JEE Main Atomic Structure 2027 — Bohr Model, Quantum Numbers, Spectra and 35 Practice MCQs

JEE Main Advanced preparation engineering entrance study material

Last Updated: May 2026

JEE Main Atomic Structure 2027 - Bohr model quantum numbers spectra

JEE Main Atomic Structure 2027 is NCERT Class XI Chapter 2 — the foundational chapter that powers every later quantum/bonding/periodicity problem. In JEE Main 2024-25 across 16 shifts, Atomic Structure contributed an average of 1.6 questions / 6.4 marks per shift, with the Bohr model + spectrum + quantum numbers triad accounting for 70% of those items. NEET-aligned reasoning makes overlap with Inorganic 1 strong, so this chapter is also a “build a complete answer for one MCQ but lock 4 chapters of trivia” deal.

This guide covers the entire JEE Main Atomic Structure syllabus — Bohr model, hydrogen spectrum, de Broglie wavelength, Heisenberg uncertainty, quantum numbers, Aufbau / Hund / Pauli, photoelectric effect, electronic configurations — followed by 35 JEE-style practice MCQs with full answer key, an interactive 10-MCQ quiz, and a JEE 2027 FAQ.

1. Sub-atomic Particles — Quick Recap

Particle Charge Mass (kg) Discoverer
Electron −1.602×10⁻¹⁹ C 9.109×10⁻³¹ J. J. Thomson (1897, e/m); Millikan (1909, e)
Proton +1.602×10⁻¹⁹ C 1.673×10⁻²⁷ Goldstein (anode rays); Rutherford
Neutron 0 1.675×10⁻²⁷ Chadwick (1932)

2. Atomic Models — Quick Hierarchy

  • Dalton (1808) — atom indivisible.
  • Thomson (1898) — plum pudding (electrons embedded in positive sphere).
  • Rutherford (1911) — gold-foil α-scattering → nuclear model. Failure: predicted unstable atom (electron should spiral into nucleus by Maxwell’s equations).
  • Bohr (1913) — quantised circular orbits; resolved hydrogen spectrum.
  • Sommerfeld (1916) — elliptical orbits; introduced ℓ.
  • Quantum mechanical (Schrödinger 1926) — orbitals, ψ², four quantum numbers.

3. Bohr Model of the Hydrogen Atom (HEAVY JEE)

Bohr’s three postulates:

  1. Electrons revolve in stationary orbits without emitting energy.
  2. Angular momentum is quantised: mvr = nh/(2π).
  3. Transition between orbits emits/absorbs energy: hν = E2 − E1.

Bohr Formulas (memorise — pure plug-and-play marks)

Quantity Hydrogen-like formula For H (Z=1, n=1)
Radius rₙ rₙ = 0.529 × n²/Z Å 0.529 Å (Bohr radius)
Velocity vₙ vₙ = 2.18×10⁶ × Z/n m/s 2.18×10⁶ m/s
Energy Eₙ Eₙ = −13.6 × Z²/n² eV −13.6 eV
Frequency of revolution fₙ ∝ Z²/n³ —

Key inferences (a JEE staple):

  • Going from n=1 to n=2 in H atom: r×4, v÷2, E×(1/4) (becomes less negative).
  • For He+ (Z=2, n=1): r=0.529/2 = 0.2645 Å, E=−54.4 eV.
  • Ionisation energy of H = 13.6 eV; Li²⁺ = 13.6×9 = 122.4 eV.

4. Hydrogen Atom Spectrum (Series)

Rydberg formula: 1/λ = RH·Z²·(1/n₁² − 1/n₂²), where RH = 1.097×10⁷ m⁻¹.

Series n₁ (lower) n₂ (upper) Region
Lyman 1 2,3,4,… UV
Balmer 2 3,4,5,… Visible
Paschen 3 4,5,6,… IR
Brackett 4 5,6,7,… Far IR
Pfund 5 6,7,8,… Far IR

Number of spectral lines when an electron drops from n=n to n=1: n(n−1)/2.

5. Limitations of Bohr Model

  • Fails for multi-electron atoms.
  • Cannot explain fine structure of spectral lines (resolved by Sommerfeld).
  • Cannot explain Zeeman / Stark effects (splitting in magnetic / electric fields).
  • Violates Heisenberg uncertainty principle (precise orbit + momentum is forbidden).

6. Wave-Particle Duality — de Broglie

λ = h / p = h / (mv). Significant only for sub-atomic particles. For an electron at 1 V acceleration: λ = 12.27/√V Å.

7. Heisenberg Uncertainty Principle

Δx · Δp ≥ h/(4π) — also expressed as Δx · Δv ≥ h/(4πm). Direct corollary: an electron in an atom cannot have a precisely defined orbit (wipes out Bohr’s classical picture).

8. Quantum Numbers — the Address of an Electron

QN Symbol Tells you Allowed values
Principal n Shell, size, energy 1, 2, 3, …
Azimuthal (Angular momentum) ℓ Subshell, shape 0 to (n−1)
Magnetic mℓ Orientation of orbital −ℓ to +ℓ
Spin ms Electron spin direction +½, −½

ℓ = 0 → s (1 orbital), ℓ = 1 → p (3 orbitals), ℓ = 2 → d (5 orbitals), ℓ = 3 → f (7 orbitals).

Total electrons in shell n: 2n². In a subshell ℓ: 2(2ℓ+1).

Nodes

  • Total nodes = n − 1.
  • Angular nodes = ℓ.
  • Radial nodes = n − ℓ − 1.

9. Rules of Filling — Electronic Configuration

  1. Aufbau principle: lower (n+ℓ) fills first; if (n+ℓ) ties, lower n fills first. Order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
  2. Pauli exclusion: no two electrons in an atom can have all four quantum numbers identical. Each orbital holds ≤2 electrons of opposite spin.
  3. Hund’s rule of maximum multiplicity: when degenerate orbitals are available, electrons occupy each singly with parallel spins before pairing.

Anomalous Configurations (JEE staples)

Element Expected Actual Why
Cr (Z=24) [Ar] 3d⁴ 4s² [Ar] 3d⁵ 4s¹ Half-filled stability
Cu (Z=29) [Ar] 3d⁹ 4s² [Ar] 3d¹⁰ 4s¹ Fully-filled stability
Mo (Z=42) [Kr] 4d⁴ 5s² [Kr] 4d⁵ 5s¹ Half-filled
Ag (Z=47) [Kr] 4d⁹ 5s² [Kr] 4d¹⁰ 5s¹ Fully-filled

10. Photoelectric Effect (Einstein 1905)

hν = W₀ + ½mv², where W₀ = work function = hν₀ (threshold frequency).

  • If ν < ν₀: no emission, regardless of intensity.
  • If ν > ν₀: kinetic energy depends on ν − ν₀, NOT on intensity.
  • Number of photoelectrons ∝ intensity.

11. Schrödinger Equation & Orbital Concept

ψ is the wavefunction; ψ² is the probability density of finding an electron. An orbital is the 3D region where probability of finding an electron is high (typically >90%). Crucially: orbitals replace orbits.

35 JEE Main Practice MCQs — Atomic Structure

  1. Energy of electron in nth orbit of H atom: (a) −13.6/n² eV (b) +13.6 eV (c) −13.6n² eV (d) 13.6/n eV
  2. Bohr’s radius for H atom is: (a) 0.529 nm (b) 0.529 Å (c) 5.29 Å (d) 1 Å
  3. Velocity of electron in 1st orbit of He⁺ is: (a) 2.18×10⁶ m/s (b) 4.36×10⁶ m/s (c) 1.09×10⁶ m/s (d) 5.45×10⁶ m/s
  4. Lyman series falls in: (a) UV (b) Visible (c) IR (d) Microwave
  5. Balmer series corresponds to n₁ = (a) 1 (b) 2 (c) 3 (d) 4
  6. Number of spectral lines from n=4 to n=1: (a) 3 (b) 4 (c) 6 (d) 10
  7. Ionisation energy of Li²⁺ is: (a) 13.6 eV (b) 27.2 eV (c) 122.4 eV (d) 54.4 eV
  8. Quantisation of angular momentum: mvr = (a) nh (b) nh/π (c) nh/(2π) (d) nh/(4π)
  9. de Broglie wavelength of electron at 100 V: (a) 12.27 Å (b) 1.227 Å (c) 0.1227 Å (d) 122.7 Å
  10. Heisenberg’s uncertainty: Δx·Δp ≥ (a) h (b) h/2 (c) h/(4π) (d) h/π
  11. For an electron, ℓ = 2 means: (a) s (b) p (c) d (d) f
  12. Number of orbitals in d subshell: (a) 1 (b) 3 (c) 5 (d) 7
  13. Maximum electrons in M shell: (a) 2 (b) 8 (c) 18 (d) 32
  14. For 3p orbital, n+ℓ = (a) 3 (b) 4 (c) 5 (d) 6
  15. Number of radial nodes in 3p: (a) 0 (b) 1 (c) 2 (d) 3
  16. Total nodes in 4s orbital: (a) 1 (b) 2 (c) 3 (d) 4
  17. Cr (Z=24) configuration: (a) [Ar] 3d⁴ 4s² (b) [Ar] 3d⁵ 4s¹ (c) [Ar] 3d⁶ 4s⁰ (d) [Ar] 4s² 3d⁴
  18. Cu (Z=29) configuration: (a) [Ar] 3d⁹ 4s² (b) [Ar] 3d¹⁰ 4s¹ (c) [Ar] 3d⁸ 4s³ (d) [Ar] 4s¹ 3d¹⁰ — same as b
  19. Pauli exclusion forbids: (a) two electrons in same orbital (b) two electrons with same spin in same orbital (c) two electrons with all four QN identical (d) odd electrons
  20. Hund’s rule applies to: (a) Filling within a shell (b) Filling of degenerate orbitals (c) Pauli pairing (d) Aufbau order
  21. The set of QN (3,2,−2,+½) corresponds to which orbital: (a) 3p (b) 3d (c) 3s (d) 4d
  22. Photoelectric current depends on: (a) Frequency only (b) Intensity (c) Work function only (d) Wavelength only
  23. KE of photoelectron depends on: (a) Frequency above threshold (b) Intensity (c) Work function only (d) Number of photons
  24. Threshold frequency relation: (a) hν₀ = W₀ (b) ν₀ = h/W₀ (c) hν₀ = mv (d) hν₀ = ½mv²
  25. Mass of photon at rest is: (a) 9.1×10⁻³¹ (b) 0 (c) 1.67×10⁻²⁷ (d) Planck’s constant
  26. Wavefunction ψ has physical meaning: (a) ψ itself = probability (b) ψ² = probability density (c) ψ³ = energy (d) ψ⁻¹ = orbital size
  27. Number of unpaired electrons in Fe³⁺ (Z=26): (a) 4 (b) 5 (c) 3 (d) 6
  28. (n+ℓ) value for 4f: (a) 5 (b) 6 (c) 7 (d) 8
  29. Order of filling using Aufbau: (a) 4s < 3d (b) 3d < 4s (c) Both equal (d) 3d before 3p
  30. Wavelength of K-α X-ray varies with Z as: (a) Moseley’s law: √ν ∝ (Z−b) (b) ν ∝ Z (c) ν ∝ Z² (d) ν ∝ 1/Z
  31. Total spectral lines from n=5 to ground: (a) 6 (b) 10 (c) 15 (d) 4
  32. Energy required to remove electron from n=2 of H: (a) 13.6 eV (b) 3.4 eV (c) 27.2 eV (d) 6.8 eV
  33. Speed of e⁻ in 2nd orbit of H is half of: (a) 1st orbit (b) Light (c) Sound (d) Itself
  34. Number of orbitals when n=3: (a) 4 (b) 9 (c) 16 (d) 18
  35. For an electron in 3d orbital, the magnetic QN cannot be: (a) −2 (b) 0 (c) +3 (d) +1

Answer Key (JEE Atomic Structure Practice)

1-a 2-b 3-b 4-a 5-b 6-c 7-c 8-c
9-b 10-c 11-c 12-c 13-c 14-b 15-b 16-c
17-b 18-b 19-c 20-b 21-b 22-b 23-a 24-a
25-b 26-b 27-b 28-c 29-a 30-a 31-b 32-b
33-a 34-b 35-c          

Interactive 10-MCQ Quiz

[cg_quiz data=”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” /]

FAQ — JEE Main Atomic Structure 2027

How many JEE Main questions come from Atomic Structure?

Average 1.6 questions / 6.4 marks per shift across 2024-25. The Bohr-energies + spectrum + quantum numbers triad accounts for ~70% of those items.

What is the most common Bohr-model question?

Energy / radius / velocity ratio between two states of a hydrogen-like ion. JEE drills the proportionality: r ∝ n²/Z, v ∝ Z/n, E ∝ −Z²/n². Memorise this row of three.

Why does Cr have 3d⁵ 4s¹ instead of 3d⁴ 4s²?

Half-filled d-subshell (3d⁵) gives extra exchange-energy stability, so an electron from 4s shifts to 3d. Cu (3d¹⁰ 4s¹) follows the same logic but for fully-filled stability.

What is the difference between an orbit and an orbital?

Orbit = fixed circular path (Bohr — disproven by uncertainty principle). Orbital = 3D probability cloud where electron is found ≥90% of the time (quantum mechanical model — Schrödinger). JEE always tests this distinction.

How is the photoelectric effect linked to Bohr’s quantum idea?

Both confirm energy is absorbed/released in discrete packets (quanta of hν). JEE asks for Einstein’s equation hν = W₀ + ½mv² and the cut-off threshold property — KE depends on frequency, not intensity.

Practise these chapters on JEE Gurukul: start with the free 7-day demo, or see plans from ₹399 a month.

Start your JEE prep free — no card needed Start Free →