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Chapter 02 · Part VI · Modern Physics
Revised on ____________________
Chapter 02 · Atoms

Every formula, with the picture that earns it

2 sheets
2 diagrams
17 results

Formulas and conditions only — no derivations, no solved numbers. The diagrams are the reader's own.

I

Rutherford scattering

ONE α IN 8000 COMES BACK
I.Closest approach, head-on  r0 = 14πε0 · 2Ze²K
II.Impact parameter  b = 14πε0 Ze²K cot θ2
III.Count law  N(θ) ∝ 1sin⁴(θ/2)
IV.Scale  nucleus ≈ 10−15 m in an atom of 10−10 m — empty by 1015 in volume
30° is ≈14× more likely than 60°: the sin−4 fall is the whole evidence. Thomson's atom gives no large-angle scattering at all.
original directionnucleus +Zeα, kinetic energy Kbimpact parameterθscattering angleb = 0 → turned straight back,closest approach r₀tan(θ/2) = (1/4πε₀) · 2Ze² / (2Kb)r₀ = (1/4πε₀) · 2Ze² / Ksmall b → large θ the nucleus is never touched, only turned
Hyperbolic path past the nucleus: a small impact parameter b gives a large deflection θ. Reader § Scattering geometry.
II

The Bohr model

HYDROGEN-LIKE, NUCLEAR CHARGE Ze
I.Quantisation  m v r = n ℏ = n h2π
II.Radius  rn = n²Z a0 a0 = 0.529 Å
III.Speed  vn = Zn α c v1 = 2.19×106 m s−1
IV.Energy  En = − 13.6 Z²n² eV
V.Virial split  En = − KE = ½ PE
VI.Revolution frequency  f = vn2π rn ∝ Z²n³
VII.Angular momentum  Ln = n ℏ independent of Z
VIII.Wave justification  2π rn = n λ with λ = hm v
III

The hydrogen spectrum

WHERE THE MARKS ACTUALLY ARE
I.Transition  h ν = Eni − Enf
II.Rydberg  1λ = R Z² ( 1nf² − 1ni² )
III.Series limit  1λ∞ = R Z²nf²
IV.Lines from level n  n(n − 1)2
V.Photon shortcut  λ(nm) = 1240E(eV)
Series
nf
Region
Limit
Lyman
1
UV
91.2 nm
Balmer
2
visible
364.6 nm
Paschen
3
IR
820 nm
Brackett · Pfund
4 · 5
far IR
1458 · 2279 nm
n → ∞0 eV · ionisedn = 1-13.60 eVn = 2-3.40 eVn = 3-1.51 eVn = 4-0.85 eVn = 5-0.54 eVn = 6-0.38 eVLyman · UVBalmer · visiblePaschen · IREₙ = −13.6 / n² eV (levels drawn equally spaced; the real energies crowd toward 0)
Jumps ending on n = 1, 2, 3 give the Lyman, Balmer and Paschen series. Reader § Energy levels.
IV

Numbers you should not have to look up

THE FOUR IN BOLD ARE NON-NEGOTIABLE
Bohr radius
a0 = 0.529 Å
Ground state
E1 = −13.6 eV
Rydberg
R = 1.097×107 m−1
Photon
hc = 1240 eV nm
Fine structure
α ≈ 1/137
First excit., H
10.2 eV
Ionisation, He+
54.4 eV
Electron rest e.
0.511 MeV
V

What the model does and does not buy

THE HONEST LIMITS
It gets right  Every hydrogen line to four figures, the ionisation energies of hydrogen-like ions, and the Z² and 1/n² scalings.
It cannot touch  Two-electron atoms, line intensities, fine structure, or why an orbiting electron does not radiate. Bohr is a correct answer reached by a wrong argument; quantum mechanics repairs the argument.
Where marks are lost in this chapter
1Dropping the Z² when the ion is not hydrogen — He+ ionises at 54.4 eV, not 13.6.
2Confusing “energy of the level” (negative) with “energy needed to remove” (positive).
3Writing tan(θ/2) for the impact parameter. It is cot.
4Mixing eV with metres. Use λ(nm) = 1240/E(eV) and stop converting.