Alpha particle scattering experiment
An alpha particle is a helium nucleus: 2 protons + 2 neutrons, charge +2e. It is fast and heavy compared with an electron.
In the experiment (by Geiger and Marsden, guided by Rutherford), a thin beam of alpha particles hit a very thin gold foil. A screen all around counted where each particle landed.
- Most particles went straight through.
- A few turned by small angles.
- A very small number (about 1 in 8000) turned by more than 90°, some almost straight back.
A particle can only bounce back if it meets something very small, very heavy and strongly positive. So the positive charge must be packed in a tiny centre.
Impact parameter and closest approach
Impact parameter (b) is the sideways distance between the line of the incoming particle and the centre of the nucleus. Small b means a big turn; b = 0 means a head-on hit and the particle comes straight back.
In a head-on hit, the particle slows down, stops for a moment, and turns back. At that point all its kinetic energy has become electric potential energy. This smallest distance is the distance of closest approach:
K = (1/4πε₀) × (2e)(Ze) / r₀, so r₀ = (1/4πε₀) × 2Ze² / K.
It comes out about 10⁻¹⁴ m for gold, so the nucleus must be even smaller than this.
Rutherford's nuclear model and its problems
Rutherford's model: the atom has a tiny nucleus (size about 10⁻¹⁵ m to 10⁻¹⁴ m) that holds all the positive charge and almost all the mass. Electrons move round it, far away (atom size about 10⁻¹⁰ m). The atom is about 99.99…% empty space.
Problem 1 – stability: an electron going in a circle is accelerating. An accelerating charge should give out energy as light. So it should lose energy, spiral in and fall into the nucleus in about 10⁻⁸ s. But atoms are stable.
Problem 2 – spectrum: a spiralling electron would give out every colour (a continuous spectrum). But hydrogen gives only certain sharp lines.
Bohr's model of the hydrogen atom
Niels Bohr (1913) kept the nucleus but added three rules (postulates):
- Stationary orbits: the electron moves only on certain orbits and does not radiate while on them.
- Quantum condition: an orbit is allowed only if the angular momentum is a whole-number multiple of h/2π: mvr = nh/2π, n = 1, 2, 3 … (n is the principal quantum number).
- Frequency rule: when the electron jumps from a higher level Eᵢ to a lower level E_f, one photon comes out with hν = Eᵢ − E_f.
n = 1 is the ground state (lowest energy). n = 2, 3 … are excited states.
Radius, speed and energy in the nth orbit
Two facts are joined: (1) the electric pull of the nucleus gives the centripetal force: mv²/r = ke²/r², where k = 1/4πε₀; (2) mvr = nh/2π. Solving them together gives:
- Radius: rₙ = n²h²ε₀ / (πme²) = 0.529 n² Å. The first radius, a₀ = 0.529 Å, is the Bohr radius.
- Speed: vₙ = e²/(2ε₀nh) = (2.19 × 10⁶)/n m/s.
- Kinetic energy: K = ke²/2r (always positive).
- Potential energy: U = −ke²/r = −2K.
- Total energy: E = K + U = −ke²/2r = −me⁴/(8ε₀²h²n²) = −13.6/n² eV.
Why negative? Zero energy means the electron is free and far away. A bound electron has less than that, so its energy is below zero. The ionisation energy of hydrogen (energy to free the electron from n = 1) is 13.6 eV.
Energy level diagram
E₁ = −13.6 eV, E₂ = −3.4 eV, E₃ = −1.51 eV, E₄ = −0.85 eV, … E∞ = 0. The levels crowd together near zero. The energy to lift the electron from n = 1 to n = 2 (10.2 eV) is called the first excitation energy.
Hydrogen spectrum (qualitative)
Hot hydrogen gas gives light of only certain wavelengths: a line spectrum. Each line is one kind of jump. Jumps that end on the same lower level form a series:
- Lyman series – ends on n = 1 – ultraviolet.
- Balmer series – ends on n = 2 – visible (red 656 nm, blue-green 486 nm, violet 434 nm, 410 nm).
- Paschen (n = 3), Brackett (n = 4), Pfund (n = 5) – infrared.
The wavelength follows 1/λ = R(1/n_f² − 1/nᵢ²), with R ≈ 1.097 × 10⁷ m⁻¹ (Rydberg constant). A quick tool: λ (in nm) ≈ 1240 / ΔE (in eV).
If the gas absorbs light instead, the same jumps happen upward, and dark lines appear at the same places: an absorption spectrum.
Limits of the Bohr model
It works for hydrogen and one-electron ions (He⁺, Li²⁺) only. It cannot explain why some lines are brighter than others, or atoms with many electrons. It also mixes old physics with a new rule. De Broglie later gave the reason for the rule: an orbit fits a whole number of electron waves, 2πr = nλ.
Try it: predict, then check
1. Before pressing Jump, guess: will a jump from 4 → 2 give a bigger or smaller photon energy than 3 → 2? Guess its colour. Now try it in the 3D.
2. Try 2 → 1, 3 → 1 and 4 → 1. Why can you not see these lines with your eyes? (Look at the wavelength.)
3. At home: look at a CD or DVD under a white LED and then under a yellow sodium street lamp. The white light spreads into a full rainbow; the sodium lamp shows mostly one yellow band. That is a line spectrum.
Key formulas and definitions
- Closest approach: r₀ = (1/4πε₀) · 2Ze² / K
- Bohr condition: mvr = nh/2π
- rₙ = 0.529 n² Å (rₙ ∝ n²)
- vₙ = 2.19 × 10⁶ / n m/s (vₙ ∝ 1/n)
- Eₙ = −13.6 / n² eV; K = −E, U = 2E
- hν = Eᵢ − E_f; λ (nm) ≈ 1240 / ΔE (eV)
- 1/λ = R (1/n_f² − 1/nᵢ²), R = 1.097 × 10⁷ m⁻¹
Worked examples
1. Find the radius of the third orbit of hydrogen.
Step 1: rₙ = 0.529 n² Å. Step 2: n = 3, so n² = 9. Step 3: r₃ = 0.529 × 9 = 4.76 Å ≈ 4.76 × 10⁻¹⁰ m.
2. Find the energy of the electron in n = 2 and n = 4 of hydrogen.
Step 1: Eₙ = −13.6 / n² eV. Step 2: E₂ = −13.6 / 4 = −3.4 eV. Step 3: E₄ = −13.6 / 16 = −0.85 eV.
3. The total energy of an electron in some orbit is −3.4 eV. Find its kinetic and potential energy.
Step 1: K = −E = +3.4 eV. Step 2: U = 2E = −6.8 eV. Check: K + U = 3.4 − 6.8 = −3.4 eV ✓.
4. Find the energy and wavelength of the photon when the electron jumps from n = 3 to n = 2.
Step 1: E₃ = −1.51 eV, E₂ = −3.40 eV. Step 2: ΔE = −1.51 − (−3.40) = 1.89 eV. Step 3: λ ≈ 1240 / 1.89 ≈ 656 nm. This is the red Balmer line (H-alpha).
5. An alpha particle of kinetic energy 7.7 MeV moves straight at a gold nucleus (Z = 79). Find the distance of closest approach. (1/4πε₀ = 9 × 10⁹ N m² C⁻², e = 1.6 × 10⁻¹⁹ C)
Step 1: K = 7.7 MeV = 7.7 × 10⁶ × 1.6 × 10⁻¹⁹ J = 1.232 × 10⁻¹² J. Step 2: r₀ = 9 × 10⁹ × 2 × 79 × (1.6 × 10⁻¹⁹)² / K. Step 3: top = 9 × 10⁹ × 158 × 2.56 × 10⁻³⁸ = 3.64 × 10⁻²⁶. Step 4: r₀ = 3.64 × 10⁻²⁶ / 1.232 × 10⁻¹² ≈ 2.95 × 10⁻¹⁴ m ≈ 30 fm.
6. Find the shortest wavelength in the Lyman series of hydrogen.
Step 1: Shortest wavelength = biggest energy jump = n = ∞ to n = 1. Step 2: ΔE = 0 − (−13.6) = 13.6 eV. Step 3: λ ≈ 1240 / 13.6 ≈ 91.2 nm (ultraviolet).
7. How many different spectral lines can appear when hydrogen atoms fall from n = 4 to the ground state?
Step 1: Each pair of levels among 1, 2, 3, 4 can give one line. Step 2: Number of pairs = n(n − 1)/2 = 4 × 3 / 2 = 6. Step 3: The lines are 4→3, 4→2, 4→1, 3→2, 3→1, 2→1.
8. Compare the speed of the electron in n = 1 with the speed of light.
Step 1: v₁ = 2.19 × 10⁶ m/s. Step 2: v₁ / c = 2.19 × 10⁶ / 3 × 10⁸ ≈ 1/137. So the electron moves at about 0.7% of the speed of light, slow enough for Bohr's simple (non-relativistic) maths.
Common mistakes
- Thinking a bigger orbit means more energy stored as a bigger negative number. E = −13.6/n²: n = 3 gives −1.51 eV, which is HIGHER than −13.6 eV.
- Writing rₙ ∝ n. The radius grows as n²: the 2nd orbit is 4 times, not 2 times, the first.
- Forgetting the sign rules: K is positive, U is negative, and U = 2E, K = −E.
- Saying the Balmer series is ultraviolet. Balmer lines end on n = 2 and are visible; Lyman (ends on n = 1) is ultraviolet.