zero kinetic energy. Far away it is zero. So the energy we must supply is E = eφ₀ (conservation of energy).
\(R=\dfrac{1.5\times1.44\ \text{eV·nm}}{13.6\ \text{eV}}=0.159\ \text{nm}\)
This is about three times the Bohr radius (0.0529 nm).
5. The potential when the electron is pushed to distance r. For light we need the electron to move, so we need the potential inside the ball away from the centre. Going out from the centre to r, the potential drops by the field integrated over that stretch:
6. The spring. The electron's potential energy is U = −eφ(r). Multiply step 5 by −e:
A spring of stiffness k has exactly this energy shape: a constant plus ½kr². Putting in numbers: \(k=\dfrac{1.44\ \text{eV·nm}}{(0.159\ \text{nm})^3}=359\ \text{eV/nm}^2=57.6\ \text{N/m}\) (using 1 eV/nm² = 0.1602 N/m).
7. The swing rate. Newton's second law gives \(m\ddot r=-dU/dr=-kr\). This is a mass on a spring, so the electron swings with
We assume the push is small (r much less than R). Then the electron stays inside the ball, where this spring law holds.
8. The wavelength. Classical electromagnetism says an oscillating charge radiates light at its own frequency. So the light has angular frequency ω. Its wavelength is λ = c/f, and f = ω/2π:
What changes and what stays the same: who moves (only the electron, both, or the ball) and the tiny mass correction μ/m change. R, ω and λ stay the same to 3 digits.
Solution by each method
Energy method. Steps 3 to 8 of the lab frame use conservation of energy twice. First E = eφ₀ gives R. Then the energy shape U = −eφ₀ + ½kr² gives ω. Then λ = 2πc/ω. Fastest, and the one the examiner expects.
Momentum method. Does not suit this problem, because nothing collides. It only helps in the CM frame.
Graph method. We plot U(r) against r. The graph is a parabola. 1) At the centre, U(0) = −13.6 eV. 2) At the ball's edge, \(U(R)=-\dfrac{e^2}{4\pi\varepsilon_0R}=-\dfrac{1.44}{0.159}=-9.07\) eV, which is two thirds of 13.6 (see the plot). 3) The parabola rises 4.53 eV over the distance R, so ½kR² = 4.53 eV and \(k=\dfrac{2\times4.53\ \text{eV}}{(0.159\ \text{nm})^2}=359\ \text{eV/nm}^2=57.6\) N/m. 4) \(\omega=\sqrt{k/m}=7.95\times10^{15}\ \text{s}^{-1}\), so λ = 237 nm.
Quantum-rule method. This is a cross-check using the photon energy E = ħω. 1) \(\hbar\omega=1.0546\times10^{-34}\times7.95\times10^{15}=8.39\times10^{-19}\ \text{J}=5.24\ \text{eV}\). 2) \(\lambda=hc/E=1239.84\ \text{eV·nm}/5.24\ \text{eV}=237\ \text{nm}\). Same answer.
Check method. 1) Units: k in N/m, and √(N/m ÷ kg) = s⁻¹ ✓. 2) R ∝ 1/E and λ ∝ 1/ω ∝ R^{3/2} ∝ E^{−3/2}. A bigger E gives a smaller atom and shorter light. 3) As E → 0, R and λ → ∞ ✓. Most reliable in an exam.
Parameter table
| Symbol | Meaning | Value | Live (slider) |
|---|---|---|---|
| E | ionization energy | 13.6 eV | |
| R | ball radius | 0.159 nm | |
| e²/4πε₀ | Coulomb constant × e² | 1.44 eV·nm | — |
| m | electron mass | 9.109×10⁻³¹ kg | — |
| k | spring stiffness | 57.6 N/m | |
| ω | swing rate | 7.95×10¹⁵ s⁻¹ | |
| λ | wavelength | 0.237 µm |
Common exam traps
- Gaussian vs SI: the book's solution leaves out 4πε₀. In SI, keep it everywhere.
- Mixing eV and joules. Convert with 1 eV = 1.602×10⁻¹⁹ J.
- The electron's total energy at rest is −E. The ionization energy is +E.
- λ = 2πc/ω, not c/ω. Do not mix ω and f.
- R is about 3 times the Bohr radius (0.529 Å). The Thomson and Bohr atoms differ in size.
- λ is about 1500 times bigger than R. Light is not the size of the atom.
- The push must be small (r ≪ R). Otherwise the spring law fails.