1🗓️ How to Study
Six sittings. This chapter is half history (who did what, which model failed why) and half numericals (energy, wavelength, quantum numbers). Learn the constants cold — most questions are one substitution away from the answer.
Day 1 — Sub-atomic particles
Cathode rays, e/m ratio (Thomson), oil-drop (Millikan), discovery of proton & neutron. Table of masses and charges.
Day 2 — Atomic models
Thomson's plum pudding → Rutherford's gold foil (all 3 observations + conclusions) → drawbacks. Atomic number, mass number, isotopes, isobars, isoelectronic.
Day 3 — Light & quantum
EM wave properties, Planck's quantum theory, photoelectric effect (work function, threshold, KE equation), atomic spectra.
Day 4 — Bohr's model
All four postulates, rₙ and Eₙ formulas, spectral series, limitations. Do 15 numericals on transitions.
Day 5 — Quantum mechanics
de Broglie, Heisenberg, Schrödinger, four quantum numbers, orbital shapes and nodes.
Day 6 — Configurations + Drill
Aufbau, (n+l) rule, Pauli, Hund; Cr & Cu exceptions; then Traps and the A-R drill.
2📚 Topic Map
| Block | What it contains | NEET priority |
|---|---|---|
| 2.1 · Sub-atomic particles | Cathode/anode rays, e/m ratio, Millikan's charge, proton & neutron discovery | High — constants asked directly |
| 2.2 · Atomic models | Thomson, Rutherford's α-scattering, atomic/mass number, isotopes & isobars, drawbacks | Highest |
| 2.3 · Towards quantum | EM waves, Planck's theory, photoelectric effect, atomic spectra (Rydberg) | Highest — numericals |
| 2.4 · Bohr's model | Four postulates, rₙ & Eₙ, line spectrum of H, limitations | Highest |
| 2.5 · Dual nature | de Broglie relation, Heisenberg uncertainty principle | High |
| 2.6 · Quantum mechanical model | Schrödinger, four quantum numbers, orbital shapes, nodes, energy order | Highest |
| 2.6.5 · Configurations | Aufbau + (n+l) rule, Pauli exclusion, Hund's rule, Cr/Cu exceptions, stability of half/fully filled | Highest |
3🧠 Concepts
1 · Discovery of the electron constants asked
Cathode rays in a discharge tube: they start at the cathode, travel in straight lines, are deflected by electric and magnetic fields (negatively charged), and — crucially — their properties are independent of the electrode material and the gas, proving electrons are universal constituents of all atoms.
J.J. Thomson measured the charge-to-mass ratio: e/mₑ = 1.758820 × 10¹¹ C kg⁻¹. Deflection increases with charge and field strength, decreases with particle mass.
Millikan's oil-drop experiment gave the charge: e = −1.6022 × 10⁻¹⁹ C, and combining the two, mₑ = 9.1094 × 10⁻³¹ kg.
2 · Proton and neutron
Anode (canal) rays: positively charged, and unlike cathode rays their charge/mass ratio depends on the gas — lightest particle from hydrogen = the proton (1.676 × 10⁻²⁷ kg, +1.6022 × 10⁻¹⁹ C). Chadwick (1932) bombarded beryllium with α-particles and found the electrically neutral neutron (1.675 × 10⁻²⁷ kg) — slightly heavier than the proton.
3 · Thomson's model and its failure
"Plum pudding" / watermelon: positive charge spread over a sphere with electrons embedded in it. It explained overall electrical neutrality but could not explain Rutherford's scattering results.
4 · Rutherford's gold foil experiment most tested
α-particles fired at a thin gold foil. Three observations → three conclusions:
- Most passed undeviated ⟹ the atom is mostly empty space.
- A few deflected through small angles ⟹ the positive charge is concentrated, not spread out.
- Very few (1 in 20,000) bounced back ⟹ that concentration is a tiny, massive nucleus.
Model: positive charge and nearly all the mass in a nucleus; electrons revolve around it at high speed in circular orbits; nucleus radius ~10⁻¹⁵ m vs atom ~10⁻¹⁰ m — the atom is 10⁵ times the nucleus.
5 · Drawbacks of Rutherford's model
By Maxwell's electromagnetic theory an accelerating charged particle radiates energy. An orbiting electron accelerates continuously, so it should spiral into the nucleus in ~10⁻⁸ s — the atom would collapse. The model also says nothing about the distribution of electrons or their energies, and cannot explain line spectra.
6 · Atomic number, isotopes, isobars
Z = protons = electrons (neutral atom) · A = protons + neutrons · neutrons = A − Z.
- Isotopes — same Z, different A (¹H, ²H, ³H; ³⁵Cl, ³⁷Cl). Same chemical behaviour (chemistry depends on electrons).
- Isobars — same A, different Z (¹⁴C and ¹⁴N).
- Isoelectronic — same number of electrons (O²⁻, F⁻, Na⁺, Mg²⁺, Ne — all 10).
- Isotones — same number of neutrons.
7 · Light: wave properties and Planck's quantum
c = νλ · ν̄ = 1/λ (wavenumber, cm⁻¹) · c = 3.0 × 10⁸ m s⁻¹
Black-body radiation and the photoelectric effect broke the wave picture. Planck (1900): energy is emitted/absorbed in discrete packets — quanta:
E = hν = hc/λ, h = 6.626 × 10⁻³⁴ J s
Energy is quantised: only integral multiples nhν are allowed.
8 · Photoelectric effect — the four facts
- Electrons eject instantaneously, with no time lag.
- Emission occurs only above a threshold frequency ν₀, however intense the light.
- Intensity controls the number of electrons; frequency controls their kinetic energy.
- Einstein's equation: hν = hν₀ + ½mₑv² ⟹ KE = h(ν − ν₀), where hν₀ = W₀ is the work function.
This is the direct proof of light's particle nature — photons.
9 · Bohr's model — the four postulates
- The electron moves in fixed circular stationary states (orbits) numbered n = 1, 2, 3… with definite energy; while in them it does not radiate.
- Radii: rₙ = n²a₀, a₀ = 52.9 pm (first Bohr orbit).
- Energy is emitted/absorbed only on a transition: ΔE = E₂ − E₁ = hν — Bohr's frequency rule.
- Angular momentum is quantised: mₑvr = n·h/2π.
Energy of the nᵗʰ state (hydrogen): Eₙ = −2.18 × 10⁻¹⁸/n² J = −13.6/n² eV. For hydrogen-like ions multiply by Z²; radius becomes n²a₀/Z.
Why the negative sign? Zero energy is defined for a free electron (n = ∞). Being bound to the nucleus lowers the energy below that reference — so all bound states are negative, and n = 1 (most negative) is the most stable.
10 · Limitations of Bohr's model
- Fails for multi-electron atoms — works only for H and H-like ions (He⁺, Li²⁺, Be³⁺).
- Cannot explain the fine structure of spectral lines, nor the Zeeman (magnetic) and Stark (electric) effects.
- Cannot explain the ability of atoms to form molecules by chemical bonds.
- Contradicts Heisenberg's uncertainty principle by assigning a definite orbit (position and momentum simultaneously).
11 · de Broglie and Heisenberg
de Broglie (1924): matter has wave character too — λ = h/mv = h/p. Significant only for very light particles: an electron's λ is measurable, a cricket ball's is absurdly small.
Heisenberg's uncertainty principle: Δx·Δp ≥ h/4π (equivalently Δx·mΔv ≥ h/4π). Position and momentum cannot both be known exactly — which is precisely why "orbits" give way to orbitals (probability regions).
12 · The four quantum numbers
- n (principal) — shell, size and energy; n = 1, 2, 3…; electrons in a shell = 2n².
- l (azimuthal) — subshell and shape; l = 0 … (n−1); s=0, p=1, d=2, f=3. Orbital angular momentum = √(l(l+1))·h/2π.
- mₗ (magnetic) — orientation; from −l to +l, so (2l+1) orbitals per subshell.
- mₛ (spin) — +½ or −½ only; not derived from the Schrödinger equation.
13 · Orbitals, nodes and shapes
s orbitals are spherical; p are dumb-bell shaped (three: pₓ, p_y, p_z); d are double dumb-bell (five). An orbital holds a maximum of 2 electrons.
radial nodes = n − l − 1 · angular nodes = l · total nodes = n − 1
The orbital is a region of high probability, not a path. A node is where the probability of finding the electron is zero.
4🧮 Formula Bank
| Quantity | Formula | Note |
|---|---|---|
| Wave relation | c = νλ, ν̄ = 1/λ | c = 3.0 × 10⁸ m s⁻¹ |
| Photon energy | E = hν = hc/λ | h = 6.626 × 10⁻³⁴ J s |
| Photoelectric effect | hν = hν₀ + ½mv² | W₀ = hν₀ (work function) |
| Bohr radius | rₙ = n²a₀ = 52.9 n²/Z pm | a₀ = 52.9 pm |
| Bohr energy | Eₙ = −2.18 × 10⁻¹⁸ Z²/n² J = −13.6 Z²/n² eV | negative ⟹ bound |
| Angular momentum | mvr = nh/2π | quantised |
| Transition energy | ΔE = 2.18 × 10⁻¹⁸ (1/n₁² − 1/n₂²) J | emission if n₂ → n₁ |
| Rydberg equation | ν̄ = 109677 (1/n₁² − 1/n₂²) cm⁻¹ | R_H = 109677 cm⁻¹ |
| de Broglie | λ = h/mv = h/√(2mKE) | with KE in joules |
| Heisenberg | Δx·Δp ≥ h/4π | also Δx·mΔv ≥ h/4π |
| Max emission lines | n(n−1)/2 | from level n back to ground |
| Electrons per shell / subshell | 2n² · 2(2l+1) | orbitals per subshell = 2l+1 |
| Nodes | radial = n−l−1, angular = l, total = n−1 | node = zero probability |
| Orbital angular momentum | √(l(l+1))·h/2π | zero for s orbitals |
5📋 Quantum Numbers — one-glance sheet
| n | l values | Subshells | mₗ range | Orbitals | Max electrons (2n²) |
|---|---|---|---|---|---|
| 1 | 0 | 1s | 0 | 1 | 2 |
| 2 | 0, 1 | 2s, 2p | 0 · −1,0,+1 | 1 + 3 = 4 | 8 |
| 3 | 0, 1, 2 | 3s, 3p, 3d | … −2 to +2 | 1 + 3 + 5 = 9 | 18 |
| 4 | 0, 1, 2, 3 | 4s, 4p, 4d, 4f | … −3 to +3 | 1 + 3 + 5 + 7 = 16 | 32 |
| Subshell | l | Orbitals (2l+1) | Max e⁻ | Shape |
|---|---|---|---|---|
| s | 0 | 1 | 2 | spherical |
| p | 1 | 3 | 6 | dumb-bell |
| d | 2 | 5 | 10 | double dumb-bell |
| f | 3 | 7 | 14 | complex |
6🧩 Problem Types
P1 · Energy / wavelength of a photon
E = hc/λ. Watch units: λ in metres, answer in joules; divide by 1.602 × 10⁻¹⁹ for eV. Shortcut: E(eV) = 1240/λ(nm).
P2 · Number of photons
n = Total energy / (hc/λ). A favourite with laser and sodium-lamp data.
P3 · Photoelectric numericals
KE = h(ν − ν₀); threshold λ₀ = hc/W₀. "Electrons emitted with zero velocity" ⟹ the light is exactly at threshold.
P4 · Bohr transitions
ΔE = 13.6(1/n₁² − 1/n₂²) eV, then λ = hc/ΔE. Ionisation from n = 1 needs 13.6 eV; from n = 5, 13.6/25 = 0.544 eV.
P5 · Rydberg / spectral series
ν̄ = 109677(1/n₁² − 1/n₂²). Longest wavelength in a series = smallest ΔE = the n₁ → n₁+1 transition.
P6 · Number of spectral lines
From level n: n(n−1)/2 lines. n = 4 gives 6, n = 6 gives 15.
P7 · de Broglie wavelength
λ = h/mv, or h/√(2mKE) when kinetic energy is given. Electron mass 9.1 × 10⁻³¹ kg.
P8 · Uncertainty principle
Δx = h/(4π·mΔv). Note Δv is often given as a percentage of v — convert first.
P9 · Quantum-number validity
Check l < n, |mₗ| ≤ l, mₛ = ±½. Any violation makes the set impossible.
P10 · Configuration & counting
Write the configuration, then count unpaired electrons, electrons with a given l, or the number in a subshell. Remember Cr and Cu.
P11 · Protons / neutrons / electrons in an ion
Z = protons; neutrons = A − Z; electrons = Z − charge. A cation has fewer electrons, an anion more.
7🌈 Spectral Series of Hydrogen
| Series | n₁ (lands on) | n₂ (comes from) | Region |
|---|---|---|---|
| Lyman | 1 | 2, 3, 4 … | Ultraviolet |
| Balmer | 2 | 3, 4, 5 … | Visible |
| Paschen | 3 | 4, 5, 6 … | Infrared |
| Brackett | 4 | 5, 6, 7 … | Infrared |
| Pfund | 5 | 6, 7, 8 … | Far infrared |
Only Balmer lies in the visible region — the single most asked fact of this section. Emission spectra are bright lines on dark; absorption spectra are dark lines on bright, at exactly the same positions. Line spectra are element-specific "fingerprints" — the basis of spectroscopy (helium was discovered in the Sun this way).
8🔢 Electronic Configuration — the four rules
Aufbau principle
Orbitals fill in order of increasing energy, lowest first: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p.
(n + l) rule energy order
Lower (n + l) fills first; if two are equal, the one with lower n fills first. So 4s (4+0 = 4) fills before 3d (3+2 = 5), and 3d (5, n = 3) before 4p (5, n = 4).
Pauli exclusion principle
No two electrons in an atom can have all four quantum numbers identical — hence at most two electrons per orbital, and they must have opposite spins.
Hund's rule of maximum multiplicity
Electrons occupy degenerate orbitals singly with parallel spins before any pairing begins. This is why nitrogen (2p³) has 3 unpaired electrons, not 1.
Stability of half-filled and fully-filled subshells
Extra stability from symmetrical distribution and greater exchange energy — the reason for the two famous exceptions:
- Cr (24) = [Ar] 3d⁵ 4s¹, not 3d⁴ 4s² — 6 unpaired electrons.
- Cu (29) = [Ar] 3d¹⁰ 4s¹, not 3d⁹ 4s² — fully-filled d.
Filling vs removing
In multi-electron atoms 4s fills before 3d — but once filled, 4s electrons are removed first during ionisation (Fe²⁺ is [Ar]3d⁶, not 3d⁴4s²). In a hydrogen atom, energy depends on n alone, so 3s = 3p = 3d.
9📐 Constants to memorise
| Constant | Value |
|---|---|
| Planck's constant h | 6.626 × 10⁻³⁴ J s |
| Speed of light c | 3.0 × 10⁸ m s⁻¹ |
| Charge on electron e | 1.6022 × 10⁻¹⁹ C |
| e/mₑ (Thomson) | 1.758820 × 10¹¹ C kg⁻¹ |
| Mass of electron | 9.1094 × 10⁻³¹ kg |
| Mass of proton | 1.6726 × 10⁻²⁷ kg |
| Mass of neutron | 1.675 × 10⁻²⁷ kg (slightly > proton) |
| Bohr radius a₀ | 52.9 pm (0.529 Å) |
| Ground-state energy of H | −2.18 × 10⁻¹⁸ J = −13.6 eV |
| Rydberg constant R_H | 109,677 cm⁻¹ |
| 1 eV | 1.602 × 10⁻¹⁹ J |
| Nucleus : atom radius | 10⁻¹⁵ m : 10⁻¹⁰ m (1 : 10⁵) |
10⚡ Shortcuts
Skips h and c entirely. λ = 620 nm ⟹ 2 eV. Works for photoelectric thresholds too.
Eₙ = −13.6 Z²/n² eV and rₙ = 0.529 n²/Z Å. Every H-like numerical is one substitution.
No need to draw transitions. From n = 5: 10 lines.
Add n and l; ties broken by smaller n. Settles every "which fills first" question without recalling the whole order.
Total nodes = n − 1 always. Split them: angular = l, radial = the rest.
IE = 13.6/n² eV (hydrogen). From n = 2 it is 3.4 eV, from n = 5 it is 0.544 eV.
11⚠️ Traps
Lyman is UV; Paschen, Brackett, Pfund are IR. The most repeated one-mark fact of the chapter.
More intense light ejects more electrons, not faster ones. Only frequency changes KE.
No emission at all — not "delayed" or "weak" emission.
H, He⁺, Li²⁺, Be³⁺ — never for He or Li atoms.
Means the electron is bound, with the free electron (n = ∞) taken as zero. More negative = more stable, so E₁ is the lowest.
Fe²⁺ = [Ar]3d⁶. Removing 3d electrons instead is the classic wrong option.
3s, 3p and 3d are degenerate in H — the (n+l) ordering applies to multi-electron atoms.
3d⁵4s¹ and 3d¹⁰4s¹. Writing 3d⁴4s² or 3d⁹4s² is the trap.
1.675 vs 1.6726 × 10⁻²⁷ kg — options claiming they are equal or that the proton is heavier are wrong.
The e/m of anode rays depends on the gas — that asymmetry is often flipped in options.
An orbit is a fixed Bohr path (violates uncertainty); an orbital is a probability region. Never call them the same.
√(l(l+1))h/2π with l = 0. But its orbital angular momentum being zero does not make the electron stationary.
Subshell has 2l+1 orbitals but 2(2l+1) electrons; shell has n² orbitals but 2n² electrons.
The other three quantum numbers are; spin is an extra, experimentally required label.
12🧵 Mnemonics
Lyman (UV) → Balmer (Visible) → Paschen → Brackett → Pfund (IR), n₁ = 1, 2, 3, 4, 5.
Balmer = Bright = visible. The other four you never see.
Empty space · concentrated positive charge · tiny massive nucleus.
n, l, mₗ, mₛ in order.
Both borrow one 4s electron to make d half- or fully-filled.
Three names, three quantities — the standard matching question.
13🚨 Exceptions
[Ar]3d⁵4s¹ and [Ar]3d¹⁰4s¹ — half-filled and fully-filled stability beats strict Aufbau.
¹H (protium) is the only atom with zero neutrons.
3s = 3p = 3d for H, but not for any multi-electron atom.
1s, 2s, 3s… all have l = 0.
Only n, l and mₗ emerge from solving Schrödinger's equation.
Spectroscopy of sunlight revealed lines belonging to no known element — the power of line spectra as fingerprints.
14🔥 Most-Asked in NEET
| Asked pattern | Frequency | Ready answer |
|---|---|---|
| Which series is visible / lies in which region | Almost every year | Balmer = visible; Lyman = UV; rest IR |
| Bohr energy or radius of an H-like ion | High | −13.6Z²/n² eV; 0.529n²/Z Å |
| Quantum-number set valid or not | High | l < n, |mₗ| ≤ l, mₛ = ±½ |
| Number of unpaired electrons / configuration | High | Hund's rule; remember Cr, Cu, and Fe²⁺ |
| Photoelectric: KE vs intensity vs frequency | High | KE = h(ν − ν₀); intensity → number only |
| de Broglie or Heisenberg substitution | Medium | λ = h/mv; Δx·mΔv ≥ h/4π |
| Nodes in a given orbital | Medium | radial n−l−1, angular l, total n−1 |
| Rutherford conclusions / Bohr limitations | Medium | Empty space, tiny nucleus; fails for multi-electron, Zeeman, Stark, uncertainty |
| Isotopes / isobars / isoelectronic | Medium | Same Z / same A / same electrons |
15🎯 Assertion–Reason Drill
Mark: (a) both true, R explains A · (b) both true, R doesn't explain A · (c) A true, R false · (d) A false, R true.
A: Rutherford's model could not explain the stability of the atom.
R: An accelerating charged particle radiates energy according to electromagnetic theory.
Answer
(a) — both true, R explains A: the orbiting electron should spiral into the nucleus in ~10⁻⁸ s.
A: Increasing the intensity of incident light increases the kinetic energy of photoelectrons.
R: Kinetic energy depends on the frequency of the incident radiation.
Answer
(d) — A false (intensity changes only the number), R true.
A: The energy of an electron in a hydrogen atom is negative.
R: The energy of a free electron at n = ∞ is taken as zero.
Answer
(a) — both true, R explains A: binding lowers the energy below the free-electron reference.
A: Chromium has the configuration [Ar]3d⁴4s².
R: Half-filled and fully-filled subshells have extra stability.
Answer
(d) — A false (it is 3d⁵4s¹), R true and is exactly why.
A: 3s, 3p and 3d orbitals of a hydrogen atom have the same energy.
R: In a one-electron system the energy depends only on the principal quantum number.
Answer
(a) — both true, R explains A.
A: Bohr's model contradicts the Heisenberg uncertainty principle.
R: Bohr assigned a definite radius and a definite momentum to the electron simultaneously.
Answer
(a) — both true, R explains A.