Chapters

Everything NCERT Class 11 Unit 2 (Structure of Atom) says that can be asked, plus what JEE Main adds beyond the book (marked Beyond NCERT). The short note stays for the last week; this is for a proper revision pass.

Formula sheet

Every formula in the chapter on one screen, for the last pass. Details and examples are in the sections below.

Quantity	Formula	Watch
Wave	c = νλ; ν̄ = 1/λ	ν̄ usually in cm⁻¹
Photon energy	E = hν = hc/λ = hcν̄	E (eV) = 1240 / λ (nm)
Photons in a beam	N = total energy / hν = P·t·λ / hc	Power in W = J s⁻¹
Photoelectric	hν = hν₀ + ½mv²; W₀ = hν₀	eV₀ = KEₘₐₓ (stopping potential)
Bohr radius	rₙ = 52.9 n²/Z pm	∝ n²/Z
Bohr energy	Eₙ = −2.18 × 10⁻¹⁸ Z²/n² J = −13.6 Z²/n² eV	∝ Z²/n²
Bohr speed	vₙ = 2.18 × 10⁶ Z/n m s⁻¹	∝ Z/n
Bohr energies	KE = −E; PE = 2E	PE = −2 KE
Angular momentum (Bohr)	mvr = nh/2π
Line in a spectrum	ν̄ = 109 677 Z² (1/n₁² − 1/n₂²) cm⁻¹	n₁ = lower level
Lines from level n down to 1	n(n − 1)/2	Δn(Δn + 1)/2 between any two levels
de Broglie	λ = h/mv = h/p = h/√(2m·KE)	Charge q through V: KE = qV
Electron through V volts	λ = 12.27/√V Å
Bohr orbit and de Broglie	2πrₙ = nλ, so λ = 2πa₀ n/Z
Heisenberg	Δx · Δp ≥ h/4π; Δx · Δv ≥ h/4πm
Orbitals in shell n	n²; electrons 2n²
Orbitals in subshell l	2l + 1; electrons 2(2l + 1)
Nodes	radial n − l − 1; angular l; total n − 1
Orbital angular momentum	√(l(l + 1)) h/2π	Zero for every s
Spin-only magnetic moment	μ = √(n(n + 2)) BM	n = unpaired electrons
Energy order (multi-electron)	lower n + l first; tie → lower n first	Not for H or H-like

0 · What JEE Main asks from this chapter

167 questions in JEE Main 2019–2026, about 1 per paper; 110 multiple-choice and 57 numerical-answer. Grouped by type (counts are approximate):

Type	Share	Typical question
Quantum numbers, orbitals, nodes, energy order, configurations	~38%	Number of orbitals with n = 5, mₗ = +2 (2021)
Photons and the photoelectric effect	~19%	Work function from λ = 4000 Å and v = 6 × 10⁵ m s⁻¹ (2019)
Hydrogen spectrum: series, line ratios, number of lines	~15%	Longest Balmer λ of He⁺ in terms of shortest Lyman λ of H (2020, 2023, 2026)
Bohr model: radius, energy, speed of H and H-like ions	~15%	Radius of 2nd orbit of Li²⁺ in terms of a₀ (2020)
de Broglie and Heisenberg	~13%	Δx for an electron at 5 × 10⁶ m s⁻¹ ± 0.02% (2021)

Statement questions ("which are correct") lift whole lines from NCERT: cathode-ray properties, black-body facts, 1s and 2s probability density, the (n + l) rule. Read NCERT's sentences closely; this note keeps them.

1 · Discovery of sub-atomic particles

Cathode rays (discharge tubes, from the mid-1850s; Faraday). Discharge passes only at very low pressure and very high voltage. A hole in the anode and a ZnS coating behind it show a bright spot where the rays strike. What NCERT lists:

1. They start at the cathode and move towards the anode.
2. They are invisible; they are seen by the glow of fluorescent or phosphorescent materials (the old TV picture tube).
3. With no field they travel in straight lines.
4. In electric or magnetic fields they behave as negatively charged particles: electrons.
5. Their properties don't depend on the electrode material or the gas. So electrons are in every atom.

(A 2023 question offered "depend on the electrode material" and "depend on the gas" as true. Both are false.)

e/mₑ of the electron (J.J. Thomson, 1897): electric and magnetic fields applied perpendicular to each other and to the beam. Electric field alone sends the beam to A, magnetic alone to C; balanced, it returns to B. Deflection grows with larger charge, smaller mass and stronger field.

Charge of the electron (Millikan, oil-drop experiment, 1906–14). X-rays ionise the air; oil drops pick up charge from the ions. From the rate of fall he found each drop's mass; adjusting the plate voltage slows, stops or speeds the drop. The charge on a drop is always a whole multiple of e: q = ne. Electron charge = −1.602176 × 10⁻¹⁹ C; with e/m this gives mₑ = 9.1094 × 10⁻³¹ kg.

Example (NCERT exercise 2.39) · A drop carries −1.282 × 10⁻¹⁸ C. Number of electrons = 1.282 × 10⁻¹⁸ / 1.602 × 10⁻¹⁹ = 8.

Canal rays (positive rays, from a perforated cathode):

• Their mass and e/m depend on the gas: they are positive gaseous ions.
• Some carry a multiple of the unit charge.
• They bend opposite to cathode rays.
• The lightest, from hydrogen, is the proton (characterised 1919).

Neutron (Chadwick, 1932): Be bombarded with α-particles gave neutral particles slightly heavier than protons.

Particle	Charge	Mass
Electron	−1.602176 × 10⁻¹⁹ C (−1)	9.109382 × 10⁻³¹ kg = 0.00054 u
Proton	+1.602176 × 10⁻¹⁹ C (+1)	1.6726216 × 10⁻²⁷ kg = 1.00727 u
Neutron	0	1.674927 × 10⁻²⁷ kg = 1.00867 u

X-rays and radioactivity (NCERT's box). Röntgen (1895): electrons striking a dense metal target make X-rays, not deflected by fields, very penetrating, λ ≈ 0.1 nm, electromagnetic. Becquerel found radioactivity; three kinds of rays:

• α: He nuclei (charge +2, mass 4 u). Least penetrating.
• β: negatively charged particles like electrons. About 100 times the α penetration.
• γ: neutral high-energy radiation, no particles. About 1000 times the α penetration.

2 · Atomic models, atomic number, isotopes

Thomson model (1898): a sphere of radius ≈ 10⁻¹⁰ m with positive charge spread uniformly; electrons embedded for the most stable arrangement (plum pudding, raisin pudding, watermelon). Mass is spread uniformly over the atom. It explained neutrality, not later experiments.

Rutherford's α-scattering (with Geiger and Marsden): α-particles on gold foil about 100 nm thick, a ZnS screen around it.

Observation	Conclusion
Most passed undeflected	Most of the atom is empty space
A small fraction deflected by small angles	Positive charge is concentrated, not spread
About 1 in 20 000 bounced back (≈ 180°)	Charge and mass sit in a tiny nucleus

Atom radius ≈ 10⁻¹⁰ m; nucleus ≈ 10⁻¹⁵ m. If the nucleus were a cricket ball, the atom would have a radius of about 5 km.

Nuclear model: a dense, tiny, positive nucleus holds nearly all the mass; electrons move round it at high speed in circular orbits (like planets); electrostatic attraction holds them together.

JEE asks (2021) · Had Thomson been right, the α-particles would have passed through deflected by small angles, with reduced speed.

Drawbacks of Rutherford's model:

1. Stability: an orbiting electron accelerates; by Maxwell's theory an accelerating charge radiates energy. The orbit would shrink and the electron would spiral into the nucleus in about 10⁻⁸ s. A stationary electron would simply fall in.
2. It says nothing about how electrons are arranged or what their energies are.

Atomic and mass number:

Protons and neutrons together are nucleons. Notation: mass number top-left, atomic number bottom-left (⁸⁰₃₅Br). For an ion, compare protons with electrons first; neutrons are always A − Z.

• Isotopes: same Z, different A (different neutrons). Hydrogen: protium ¹H (99.985%), deuterium ²H (0.015%), tritium ³H (trace). Carbon ¹²C, ¹³C, ¹⁴C; chlorine ³⁵Cl, ³⁷Cl.
• Isotopes have the same chemical properties: electrons decide chemistry, and the protons decide the electrons.
• Isobars: same A, different Z (¹⁴C and ¹⁴N).
• Isotones · Beyond NCERT: same number of neutrons (¹⁴C and ¹⁶O, 8 each). Isoelectronic: same number of electrons (Na⁺, Mg²⁺, F⁻, Ne).

Example (NCERT Problem 2.2) · 18 electrons, 16 protons, 16 neutrons → Z = 16 (S), A = 32, two extra electrons → ³²₁₆S²⁻.

Example (2022) · Imaginary ion ⁴⁸₂₂X³⁻: neutrons = 26, electrons = 25. Neutrons exceed electrons by 1/25 = 4%.

3 · Electromagnetic radiation: the wave picture

Maxwell (1870): an accelerating charge produces oscillating electric and magnetic fields, transmitted as electromagnetic waves. Hertz confirmed it experimentally.

• The electric and magnetic fields are perpendicular to each other and to the direction of travel.
• They need no medium; they travel in vacuum.
• All travel at c = 3.0 × 10⁸ m s⁻¹ in vacuum (2.997925 × 10⁸).
• Frequency ν: waves passing a point per second, unit hertz (Hz = s⁻¹). Wavelength λ in m, nm or Å. Wavenumber ν̄: wavelengths per unit length, m⁻¹ (SI) but usually cm⁻¹.

The spectrum, low to high frequency: radio (~10⁶ Hz, broadcasting) → microwave (~10¹⁰ Hz, radar) → infrared (~10¹³ Hz, heating) → visible (~10¹⁵ Hz) → ultraviolet (~10¹⁶ Hz) → X-rays → γ-rays → cosmic rays. Visible light runs from violet 400 nm (7.50 × 10¹⁴ Hz) to red 750 nm (4.00 × 10¹⁴ Hz).

Example (NCERT Problems 2.3, 2.5) · 1368 kHz radio: λ = 3 × 10⁸ / 1.368 × 10⁶ = 219.3 m. Yellow light 5800 Å: ν̄ = 1/(5800 × 10⁻⁸ cm) = 1.724 × 10⁴ cm⁻¹ (asked in 2024 as "x × 10 cm⁻¹", x = 1724); ν = 5.17 × 10¹⁴ Hz.

The wave picture explains diffraction (bending round an obstacle) and interference. It could not explain four things: black-body radiation, the photoelectric effect, how a solid's heat capacity varies with temperature, and the line spectra of atoms.

4 · Planck's quantum theory and the photoelectric effect

Black body: an ideal body that absorbs and emits all frequencies. Carbon black is close; a cavity with a tiny hole is a good model. It is in thermal equilibrium with its surroundings. Its emission depends only on its temperature:

• At a given T, intensity rises with λ, reaches a maximum, then falls.
• As T rises, the maximum shifts to shorter wavelength (an iron rod goes dull red → red → white → blue).

(Two 2023 statement questions on these facts. "The black body consists of particles performing simple harmonic motion" was marked false.)

Planck (1900): atoms emit or absorb energy only in discrete packets, quanta. A quantum's energy is proportional to frequency. Energy comes as 0, hν, 2hν … nhν, like standing on steps, never between them.

Example (NCERT Problems 2.6, 2.7) · 1 mol of photons at 5 × 10¹⁴ Hz: 3.313 × 10⁻¹⁹ J × 6.022 × 10²³ = 199.5 kJ mol⁻¹. A 100 W bulb at 400 nm: one photon = 4.969 × 10⁻¹⁹ J, so 100 / 4.969 × 10⁻¹⁹ = 2.012 × 10²⁰ photons s⁻¹.

Photoelectric effect (Hertz, 1887; K, Rb, Cs in light). Observations:

1. Electrons leave at once; no time lag.
2. The number of electrons is proportional to the light's intensity.
3. Each metal has a threshold frequency ν₀; below it, no electrons at any intensity. Above it, the electrons' KE rises with frequency.

Classical physics predicted the KE should rise with brightness. It doesn't: red light (4.3–4.6 × 10¹⁴ Hz) of any brightness ejects nothing from potassium; weak yellow (5.1–5.2 × 10¹⁴ Hz) does, because ν₀(K) = 5.0 × 10¹⁴ Hz.

Einstein (1905): light is a stream of photons. One photon gives all its energy to one electron. hν₀ (the work function W₀) frees it; the rest is KE. Brighter light = more photons = more electrons, not faster ones.

Work functions (NCERT Table 2.2, eV): Li 2.42, Na 2.3, K 2.25, Mg 3.7, Cu 4.8, Ag 4.3.

(Asked as this table in 2023: 400 nm light carries 1240/400 = 3.1 eV, so Li, Na and K respond: 3 metals.)

Graphs JEE draws (2019): KE vs ν is a straight line with slope h starting at ν₀; KE vs intensity is flat; number of electrons vs ν is zero below ν₀; photocurrent vs intensity rises linearly.

Example (NCERT Problem 2.8) · 300 nm on sodium gives electrons of KE 1.68 × 10⁵ J mol⁻¹. Photon: 6.626 × 10⁻¹⁹ J = 3.99 × 10⁵ J mol⁻¹. Work function = 3.99 − 1.68 = 2.31 × 10⁵ J mol⁻¹ = 3.84 × 10⁻¹⁹ J per electron. Threshold λ = hc/W₀ = 517 nm (green).

Example (2020) · W₀ = 4.41 × 10⁻¹⁹ J, λ = 300 nm. Photon = 6.63 × 10⁻¹⁹ J. KE = 2.22 × 10⁻¹⁹ J = 222 × 10⁻²¹ J.

Example (2026) · Work functions in ratio 1 : 2, photons of 6 eV, KE ratio 2.642 : 1. (6 − W)/(6 − 2W) = 2.642 → W = 2.3 eV and 4.6 eV.

Example (2024, NCERT exercise 2.10) · 242 nm just ionises Na: E = 8.21 × 10⁻¹⁹ J × 6.022 × 10²³ = 494 kJ mol⁻¹.

Dual behaviour of radiation: light acts as particles when it meets matter (black body, photoelectric effect) and as waves while travelling (interference, diffraction).

5 · Atomic spectra and the hydrogen spectrum

A prism bends short wavelengths most: violet most, red least. White light gives a continuous spectrum (one colour merges into the next, as in a rainbow).

• Emission spectrum: the light an excited sample gives out as it returns to lower energy.
• Absorption spectrum: white light through a sample; the absorbed wavelengths are dark lines. It is the photographic negative of the emission spectrum.
• Atoms in the gas phase give line spectra (atomic spectra): bright lines at fixed wavelengths with dark gaps. Each element's is unique, a fingerprint. Bunsen used them; Rb, Cs, Tl, In, Ga and Sc were found this way, and helium was discovered in the sun.

(A 2023 question: "atoms in the gas phase show a continuous spread from red to violet" is the one false statement.)

The hydrogen spectrum. A discharge breaks H₂ into excited H atoms, which emit discrete frequencies. Balmer (1885) fitted the visible lines; Rydberg generalised it to every series:

109 677 cm⁻¹ is the Rydberg constant for hydrogen (R = 1.097 × 10⁷ m⁻¹). For an H-like ion multiply by Z².

Series	n₁ → (n₂ = n₁ + 1, n₁ + 2 …)	Region
Lyman	1	Ultraviolet
Balmer	2	Visible (the only visible series)
Paschen	3	Infrared
Brackett	4	Infrared
Pfund	5	Infrared

First line (longest λ, lowest energy) of a series: n₂ = n₁ + 1. Series limit (shortest λ, highest energy): n₂ = ∞. Lines crowd together (converge) towards the series limit, at shorter wavelength.

Number of lines · Beyond NCERT · Electrons falling from level n₂ to level n₁ in a sample of many atoms: Δn(Δn + 1)/2 lines, where Δn = n₂ − n₁. From n to the ground state that is n(n − 1)/2. One single atom gives at most Δn lines.

Example (NCERT exercise 2.15, asked 2022) · From n = 5 to ground: 5 × 4 / 2 = 10 lines. 2024: He⁺ from the fifth excited state (n = 6) to the first excited state (n = 2): Δn = 4 → 10.

Example (2020, 2023, 2026) · Shortest Lyman line of H: ν̄ = R (n₂ = ∞), so λ₁ = 1/R. Longest Balmer line of He⁺: ν̄ = 4R(1/4 − 1/9) = 5R/9, so λ = 9λ₁/5.

Example (2023) · First Paschen line 720 nm. First: 1/9 − 1/16 = 7/144. Second: 1/9 − 1/25 = 16/225. λ₂ = 720 × (7/144)/(16/225) = 492 nm.

Example (2019, 2025) · A 900 nm line: R/9 = 10⁵/9 cm⁻¹ gives λ = 9 × 10⁻⁵ cm = 900 nm → Paschen, ∞ → 3.

6 · Bohr's model of the hydrogen atom

Bohr (1913) applied Planck's quantisation to the atom. Postulates:

1. The electron moves round the nucleus in circular orbits of fixed radius and energy, called stationary states or allowed energy states, arranged concentrically.
2. An electron's energy in an orbit doesn't change with time. It moves up by absorbing, or down by emitting, exactly the energy difference; never continuously.
3. Bohr's frequency rule: the radiation absorbed or emitted between two states differing by ΔE has

4. Angular momentum is quantised: only orbits with mₑvr a whole multiple of h/2π are allowed. (Angular momentum = Iω = mₑr² × v/r = mₑvr.)

Results (n is the principal quantum number; the formulas hold for H and for H-like ions: He⁺, Li²⁺, Be³⁺, one electron each):

• a₀ = 52.9 pm is the radius of the first orbit of H, the Bohr orbit.
• R(H) = 2.18 × 10⁻¹⁸ J here is the Rydberg constant in energy units. NCERT uses the same name for 109 677 cm⁻¹; check the unit.
• E₁ = −2.18 × 10⁻¹⁸ J (−13.6 eV); E₂ = −0.545 × 10⁻¹⁸ J (−3.4 eV).
• Why negative: a free electron at rest, infinitely far away (n = ∞), has energy zero. A bound electron has less, so its energy is negative. The more negative, the more stable; n = 1 is the ground state.
• Larger Z → more negative energy and smaller radius: the electron is held more tightly.
• Speed rises with Z and falls with n (NCERT says this qualitatively). The formula · Beyond NCERT:

Energy level diagram (H atom; not to scale):

The gaps shrink as n rises, so a jump down to n = 1 always carries more energy than any jump down to n = 2.

Example (2019) · Second excited state of He⁺ means n = 3: E = −13.6 × 4/9 = −6.04 eV.

Example (2020, 2023) · Li²⁺, n = 2: r = a₀ × 4/3 = 4a₀/3. Be³⁺, n = 3: r = 9a₀/4. If the first is x, the second is (9/4)/(4/3) x = 27x/16.

Example (2026) · A radius of 70.53 pm = 52.9 × n²/Z gives n²/Z = 4/3 → Li²⁺, n = 2.

Example (NCERT Problem 2.11, exercise 2.34) · He⁺, n = 1: E = −2.18 × 10⁻¹⁸ × 4 = −8.72 × 10⁻¹⁸ J; r = 0.0529/2 = 0.02645 nm. Energy to ionise He⁺ = +8.72 × 10⁻¹⁸ J; for Li²⁺ (2026), × 9 = 1.962 × 10⁻¹⁷ J.

Explaining the line spectrum. Absorption when the electron goes up (n(final) > n(initial)), emission when it comes down. Each line is one transition; many atoms make many transitions at once, so many lines. A line's brightness depends on how many photons of that wavelength are emitted or absorbed.

For emission ΔE comes out negative (energy released). Bohr's expression matches Rydberg's empirical one, 1.09677 × 10⁷ m⁻¹ = 109 677 cm⁻¹.

Example (NCERT Problem 2.10) · H, n = 5 → 2: ΔE = 2.18 × 10⁻¹⁸ (1/25 − 1/4) = −4.58 × 10⁻¹⁹ J; ν = 6.91 × 10¹⁴ Hz; λ = 434 nm (a Balmer line, visible).

Example (2026) · Li²⁺ between levels whose sum is 4 and difference 2 (n = 3 and 1): ν̄ = 109 677 × 9 × (1 − 1/9) = 8 × 109 677 cm⁻¹ → λ = 1.14 × 10⁻⁶ cm.

Limitations of Bohr's model:

1. It can't explain the fine structure (doublets) of the H spectrum, nor the spectrum of any atom with more than one electron (not even He; nor Li⁺, a 2021 trap).
2. It can't explain splitting of lines in a magnetic field (Zeeman effect) or an electric field (Stark effect).
3. It can't explain how atoms form molecules by chemical bonds.
4. The deeper reasons (section 7): it ignores the electron's wave character and contradicts Heisenberg's principle, since a fixed orbit needs exact position and velocity together.

7 · Dual behaviour of matter and the uncertainty principle

de Broglie (1924): matter, like radiation, has both particle and wave properties.

• Confirmed when an electron beam was diffracted. The electron microscope uses it (magnification about 15 million times).
• Every moving object has a wave, but for ordinary masses λ is far too small to detect.

Example (NCERT Problems 2.12, 2.13) · A 0.1 kg ball at 10 m s⁻¹: λ = 6.626 × 10⁻³⁴ / 1 = 6.626 × 10⁻³⁴ m. An electron with KE 3.0 × 10⁻²⁵ J: v = √(2KE/m) = 812 m s⁻¹, λ = 896.7 nm.

Charged particle through a potential V · Beyond NCERT: KE = qV, so

Example (2021) · Electron through 40 kV: λ = 12.27/200 = 0.0614 Å = 6 × 10⁻¹² m. Proton vs Li³⁺ (8.3 times heavier, 3 times the charge) through the same V: λ(Li)/λ(p) = √(1/(8.3 × 3)) = 0.20 → 2 × 10⁻¹.

Example (2021) · Light of 248 nm (5.0 eV) on a metal with W₀ = 3.0 eV: KE = 2.0 eV = 3.2 × 10⁻¹⁹ J. λ = h/√(2m·KE) = 9 Å.

de Broglie and Bohr orbits · Beyond NCERT (NCERT exercise 2.32): a stable orbit holds a whole number of waves, 2πr = nλ. With rₙ = a₀n²/Z this gives λ = 2πa₀n/Z. Fourth orbit of H: 8πa₀ (2020, 2024); third: 6πa₀ (2023).

Heisenberg's uncertainty principle (1927): the exact position and the exact momentum (or velocity) of an electron can't both be known at the same time.

• Locating an electron needs light of very short wavelength, whose photons (p = h/λ) knock it and change its velocity.
• Significance: it rules out definite paths or trajectories for electrons. Exact statements are replaced by probabilities.
• It matters only for microscopic particles. For a milligram object ΔvΔx ≈ 10⁻²⁸ m² s⁻¹, negligible; for an electron it is ≈ 10⁻⁴ m² s⁻¹.

Example (NCERT Problem 2.15) · Electron located within 0.1 Å: Δv = h/(4πmΔx) = 6.626 × 10⁻³⁴ / (4π × 9.11 × 10⁻³¹ × 10⁻¹¹) = 5.79 × 10⁶ m s⁻¹.

Example (NCERT Problem 2.16) · Golf ball 40 g at 45 m s⁻¹ ± 2%: Δv = 0.9 m s⁻¹; Δx = 1.46 × 10⁻³³ m, far below any measurable size.

Example (2021) · Electron at 5 × 10⁶ m s⁻¹ ± 0.02%: Δv = 1000 m s⁻¹. Δx = 6.63 × 10⁻³⁴ / (4 × 3.14 × 9.1 × 10⁻³¹ × 1000) = 5.8 × 10⁻⁸ m = 58 × 10⁻⁹ m.

8 · The quantum mechanical model

Classical mechanics (Newton) works for large objects but fails for electrons, because it ignores wave–particle duality and the uncertainty principle. Quantum mechanics was developed in 1926, independently by Heisenberg and Schrödinger (Nobel 1933).

Ĥ, the Hamiltonian, is built from the total energy: kinetic energy of all particles, electron–nucleus attraction, electron–electron and nucleus–nucleus repulsion. Solving gives the allowed energies E and wave functions ψ.

• For H, the solutions come with three quantum numbers n, l, mₗ, arising naturally. It predicts every feature of the H spectrum, including those Bohr couldn't.
• For atoms with more electrons it can't be solved exactly; approximate solutions show orbitals like hydrogen's, but contracted by the larger nuclear charge, and with energies depending on n and l (only on n for H).

Features of the model:

1. Electron energies in atoms are quantised.
2. Quantised levels come from the wave nature of electrons: they are the allowed solutions of the equation.
3. Exact position and velocity can't both be known, so only the probability of finding the electron is given.
4. An atomic orbital is a wave function ψ for one electron. The electron in it has a definite energy. An orbital holds at most two electrons. ψ itself has no physical meaning.
5. |ψ|² is the probability density, always positive. |ψ|² × a small volume = probability of finding the electron there (Max Born).

Orbit vs orbital:

	Orbit (Bohr)	Orbital (quantum mechanics)
What	A circular path	A one-electron wave function ψ
Defined by	n	n, l, mₗ
Real?	Can't be shown to exist (uncertainty principle)	ψ² (probability density) gives the electron's probable distribution

(2022: "an atomic orbital is characterised by n and l only" is the incorrect statement; it needs mₗ too.)

9 · Quantum numbers

Number	Values	Tells you
Principal n	1, 2, 3 … (K, L, M, N …)	Shell; size; mostly the energy
Azimuthal l	0 to n − 1	Subshell; shape; energy too, in multi-electron atoms
Magnetic mₗ	−l … 0 … +l (2l + 1 values)	Orientation in space
Spin mₛ	+½ or −½ (↑ or ↓)	Spin direction

l = 0, 1, 2, 3, 4, 5 are called s, p, d, f, g, h, with 1, 3, 5, 7, 9, 11 orbitals.

• Shell n has n subshells, n² orbitals and at most 2n² electrons.
• Subshells: n = 1 → 1s; n = 2 → 2s, 2p; n = 3 → 3s, 3p, 3d; n = 4 → 4s, 4p, 4d, 4f. So 1p, 2d and 3f don't exist; g first appears at n = 5.
• mₗ comes from l, and l comes from n.
• Spin (Uhlenbeck and Goudsmit, 1925) was needed because lines of multi-electron atoms appear as doublets and triplets. The electron spins on its own axis; its spin angular momentum has two orientations. Two electrons in one orbital must have opposite spins.

Example (NCERT exercise 2.30) · Not allowed: n = 0 (n starts at 1); n = 1, l = 1 (l ≤ n − 1); n = 3, l = 3. Allowed: n = 1, l = 0, mₗ = 0, mₛ = −½; n = 2, l = 1, mₗ = 0; n = 3, l = 1, mₗ = 0.

Counting with quantum numbers (about one question a year):

Example (2021, 2024) · n = 5, mₗ = +2: l can be 2, 3 or 4 → 3 orbitals. n = 4, mₗ = 0: l = 0, 1, 2, 3 → 4 orbitals.

Example (2026) · n = 5, mₗ = −1: l = 1, 2, 3, 4 → 4 orbitals → 8 electrons. n = 3, l = 2, mₗ = −1, mₛ = +½ → 1 electron.

Example (2024) · n = 4, |mₗ| = 1, mₛ = −½: l = 1, 2, 3 each have mₗ = ±1 → 6 orbitals → 6 electrons. n = 5, mₛ = +½ (2020): one electron per orbital → 25.

Orbital angular momentum · Beyond NCERT: √(l(l + 1)) h/2π. Zero for every s orbital, √2 h/2π for p, √6 h/2π for d. Bohr's ground-state value h/2π is wrong in quantum mechanics: the 1s value is 0 (2019, 2023).

10 · Shapes of orbitals and nodes

Probability density (Fig. 2.12, a favourite of statement questions):

• 1s: |ψ|² is maximum at the nucleus and falls sharply outward.
• 2s: |ψ|² falls sharply to zero, rises to a small maximum, then falls towards zero.
• A region where |ψ|² is zero is a node (nodal surface).
• ψ can be positive or negative; |ψ|² can't be negative. Signs on lobes are signs of ψ, not charges (2025).

Boundary surface diagram: the surface of constant |ψ|² enclosing about 90% probability. Never 100%: |ψ|² has some value at every finite distance.

• s orbitals: spherical (a circle in 2-D); size 4s > 3s > 2s > 1s.
• p orbitals: two lobes on either side of a plane through the nucleus, where |ψ|² is zero. Three of them, pₓ, pᵧ, pz, mutually perpendicular, same size, shape and energy. There is no simple link between mₗ = −1, 0, +1 and x, y, z. Size and energy 4p > 3p > 2p.
• d orbitals (from n = 3): dₓᵧ, dᵧz, dₓz, dₓ²₋ᵧ² and dz². The first four look alike; dz² is different. All five 3d are equal in energy. 4d, 5d have similar shapes, larger size and energy.
• Lobes on the axes (2023): pₓ, pᵧ, pz, dₓ²₋ᵧ², dz² = 5. dₓᵧ, dᵧz, dₓz lie between the axes.

Nodes:

• Angular nodes are planes (or cones) through the nucleus: pz has the xy-plane; dₓᵧ has two planes (the xz and yz planes).
• ns has n − 1 radial nodes: 2s has 1, 3s has 2.

Orbital	Radial	Angular
2s	1	0
3s	2	0
3p	1	1
4d	1	2
5d	2	2
4f	0	3

Example (2021) · No angular node and two radial nodes → l = 0, n − 1 = 2 → 3s. n = 4, mₗ = −3 forces l = 3 (4f): radial = 4 − 3 − 1 = 0.

Example (2023) · Five radial nodes among 7s (6), 7p (5), 6s (5), 8p (6), 8d (5) → 3.

Radial node from ψ · Beyond NCERT: set the part of ψ that can change sign to zero. ψ₂ₛ contains (2 − r/a₀), so the node is at r = 2a₀ (2023).

Radial probability · Beyond NCERT: the chance of finding the electron in a thin shell at distance r is 4πr²ψ²dr. For 1s it peaks at r = a₀ (the Bohr radius), although ψ² itself peaks at the nucleus. The number of peaks is n − l. The electron can still be found at 2a₀ (2019, 2025).

11 · Energies of orbitals

Hydrogen (and H-like ions): energy depends on n only. Orbitals of the same n are degenerate (same energy).

An electron in 1s is in the ground state, held most strongly; in 2s, 2p or higher it is excited. (2025: for H, 3pₓ, 3dₓ²₋ᵧ² and 3dz² are equal and lower than 4s and 4pz.)

Multi-electron atoms: energy depends on n and l. Within a shell s < p < d < f. At higher levels they overlap: 4s < 3d, 6s < 5d, 4f < 6p.

• Why: besides nuclear attraction there is repulsion between electrons. Inner electrons shield outer ones, which feel only the effective nuclear charge Z(eff).
• Shielding power within a shell: s > p > d > f. An s electron also spends more time near the nucleus (it penetrates). So for a given n, Z(eff) falls as l rises: s is bound most tightly, then p, then d.
• Energy of the same orbital falls as Z rises: E₂ₛ(H) > E₂ₛ(Li) > E₂ₛ(Na) > E₂ₛ(K). So 2s is lowest in K (2019, 2022).

The (n + l) rule (Aufbau order):

1. Lower n + l → lower energy.
2. Same n + l → lower n has lower energy (3p < 4s, both 4; 3d < 4p < 5s, all 5).

Example (2024) · 4p, 4d, 3p, 3d, 4s: n + l = 5, 6, 4, 5, 4 → 3p < 4s < 3d < 4p < 4d.

Example (2026) · 3d, 4s, 6p, 5p, 2p (n + l = 5, 4, 7, 6, 3) → 2p < 4s < 3d < 5p < 6p.

12 · Filling orbitals and electronic configuration

Aufbau principle ("building up"): in the ground state, orbitals fill in order of increasing energy.

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s …

Pauli exclusion principle (1926): no two electrons in an atom have the same set of all four quantum numbers. Equivalently: an orbital holds at most two electrons, with opposite spins. So s holds 2, p 6, d 10, f 14, and shell n holds 2n².

Hund's rule of maximum multiplicity: in a subshell of equal-energy (degenerate) orbitals, electrons don't pair until every orbital has one, and the single electrons have parallel spins. Pairing begins with the 4th electron in p, 6th in d, 8th in f.

Writing configurations. Two ways: sᵃpᵇdᶜ notation (1s²2s²2p⁶), or an orbital diagram with a box per orbital and ↑↓ arrows (this shows all four quantum numbers).

• H 1s¹; He 1s²; Li 1s²2s¹; Be 1s²2s²; B to Ne fill 2p (Ne 1s²2s²2p⁶).
• Na to Ar repeat the pattern in 3s, 3p: Na [Ne]3s¹ … Ar [Ne]3s²3p⁶.
• Core electrons fill the inner shells; valence electrons are in the outermost shell.
• K and Ca put electrons in 4s (lower than 3d). Sc to Zn fill 3d; Ga to Kr fill 4p. Rb to Xe repeat with 5s, 4d, 5p. Cs, Ba: 6s. La to Hg: 4f and 5d. Then 6p, 7s, 5f, 6d. Elements after U are short-lived and made artificially.

Extra stability of half-filled and full subshells (p³, p⁶, d⁵, d¹⁰, f⁷, f¹⁴). When 4s and 3d are close in energy, an electron moves up to complete or half-complete the d set:

• Cr: [Ar] 3d⁵ 4s¹, not 3d⁴4s². Cu: [Ar] 3d¹⁰ 4s¹, not 3d⁹4s².
• Reasons: (1) symmetrical distribution, with small mutual shielding, so electrons are pulled harder by the nucleus; (2) exchange energy: same-spin electrons in degenerate orbitals can swap places, releasing energy. Swaps are most numerous in half-filled and full subshells. Exchange energy is the basis of Hund's rule.
• Overall the gain is: smaller shielding, smaller coulombic repulsion, larger exchange energy.
• Exchanges for k parallel electrons · Beyond NCERT: k(k − 1)/2. For d⁵, 10.

Other exceptions in NCERT's Table 2.6 (learn the first row): Mo 4d⁵5s¹, Ag 4d¹⁰5s¹, Au 5d¹⁰6s¹, Pd 4d¹⁰5s⁰, Pt 5d⁹6s¹; also Nb 4d⁴5s¹, Ru 4d⁷5s¹, Rh 4d⁸5s¹; La 5d¹6s², Gd 4f⁷5d¹6s².

Example (2025) · Cr: l = 1 electrons = 2p⁶ + 3p⁶ = 12; l = 2 electrons = 3d⁵ = 5. V (Z = 23) has 12 p electrons too (2021).

Example (2021) · Ge (Z = 32), completely filled orbitals with mₗ = 0: 1s, 2s, 2p₀, 3s, 3p₀, 3d₀, 4s = 7 (4p² puts its two electrons in separate orbitals).

Example (2026) · The 19th electron of Cr, by Aufbau, enters 4s: n = 4, l = 0, m = 0, s = +½. K's valence electron (2024) and Rb's (5, 0, 0, +½) are the same type.

Ions · Beyond NCERT: remove electrons from the highest n first, so 4s goes before 3d. Fe [Ar]3d⁶4s² → Fe³⁺ [Ar]3d⁵ (5 unpaired), not 3d³4s². Ga⁺ is [Ar]3d¹⁰4s², so its valence l = 0 (2021). Zn⁺ is [Ar]3d¹⁰4s¹, so its outermost mₗ = 0.

Spin-only magnetic moment · Beyond NCERT: μ = √(n(n + 2)) BM for n unpaired electrons. n = 1: 1.73; 2: 2.83; 3: 3.87; 4: 4.90; 5: 5.92.

13 · Traps

Trap	What goes wrong	Do this
Excited-state count	Taking "second excited state" as n = 2	Ground n = 1, first excited n = 2, second excited n = 3
Number of lines	Using n(n − 1)/2 for one atom	Many atoms: Δn(Δn + 1)/2; one atom: Δn at most
He⁺, Li²⁺ lines	Forgetting Z² in ν̄	Multiply R by Z²
Two Rydberg constants	2.18 × 10⁻¹⁸ J vs 109 677 cm⁻¹	Read the unit in the question
"First line" of a series	Taking n₂ = ∞	First line: n₂ = n₁ + 1, longest λ; limit: n₂ = ∞
Photoelectric intensity	Thinking brighter light gives faster electrons	Intensity → number; frequency → KE
Work function units	Mixing J, eV and J mol⁻¹	Convert to one; E (eV) = 1240/λ (nm)
Nodes	Giving total nodes when radial are asked	Radial n − l − 1; angular l
Energy in H	Using the n + l rule for hydrogen	H depends on n only: 3s = 3p = 3d
Ion configurations	Removing 3d before 4s	Highest n leaves first
Cr and Cu	Writing 3d⁴4s² or 3d⁹4s²	Half-filled and full d win
Probability density	"ψ² is never zero for 2s"; "boundary holds 100%"	2s has a radial node; boundary ≈ 90%
Heisenberg	Using % of v as Δv without converting	Δv = (%/100) × v; mass in kg
Orbital count	Counting electrons when orbitals are asked	Orbitals first; × 2 only for electrons

14 · Numbers worth knowing

• h = 6.626 × 10⁻³⁴ J s; c = 3.0 × 10⁸ m s⁻¹; hc = 1.988 × 10⁻²⁵ J m = 1240 eV nm.
• e = 1.602 × 10⁻¹⁹ C; mₑ = 9.109 × 10⁻³¹ kg; 1 eV = 1.602 × 10⁻¹⁹ J; 1 eV per particle = 96.5 kJ mol⁻¹; 1 Å = 0.1 nm.
• a₀ = 52.9 pm = 0.529 Å; E₁(H) = −13.6 eV = −2.18 × 10⁻¹⁸ J = −1312 kJ mol⁻¹; v₁(H) = 2.18 × 10⁶ m s⁻¹.
• R = 109 677 cm⁻¹ = 1.097 × 10⁷ m⁻¹; 1/R = 91.2 nm (the Lyman limit).

Line of H	λ (nm)
Lyman first (2 → 1)	121.6
Lyman limit (∞ → 1)	91.2
Balmer first (3 → 2)	656.3
Balmer limit (∞ → 2)	364.6
Paschen limit (∞ → 3)	820.6

15 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• "Atom" comes from the Greek a-tomio, "uncut-able". Early Indian and Greek philosophers (about 400 BC) proposed atoms; Dalton (1808) put atomic theory on a scientific basis. It failed to explain why glass or ebonite rubbed with silk or fur becomes charged.
• Faraday (1830) showed electrolysis deposits matter at electrodes, suggesting electricity is particulate.
• Nobel Prizes NCERT mentions: Thomson 1906, Planck 1918, Einstein 1921 (photoelectric effect), Bohr 1922, de Broglie 1929, Heisenberg 1932, Schrödinger and Dirac 1933.
• Configurations explain why atoms bond, why some elements are metals and why He and Ar are inert; Dalton's model could not.