Atomic Structure

Atomic structure underpins periodicity and bonding. Advanced probes the Bohr model for hydrogen-like species, quantum numbers and orbital shapes, the Heisenberg and de Broglie relations, and electronic configurations with exchange-energy stability. Numerical work on spectra and quantum numbers is common.

Key results

• Bohr (H-like): E_n = −13.6 Z²/n² eV; r_n = 0.529 n²/Z Å; v_n = 2.18×10⁶ Z/n m/s.
• Spectra: 1/λ = R Z²(1/n1² − 1/n2²); series Lyman (UV, n1=1), Balmer (visible, n1=2), Paschen (IR, n1=3).
• Quantum numbers: n (shell), l (0..n−1, subshell shape), m (−l..+l), s (±½). Orbitals per subshell = 2l+1.
• de Broglie/Heisenberg: λ = h/mv; Δx·Δp ≥ h/4π.
• Configuration rules: Aufbau (increasing n+l), Pauli, Hund; half/fully filled subshells (d⁵, d¹⁰) are extra stable (Cr, Cu anomalies).

Species	Relation
Energy	−13.6 Z²/n² eV
Radius	0.529 n²/Z Å
Orbitals in subshell	2l + 1
Electrons in shell	2n²

Exam Tricks & Tips

• 🎯 Everything for H-like ions scales as Z²/n² (energy) or n²/Z (radius) — He⁺, Li²⁺ energies follow instantly from hydrogen's −13.6 eV.
• 🎯 Half-filled and fully-filled subshells are extra stable due to exchange energy and symmetry — hence Cr is [Ar]3d⁵4s¹ and Cu is [Ar]3d¹⁰4s¹.
• 🎯 Number of spectral lines from level n = n(n−1)/2; identify the series from n1 (Lyman UV, Balmer visible).
• 🎯 Aufbau fills by increasing (n+l); ties break by lower n — this explains 4s before 3d.
• 🎯 de Broglie wavelength is significant only for light, fast particles; for macroscopic bodies it is immeasurably small.
• ❌ Common mistake: applying the Bohr model (which is exact only for one-electron species) to multi-electron atoms for quantitative energies.

Expected exam pattern

1–2 questions: a Bohr-spectrum or quantum-number numerical and a configuration/stability conceptual (multiple-correct). Ionisation-energy comparisons recur.

Quick recap

H-like: E = −13.6Z²/n², r = 0.529n²/Z. Spectral lines n(n−1)/2; know the series. Quantum numbers set orbital count 2l+1, shell capacity 2n². Half/full subshells are stable (Cr, Cu). Bohr is exact only for one electron.

Cover the answer, recall, then check. 12 cards on atomic structure for JEE Advanced.

Q1. Bohr energy of a hydrogen-like ion.
A1. E_n = −13.6 Z²/n² eV.

Q2. Radius of the nth Bohr orbit.
A2. r_n = 0.529 n²/Z Å.

Q3. Rydberg formula for a hydrogen-like spectrum.
A3. 1/λ = R Z²(1/n1² − 1/n2²).

Q4. Which spectral series lies in the visible region?
A4. Balmer series (transitions to n = 2).

Q5. Allowed values of the azimuthal quantum number l.
A5. 0 to n − 1; it sets the subshell (s, p, d, f).

Q6. Number of orbitals in a subshell of azimuthal number l.
A6. 2l + 1.

Q7. Electronic configuration of chromium and why.
A7. [Ar]3d⁵4s¹ — a half-filled d subshell is extra stable.

Q8. Heisenberg uncertainty principle.
A8. Δx·Δp ≥ h/4π.

Q9. Aufbau filling order rule.
A9. Fill by increasing (n + l); for a tie, the lower n fills first (e.g. 4s before 3d).

Q10. Maximum electrons in a shell of principal quantum number n.
A10. 2n².

Q11. Configuration of copper.
A11. [Ar]3d¹⁰4s¹ — a fully filled d subshell is favoured.

Q12. Number of spectral lines emitted from level n to ground.
A12. n(n − 1)/2.

Atomic structure is the quantum foundation of all chemistry — it explains why the periodic table has the shape it does. JEE Advanced tests the Bohr model quantitatively, the quantum numbers and orbital picture qualitatively, and the subtle exceptions (electronic configurations of Cr and Cu) that examiners love.

Core concept: electrons occupy quantised energy levels described by four quantum numbers; the Bohr model works for one-electron species, while multi-electron atoms fill orbitals by the Aufbau, Pauli and Hund rules.

Core ideas

Beginner — models and spectra

Bohr's model (for H and hydrogen-like ions) quantises angular momentum mvr = nh/2π, giving energy E_n = −13.6 Z²/n² eV and radius r_n = 0.529 n²/Z Å. Transitions produce line spectra: the Rydberg formula 1/λ = R Z²(1/n₁² − 1/n₂²) generates the Lyman (UV), Balmer (visible) and Paschen (IR) series. The number of spectral lines from level n to ground = n(n−1)/2.

Intermediate — quantum numbers and orbitals

Four quantum numbers label each electron: principal n (shell/size), azimuthal l (subshell/shape, 0→s,1→p,2→d,3→f), magnetic m_l (orientation, −l…+l), spin m_s (±½). Pauli's principle: no two electrons share all four. Number of orbitals in a shell = n²; maximum electrons = 2n². Orbital shapes: s spherical, p dumbbell (with a node through the nucleus), d cloverleaf. Radial nodes = n − l − 1; angular nodes = l; total nodes = n − 1.

Advanced — filling, exceptions, and quantum weirdness

Aufbau fills lowest (n+l) first (and lower n if n+l ties); Hund's rule maximises unpaired spins. Anomalous configurations: Cr is [Ar]3d⁵4s¹ and Cu is [Ar]3d¹⁰4s¹ — half-filled and fully-filled d subshells give extra exchange-energy stability. The de Broglie wavelength λ = h/p and Heisenberg's uncertainty Δx·Δp ≥ h/4π mark the limits of the classical picture and appear as short numericals.

Worked example

Calculate the wavelength of the second line of the Balmer series for hydrogen (R = 1.097×10⁷ m⁻¹). Balmer series has n₁ = 2; the second line is the transition n₂ = 4 → 2. 1/λ = R(1/4 − 1/16) = R(4−1)/16 = 3R/16 = 3×1.097×10⁷/16 ≈ 2.056×10⁶ m⁻¹. So λ ≈ 4.86×10⁻⁷ m = 486 nm (blue-green), the Balmer-β line. Identifying which n₁, n₂ the "series" and "line number" refer to is the whole skill.

How JEE Advanced tests this

Bohr-model numericals for hydrogen-like ions (energy, radius, ionisation energy, spectral wavelength); counting spectral lines; assigning valid sets of quantum numbers and counting nodes; writing/identifying anomalous electronic configurations; de Broglie and uncertainty-principle short calculations; and photoelectron/ionisation-energy trends.

Exam tricks & shortcuts

• E ∝ Z²/n², r ∝ n²/Z — solve hydrogen-like problems by ratio.
• Radial nodes = n − l − 1; total nodes = n − 1.
• Lines from level n = n(n−1)/2.
• Mnemonic: half-filled/full d is extra stable → "Cr and Cu steal an s-electron."

Key formulas

• Planck: E = hν = hc/λ; photoelectric KE = hν − W₀.
• Bohr: Eₙ = −13.6 Z²/n² eV; rₙ = 0.529 n²/Z Å; v = 2.18×10⁶ Z/n m/s.
• Rydberg: 1/λ = R_H Z²(1/n₁² − 1/n₂²).
• de Broglie: λ = h/mv; Heisenberg: Δx·Δp ≥ h/4π.
• Quantum numbers: l = 0…n−1; mₗ = −l…+l (2l+1); electrons per shell 2n²; orbital nodes = n−1.

Bohr's model works for one-electron species; quantum numbers (Aufbau, Pauli, Hund) fix electron configurations.

Worked Example

Problem: (a) Find the wavelength of the third line of the Balmer series of hydrogen (transition n = 5 → 2). (b) How many distinct spectral lines can appear when electrons de-excite from n = 5 to the ground state? (Rydberg constant R = 1.097 × 10⁷ m⁻¹.)

Solution:
(a) The Balmer series has the lower level n = 2. The third line is n = 5 → 2:
1/λ = R(1/2² − 1/5²) = R(1/4 − 1/25) = R(25 − 4)/100 = R(21/100).
1/λ = 1.097 × 10⁷ × 0.21 = 2.304 × 10⁶ m⁻¹.
λ = 1/(2.304 × 10⁶) ≈ 4.34 × 10⁻⁷ m = 434 nm (violet).

(b) The number of spectral lines from level n down to the ground state is:
N = n(n − 1)/2 = 5 × 4 / 2 = 10 lines.

Answer: (a) λ ≈ 434 nm; (b) 10 spectral lines.