Every definition, experiment, model and formula for the BIEK / Sindh Board paper — from the discovery of the electron to quantum numbers and electronic configuration. Read it straight through, or open the interactive lecture to watch each idea animate.
1 — Sub-atomic particles & the electron
Dalton thought the atom was a solid, indivisible ball. A series of discharge-tube experiments at the end of the 19th century proved otherwise — the atom is built from three fundamental sub-atomic particles: the electron, the proton and the neutron.
Cathode rays → the electron
When a high voltage is applied across a gas at very low pressure in a discharge tube, rays travel from the cathode (−) to the anode (+). These cathode rays:
- travel in straight lines and cast a sharp shadow;
- are deflected by electric and magnetic fields towards the positive plate → they are negatively charged;
- are independent of the gas used → they are a universal constituent of all matter.
J. J. Thomson measured their charge-to-mass ratio e/m = 1.76 × 10¹¹ C kg⁻¹; Millikan's oil-drop experiment fixed the charge. The particle is the electron.
- Electron — charge −1.602 × 10⁻¹⁹ C (relative −1), mass 9.11 × 10⁻³¹ kg (≈ 1/1836 a.m.u).
2 — The proton & the neutron
Positive (canal) rays → the proton
Goldstein used a perforated cathode and saw rays travelling the opposite way (towards the cathode) — positive rays. The lightest, formed from hydrogen, is the proton.
- Proton — charge +1.602 × 10⁻¹⁹ C (relative +1), mass 1.0073 a.m.u (1.67 × 10⁻²⁷ kg).
The neutron
In 1932 James Chadwick bombarded beryllium with α-particles and detected a neutral particle of almost proton mass — the neutron.
- Neutron — no charge, mass 1.0087 a.m.u (slightly heavier than a proton).
| Particle | Symbol | Relative charge | Relative mass (a.m.u) | Discovered by |
|---|
| Electron | e⁻ | −1 | 1/1836 (≈ 0.000549) | J. J. Thomson |
| Proton | p⁺ | +1 | 1.0073 | Goldstein / Rutherford |
| Neutron | n⁰ | 0 | 1.0087 | Chadwick |
Nucleons: protons and neutrons sit in the nucleus and are together called nucleons. Electrons occupy the space around it.
3 — Rutherford's atomic model (gold-foil)
Rutherford fired a beam of α-particles (He²⁺) at a very thin gold foil and recorded where they struck a fluorescent screen.
Observations
- Most α-particles passed straight through undeflected.
- A small number were deflected through large angles.
- About 1 in 20,000 bounced almost straight back.
Conclusions
- The atom is mostly empty space (most pass straight).
- All the positive charge and nearly all the mass are concentrated in a tiny, dense nucleus.
- Electrons revolve around the nucleus, like planets around the sun.
Defects: a revolving (accelerating) electron must continuously radiate energy and spiral into the nucleus — so the atom would collapse. The model also could not explain the atom's line spectrum.
4 — Atomic number, mass number & isotopes
- Atomic number (Z) — number of protons in the nucleus (= number of electrons in a neutral atom).
- Mass number (A) — number of protons + neutrons. So number of neutrons = A − Z.
Nuclide symbolAZX — e.g. 2311Na has 11 p, 11 e, 12 n
- Isotopes — atoms of the same element (same Z) with different mass numbers (different neutrons). e.g. ¹H, ²H, ³H or ³⁵Cl and ³⁷Cl.
- Isobars — different elements with the same A (e.g. ⁴⁰Ar, ⁴⁰K, ⁴⁰Ca).
- Isotones — different elements with the same number of neutrons.
relative atomic mass of chlorine
Chlorine is 75% ³⁵Cl and 25% ³⁷Cl. Find its relative atomic mass.
A_r = (75×35 + 25×37)/100 = (2625 + 925)/100 = 35.5 a.m.u
5 — Electromagnetic radiation & Planck's quantum theory
Light is an electromagnetic wave described by its wavelength λ and frequency ν, related to the speed of light c:
Wave relationc = ν λ (c = 3 × 10⁸ m/s)
Max Planck proposed that energy is emitted or absorbed not continuously but in tiny discrete packets called quanta (a quantum of light = a photon). The energy of one quantum is:
Planck's equationE = h ν = h c / λ (h = 6.63 × 10⁻³⁴ J·s)
energy of a photon
Find the energy of a photon of frequency 5 × 10¹⁴ Hz.
E = hν = (6.63×10⁻³⁴)(5×10¹⁴) = 3.315 × 10⁻¹⁹ J
6 — Bohr's atomic model (hydrogen)
Niels Bohr (1913) fixed Rutherford's stability problem by adding quantisation.
Postulates
- The electron revolves only in certain fixed circular orbits (stationary states) without radiating energy.
- Allowed orbits are those where the angular momentum is quantised: mvr = nh/2π (n = 1, 2, 3…).
- Energy is absorbed or emitted only when the electron jumps between orbits: E₂ − E₁ = hν.
Radius & energy of the nth orbit (hydrogen)rₙ = 0.529 × n² Å
Eₙ = −1312 / n² kJ·mol⁻¹ (= −13.6/n² eV)
The negative sign means the electron is bound — energy must be supplied to remove it. n = 1 (ground state) is the lowest, most stable energy.
7 — The hydrogen emission spectrum
When hydrogen is energised, its electron jumps to a higher orbit, then falls back, emitting a photon of a definite wavelength. Because only certain jumps are allowed, the result is a line spectrum — sharp coloured lines, not a continuous band.
| Series | Electron falls to | Region |
|---|
| Lyman | n = 1 | Ultraviolet |
| Balmer | n = 2 | Visible |
| Paschen | n = 3 | Infrared |
| Brackett / Pfund | n = 4 / 5 | Infrared |
Rydberg equation1/λ = R_H ( 1/n₁² − 1/n₂² ) (R_H = 1.09 × 10⁷ m⁻¹)
8 — Defects of Bohr's model & the wave picture
Bohr's model works beautifully for hydrogen but fails for atoms with more than one electron, cannot explain the fine structure of lines, or the splitting of lines in magnetic (Zeeman) and electric (Stark) fields. Two new ideas replaced fixed orbits with orbitals:
- de Broglie — dual nature — a moving particle also behaves as a wave: λ = h/mv.
- Heisenberg uncertainty principle — it is impossible to know an electron's exact position and momentum simultaneously: Δx · Δp ≥ h/4π.
- Orbital — a region of space around the nucleus where the probability of finding the electron is maximum (≈ 95%). Orbits are paths; orbitals are probability clouds.
9 — Quantum numbers
Four quantum numbers describe each electron completely — like a full postal address.
| Quantum number | Symbol | Tells you | Values |
|---|
| Principal | n | shell — size & energy | 1, 2, 3, … |
| Azimuthal (subsidiary) | ℓ | subshell — shape | 0 … (n−1) → s, p, d, f |
| Magnetic | m | orbital orientation | −ℓ … 0 … +ℓ |
| Spin | s | electron spin direction | +½ or −½ |
Subshell capacity = 2(2ℓ+1): s = 2, p = 6, d = 10, f = 14. Shell capacity = 2n².
10 — Atomic orbitals & their shapes
- s-orbital — spherical; one orientation; holds 2 electrons.
- p-orbital — dumb-bell shaped; three orientations (pₓ, p_y, p_z); holds 6 electrons.
- d-orbital — cloverleaf shaped; five orientations; holds 10 electrons.
Each orbital holds a maximum of 2 electrons of opposite spin.
11 — Filling the orbitals: the three rules
- Aufbau principle — electrons fill the lowest-energy orbital first.
- (n + ℓ) rule — lower (n+ℓ) fills first; if equal, the lower n fills first. This gives the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p…
- Pauli exclusion principle — no two electrons in an atom can have all four quantum numbers the same (an orbital holds 2, opposite spins).
- Hund's rule — within a subshell, electrons occupy orbitals singly first, with parallel spins, before pairing.
Order of filling1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p …
12 — Electronic configurations
| Element | Z | Configuration |
|---|
| Carbon | 6 | 1s² 2s² 2p² |
| Oxygen | 8 | 1s² 2s² 2p⁴ |
| Sodium | 11 | 1s² 2s² 2p⁶ 3s¹ = [Ne] 3s¹ |
| Chlorine | 17 | [Ne] 3s² 3p⁵ |
| Potassium | 19 | [Ar] 4s¹ |
| Iron | 26 | [Ar] 3d⁶ 4s² |
Exam anomalies (extra stability of half/fully-filled d): chromium is [Ar] 3d⁵ 4s¹ and copper is [Ar] 3d¹⁰ 4s¹, not the expected d⁴/d⁹.
13 — Worked numericals
de Broglie wavelength
Find the wavelength of an electron (m = 9.1 × 10⁻³¹ kg) moving at 2.2 × 10⁶ m/s.
λ = h/mv = (6.63×10⁻³⁴)/[(9.1×10⁻³¹)(2.2×10⁶)] = 3.31 × 10⁻¹⁰ m
Bohr energy of a level
Energy of the n = 2 orbit of hydrogen?
E₂ = −1312/2² = −1312/4 = −328 kJ·mol⁻¹
spectral line (Balmer)
Wavelength for an electron falling from n = 3 → n = 2.
1/λ = R_H(1/2² − 1/3²) = 1.09×10⁷ (1/4 − 1/9) = 1.09×10⁷ (0.1389)
1/λ = 1.514×10⁶ → λ = 6.6 × 10⁻⁷ m (656 nm, red)
14 — Exam recap
- The three sub-atomic particles, their charge/mass and who discovered them.
- Rutherford's experiment, conclusions and two defects.
- Z, A, neutrons = A − Z; isotopes / isobars / isotones; relative atomic mass.
- Planck E = hν; Bohr's postulates, radius & energy formulas.
- The hydrogen spectrum series and the Rydberg equation.
- The four quantum numbers; orbital shapes; Aufbau, Pauli, Hund and electronic configuration.