Alpha-particle scattering and Rutherford's nuclear model
A ball thrown into a thick fog that sometimes bounces back tells you something solid is hidden there; alpha particles told Rutherford about the nucleus. Link to earlier classes : Thomson's plum-pudding model from Class 9 and Coulomb's law from Chapter 1.
In the experiment of Geiger and Marsden (under Rutherford), a beam of alpha particles was fired at a thin gold foil. Most went straight through, a few were deflected through large angles, and about one in eight thousand bounced almost straight back. Rutherford concluded that an atom has a tiny, dense, positively charged nucleus that holds almost all of the mass, with the electrons around it.
Rule 1: Distance of closest approach — For a head-on approach, the whole kinetic energy of the alpha particle changes into electrical potential energy: K = k (2e)(Ze) ÷ r₀, so r₀ = 2 k Z e² ÷ K.
- Easy : A 5.5 MeV alpha particle (K = 5.5 × 1.6 × 10⁻¹³ J) approaching gold (Z = 79) comes within r₀ = 9 × 10⁹ × 2 × 79 × (1.6 × 10⁻¹⁹)² ÷ (5.5 × 1.6 × 10⁻¹³) = 4.1 × 10⁻¹⁴ m = 41 fm.
- Medium : With twice the energy, r₀ would be halved to about 21 fm.
- Exam : The nucleus is smaller than r₀, so it is at most about 10⁻¹⁴ m across, against 10⁻¹⁰ m for the atom, a ratio of about 10⁴ to 10⁵.
Rule 2: What the observations show — Large deflections happen only for rare close approaches to the nucleus. The atom is mostly empty space.
- Easy : Most alpha particles pass undeflected because they never come near a nucleus.
- Medium : The scattering follows the inverse-square force law of Coulomb's law.
- Exam : The model fails because an electron going round the nucleus is accelerated and would radiate, spiral inwards and collapse, and it cannot explain line spectra.
Nucleus: tiny, dense, positive. Atom: mostly empty. r₀ = 2kZe²/K.
Believing that the electrons scatter the alpha particles. Their mass is far too small.
Memory clue : "Rare big bounces mean a small hard core."
Spot the difference
| Model | Positive charge | Problem |
|---|---|---|
| Thomson | spread through the atom | cannot give large-angle scattering |
| Rutherford | concentrated in a nucleus | atom would collapse |
Practice questions
- 1Who performed the alpha-scattering experiment?
- 2What did most alpha particles do?
- 3What does the experiment show about the atom?
- 4Find the distance of closest approach of a 5.5 MeV alpha particle to a gold nucleus (Z = 79).
- 5What is the approximate ratio of the size of an atom to that of its nucleus?
- 6Why do a few alpha particles scatter through large angles?
- 7What is the closest approach if the alpha particle has twice the energy?
- 8Why is Rutherford's model unstable according to classical physics?
Show answers (try them yourself first)
- 1Geiger and Marsden, under Rutherford.
- 2They passed straight through.
- 3The atom has a tiny dense positive nucleus, and the rest is mostly empty space.
- 4r₀ ≈ 4.1 × 10⁻¹⁴ m (41 fm).
- 5About 10⁴ to 10⁵.
- 6They come very close to a nucleus and are strongly repelled.
- 7About 21 fm, half of the earlier value.
- 8The orbiting electron is accelerated, so it radiates energy and spirals into the nucleus.
Atomic spectra and the Rydberg formula
A neon sign glows red-orange and a sodium street lamp glows yellow because each element gives out its own particular colours. Link to earlier classes : the electromagnetic spectrum of Chapter 8.
A heated solid gives a continuous spectrum, but a gas excited by heat or a discharge gives a line spectrum: only certain wavelengths, characteristic of the element. Hydrogen's lines fall into series, named after their discoverers: Lyman (ultraviolet), Balmer (visible), Paschen, Brackett and Pfund (infrared).
Rule 1: Rydberg formula — 1 ÷ λ = R (1 ÷ n_f² − 1 ÷ n_i²) with R = 1.097 × 10⁷ m⁻¹ (n_f = final level, n_i = initial level, n_i > n_f). Lyman : n_f = 1; Balmer : n_f = 2; Paschen : n_f = 3.
- Easy : The first Balmer line (3 → 2) has 1/λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.524 × 10⁶, so λ = 656 nm (red).
- Medium : The first Lyman line (2 → 1) has λ = 1 ÷ (1.097 × 10⁷ × 0.75) = 121.5 nm.
- Exam : The Balmer series limit (n_i → ∞) is λ = 4 ÷ R = 364.6 nm.
Rule 2: Reading a spectrum — In emission, an excited atom gives out light of these wavelengths; in absorption, a gas takes in exactly the same wavelengths from white light, giving dark lines.
- Easy : The Paschen line (4 → 3) has λ = 1875 nm, in the infrared.
- Medium : The Balmer series is the only one lying in the visible range.
- Exam : Spectral lines identify the elements in distant stars.
1/λ = R(1/n_f² − 1/n_i²); Lyman n_f = 1 (UV); Balmer n_f = 2 (visible).
Swapping the two terms, which gives a negative 1/λ.
Memory clue : "Lyman Ultraviolet; Balmer Brilliant (visible); Paschen and after: Infrared."
Spot the difference
| Series | Ends at n_f | Region |
|---|---|---|
| Lyman | 1 | ultraviolet |
| Balmer | 2 | visible |
| Paschen | 3 | infrared |
Practice questions
- 1What is a line spectrum?
- 2Which hydrogen series lies in the visible region?
- 3Which series ends on the level n = 1?
- 4Find the wavelength of the first Balmer line.
- 5Find the wavelength of the first Lyman line.
- 6Find the wavelength of the Paschen line for 4 → 3.
- 7Find the series limit of the Balmer series.
- 8In which region of the spectrum does the Balmer series lie?
Show answers (try them yourself first)
- 1A spectrum made of separate bright lines of definite wavelengths.
- 2The Balmer series.
- 3The Lyman series.
- 4λ ≈ 656 nm.
- 5λ ≈ 121.5 nm.
- 6λ ≈ 1875 nm.
- 7λ = 4 ÷ R ≈ 364.6 nm.
- 8Visible (and a bit of near ultraviolet at the limit).
The Bohr model of the hydrogen atom
A ladder has steps but no positions in between; in the same way an electron in an atom can occupy only certain energy levels. Link to earlier classes : Rutherford's model and spectra from the earlier sections.
Niels Bohr used three postulates to rescue the nuclear atom. (1) The electron moves in certain stationary orbits without radiating. (2) Only orbits with angular momentum L = n h ÷ 2π are allowed (n = 1, 2, 3 … the principal quantum number). (3) The atom emits or absorbs a photon only when an electron jumps between orbits, with h f = E_i − E_f.
Rule 1: Orbit radius and speed — r_n = n² a₀ with a₀ = 0.529 Å, and v_n = (2.19 × 10⁶ ÷ n) m s⁻¹. The speed falls and the radius grows as n increases.
- Easy : r₃ = 9 × 0.529 = 4.76 Å.
- Medium : In the second orbit the speed is 2.19 × 10⁶ ÷ 2 = 1.1 × 10⁶ m s⁻¹.
- Exam : The orbit radii are in the ratio 1 : 4 : 9 : 16 …
Rule 2: Energy levels — E_n = − 13.6 ÷ n² eV. The negative sign shows that the electron is bound. The energy needed to free it from the ground state is the ionisation energy, 13.6 eV.
- Easy : E₁ = −13.6 eV, E₂ = −3.4 eV, E₃ = −1.51 eV.
- Medium : To lift the electron from n = 1 to n = 2 needs 13.6 − 3.4 = 10.2 eV.
- Exam : An electron in n = 2 needs 3.4 eV to be removed from the atom.
In the figure : Energy levels of hydrogen, with the Lyman, Balmer and Paschen transitions drawn as arrows.
r ∝ n²; E = −13.6/n² eV; hf = E_i − E_f.
Dropping the negative sign of the bound-state energy.
Memory clue : "Thirteen-point-six over n squared, always negative."
Spot the difference
| n | Radius | Energy |
|---|---|---|
| 1 | a₀ | −13.6 eV |
| 2 | 4a₀ | −3.4 eV |
| 3 | 9a₀ | −1.51 eV |
Practice questions
- 1State Bohr's quantum condition for the angular momentum.
- 2Write the formula for the energy of the n-th level of hydrogen.
- 3Write the formula for the radius of the n-th orbit.
- 4Find the radius of the third orbit.
- 5Find the energy of the second level.
- 6Find the speed of the electron in the second orbit.
- 7Find the energy needed to ionise hydrogen from the second level.
- 8Find the energy needed to excite the electron from n = 1 to n = 2.
Show answers (try them yourself first)
- 1L = nh ÷ 2π.
- 2E_n = −13.6 ÷ n² eV.
- 3r_n = n²a₀, with a₀ = 0.529 Å.
- 4r₃ ≈ 4.76 Å.
- 5E₂ = −3.4 eV.
- 6v₂ ≈ 1.1 × 10⁶ m s⁻¹.
- 73.4 eV.
- 810.2 eV.
Transitions and the hydrogen spectrum
A neon tube glows because its excited atoms drop to lower levels and give out the energy as light. Link to earlier classes : energy levels from the previous section and photon energy from Chapter 11.
When the electron falls from a higher level n_i to a lower level n_f, the atom emits a photon of energy equal to the difference of the two level energies. This energy gives the wavelength of the line, which agrees with the Rydberg formula.
Rule 1: Photon from a transition — E = E_i − E_f = 13.6 (1 ÷ n_f² − 1 ÷ n_i²) eV and λ (nm) = 1240 ÷ E (eV).
- Easy : 3 → 2 : E = 13.6 × (1/4 − 1/9) = 1.89 eV, so λ = 1240 ÷ 1.89 = 656 nm.
- Medium : 2 → 1 : E = 10.2 eV and λ = 121.6 nm.
- Exam : 4 → 3 : E = 0.66 eV and λ = 1876 nm, so this Paschen line is in the infrared.
Rule 2: Counting lines — An electron starting in level n can return to the ground state through n (n − 1) ÷ 2 different transitions, giving that many spectral lines.
- Easy : From n = 4 there are 4 × 3 ÷ 2 = 6 possible lines.
- Medium : The Lyman series limit is 1240 ÷ 13.6 = 91 nm.
- Exam : The longest wavelength of the Balmer series is for the 3 → 2 jump.
hf = E_i − E_f; lines from level n: n(n−1)/2; longest wavelength means the smallest energy gap.
Taking the photon energy to be E_i + E_f. It is the difference.
Memory clue : "Small gap, long wave; big gap, short wave."
Spot the difference
| Jump | Energy | Region |
|---|---|---|
| 2 → 1 | 10.2 eV | ultraviolet |
| 3 → 2 | 1.89 eV | visible |
| 4 → 3 | 0.66 eV | infrared |
Practice questions
- 1Write the relation between a photon's energy and the two levels of the transition.
- 2Which jump gives the longest wavelength in the Balmer series?
- 3How is wavelength related to the energy gap?
- 4Find the energy and wavelength of the 3 → 2 transition.
- 5Find the energy and wavelength of the 2 → 1 transition.
- 6Find the energy and wavelength of the 4 → 3 transition.
- 7Find the Lyman series limit.
- 8How many different spectral lines can be emitted by atoms excited to n = 4?
Show answers (try them yourself first)
- 1hf = E_i − E_f.
- 23 → 2.
- 3λ is inversely proportional to the gap.
- 41.89 eV; 656 nm.
- 510.2 eV; 121.6 nm.
- 60.66 eV; 1876 nm.
- 7λ = 1240 ÷ 13.6 ≈ 91 nm.
- 86 lines.
de Broglie's explanation of Bohr's quantisation
A guitar string vibrates only at frequencies for which a whole number of half-waves fit on it; electron waves in an orbit behave in a similar way. Link to earlier classes : the de Broglie wavelength from Chapter 11 and standing waves from Class 11.
De Broglie explained Bohr's rule by treating the orbiting electron as a standing wave. The wave closes on itself smoothly only if the circumference of the orbit holds a whole number of wavelengths. This gives exactly Bohr's condition for the angular momentum.
Rule 1: Standing-wave condition — 2π r = n λ with λ = h ÷ (m v). Together they give m v r = n h ÷ 2π.
- Easy : For n = 1, L = h ÷ 2π = 6.63 × 10⁻³⁴ ÷ 6.28 = 1.05 × 10⁻³⁴ J s.
- Medium : For n = 3, L = 3 × 1.05 × 10⁻³⁴ = 3.2 × 10⁻³⁴ J s.
- Exam : An orbit that does not fit a whole number of wavelengths would destroy itself by interference.
Rule 2: Check with n = 2 — Here r₂ = 4 × 0.529 = 2.12 Å, so the circumference is 13.3 Å. The electron's speed is 1.1 × 10⁶ m s⁻¹, so λ = 6.63 × 10⁻³⁴ ÷ (9.1 × 10⁻³¹ × 1.1 × 10⁶) = 6.65 Å.
- Easy : The circumference 13.3 Å is exactly 2λ = 13.3 Å.
- Medium : Bohr's model works well for hydrogen and hydrogen-like ions only.
- Exam : It cannot explain the relative brightness of the lines, fine structure or atoms with more than one electron; the full explanation comes from quantum mechanics.
2πr = nλ; L = nh/2π; the Bohr model suits one-electron atoms only.
Thinking that the electron literally moves in a circle like a planet. That is a picture of the model.
Memory clue : "A whole number of waves must fit round the orbit."
Spot the difference
| Idea | Statement |
|---|---|
| Bohr | assumed L = nh/2π |
| de Broglie | derived it from 2πr = nλ |
Practice questions
- 1State the standing-wave condition for an electron orbit.
- 2Write the quantised angular momentum.
- 3Why can only certain orbits exist?
- 4Find the angular momentum for n = 1.
- 5Show that the circumference of the n = 2 orbit is 2λ (r₂ = 2.12 Å).
- 6Find the de Broglie wavelength in the n = 2 orbit.
- 7Find the angular momentum of the electron in n = 3.
- 8Name one limitation of the Bohr model.
Show answers (try them yourself first)
- 12πr = nλ.
- 2L = nh ÷ 2π.
- 3The orbit must hold a whole number of wavelengths.
- 4L = 1.05 × 10⁻³⁴ J s.
- 52πr₂ = 13.3 Å and λ = 6.65 Å, so 2πr₂ = 2λ.
- 6λ ≈ 6.65 Å.
- 7L = 3.2 × 10⁻³⁴ J s.
- 8It fails for atoms with more than one electron (or does not explain line intensities).