Last Updated: May 2026
JEE 2027 — Atomic Structure overview
For JEE Main 2027 Atomic Structure, expect 2-3 questions covering Bohr’s model numerics, quantum numbers, orbital configuration, photoelectric effect formulas and de Broglie wavelength. The chapter sits at the foundation of physical chemistry and bridges into modern physics in JEE Physics — making formula-recall here doubly valuable.
Bohr Model — Master Formulas
| Quantity | Formula | Notes |
|---|---|---|
| Radius of nth orbit (H) | rn = 0.529 × n² /Z Å | 0.529 Å = Bohr radius |
| Velocity in nth orbit | vn = 2.18 × 10⁶ × Z/n m/s | — |
| Energy of nth orbit | En = –13.6 × Z²/n² eV | For H: −13.6, −3.4, −1.51, −0.85 eV |
| Energy in joules | En = –2.18 × 10⁻¹⁸ × Z²/n² J | — |
| Frequency of transition | ν = (Ei – Ef)/h | h = 6.626 × 10⁻³⁴ J·s |
| Rydberg formula | 1/λ = RH·Z²(1/n₁² – 1/n₂²) | RH = 1.097 × 10⁷ m⁻¹ |
Hydrogen Spectrum Series
| Series | Final n (n₁) | Region | First line λ |
|---|---|---|---|
| Lyman | 1 | Ultraviolet | 121.6 nm |
| Balmer | 2 | Visible | 656.3 nm (Hα, red) |
| Paschen | 3 | Infrared | 1875 nm |
| Brackett | 4 | Far infrared | 4050 nm |
| Pfund | 5 | Far infrared | 7460 nm |
Photoelectric Effect
- Einstein’s equation: hν = φ + KEmax
- φ = work function = hν₀ (threshold frequency)
- Stopping potential: eVs = KEmax = h(ν − ν₀)
- KEmax depends on frequency, not intensity
- Number of electrons emitted depends on intensity, not frequency
de Broglie Wavelength
- λ = h/p = h/(mv)
- For an electron accelerated through V volts: λ = 12.27/√V Å (where V is in volts)
- Photon: p = h/λ; E = pc
Heisenberg Uncertainty Principle
Δx · Δp ≥ h/(4π) — applies to conjugate variables (position-momentum, energy-time). Cannot simultaneously know exact position and momentum.
Quantum Numbers
| Quantum No. | Symbol | Range | Determines |
|---|---|---|---|
| Principal | n | 1, 2, 3,… | Energy, size of orbit |
| Azimuthal | l | 0 to (n−1) | Subshell shape (s, p, d, f) |
| Magnetic | ml | –l to +l | Orbital orientation |
| Spin | ms | +½, –½ | Electron spin |
Aufbau: Fill orbitals in order of increasing (n+l), with lower n first when (n+l) is equal. Pauli’s exclusion: No two electrons in an atom can have all four quantum numbers identical. Hund’s rule: Maximum unpaired electrons in degenerate orbitals before pairing.
Number of Orbitals and Electrons
- Total orbitals in nth shell: n²
- Total electrons in nth shell: 2n²
- s subshell: 1 orbital, 2 electrons
- p subshell: 3 orbitals, 6 electrons
- d subshell: 5 orbitals, 10 electrons
- f subshell: 7 orbitals, 14 electrons
Number of Radial and Angular Nodes
- Radial nodes = n − l − 1
- Angular nodes = l
- Total nodes = n − 1
35 Practice MCQs
- Bohr radius for H is — (A) 0.529 Å (B) 1.058 Å (C) 5.29 Å (D) 0.0529 Å
- Energy of H atom in n=1 — (A) −13.6 eV (B) −3.4 eV (C) +13.6 eV (D) 0
- Energy of H atom in n=2 — (A) −13.6 eV (B) −3.4 eV (C) −1.51 eV (D) −0.85 eV
- Lyman series lies in — (A) UV (B) visible (C) IR (D) far IR
- Balmer series lies in — (A) UV (B) visible (C) IR (D) microwave
- Hα line is in — (A) Lyman (B) Balmer (C) Paschen (D) Pfund
- Number of orbitals in n=3 — (A) 1 (B) 4 (C) 9 (D) 16
- Maximum electrons in n=3 — (A) 2 (B) 8 (C) 18 (D) 32
- Number of f orbitals — (A) 1 (B) 3 (C) 5 (D) 7
- Maximum value of ml for l=2 — (A) ±1 (B) ±2 (C) ±3 (D) 0
- 3p subshell electrons — (A) 2 (B) 6 (C) 10 (D) 14
- de Broglie wavelength of an electron at 100 V is — (A) 12.27 Å (B) 1.227 Å (C) 0.1227 Å (D) 12.27 nm
- Heisenberg’s principle: Δx·Δp ≥ — (A) h (B) h/2 (C) h/(4π) (D) h/π
- Threshold frequency for K (work function 2.3 eV) — (A) 5.55×10¹⁴ Hz (B) 1.0×10¹⁵ Hz (C) 1.5×10¹⁵ Hz (D) 7.0×10¹⁴ Hz
- Stopping potential depends on — (A) frequency (B) intensity (C) both (D) neither
- Photoelectric current depends on — (A) frequency (B) intensity (C) both (D) neither
- The Aufbau order: 4s vs 3d — fills first — (A) 3d (B) 4s (C) simultaneously (D) depends on element
- Cu electronic config — (A) [Ar]3d⁹4s² (B) [Ar]3d¹⁰4s¹ (C) [Ar]3d¹⁰4s² (D) [Ar]3d⁸4s²
- Cr electronic config — (A) [Ar]3d⁴4s² (B) [Ar]3d⁵4s¹ (C) [Ar]3d⁶4s⁰ (D) [Ar]3d³4s³
- Total nodes in 3p orbital — (A) 0 (B) 1 (C) 2 (D) 3
- Radial nodes in 3p — (A) 0 (B) 1 (C) 2 (D) 3
- Angular nodes in 4d — (A) 0 (B) 1 (C) 2 (D) 4
- Quantum number determining orbital shape — (A) n (B) l (C) ml (D) ms
- For 4s subshell: l = (A) 0 (B) 1 (C) 2 (D) 3
- Pauli’s exclusion principle states — (A) no two electrons same quantum numbers (B) electrons fill lowest energy first (C) maximize unpaired (D) all are random
- Speed of electron in n=2 of H = — (A) 1.09×10⁶ m/s (B) 2.18×10⁶ m/s (C) 4.36×10⁶ m/s (D) 1.5×10⁷ m/s
- Hund’s rule applies to — (A) degenerate orbitals (B) different shells (C) all atoms (D) noble gases
- Wavelength of first line of Lyman series of H ≈ — (A) 121.6 nm (B) 656.3 nm (C) 91.2 nm (D) 410 nm
- The number of electrons with n=3, l=2 in iron — (A) 6 (B) 5 (C) 4 (D) 0 (no 3d in Fe? actually Fe = [Ar]3d⁶4s²)
- Photon momentum p = (A) hν/c (B) hλ (C) h/c (D) hc
- Compton effect demonstrated — (A) wave nature of light (B) particle nature of light (C) wave nature of matter (D) zero rest mass
- Davisson-Germer experiment proved — (A) wave nature of electrons (B) particle nature (C) Bohr’s model (D) Pauli’s principle
- Schrödinger equation gives — (A) wave function ψ (B) trajectory (C) classical orbit (D) field
- The probability density is — (A) ψ (B) ψ² (C) ψ⁻¹ (D) ψ + ψ*
- Zeeman effect is — (A) splitting of spectral lines in magnetic field (B) photoelectric (C) Compton (D) Doppler
Answer Key
1-A, 2-A, 3-B, 4-A, 5-B, 6-B, 7-C, 8-C, 9-D, 10-B, 11-B, 12-B, 13-C, 14-A, 15-A, 16-B, 17-B, 18-B, 19-B, 20-C, 21-B, 22-C, 23-B, 24-A, 25-A, 26-A, 27-A, 28-A, 29-A, 30-A, 31-B, 32-A, 33-A, 34-B, 35-A
FAQ
How many questions from Atomic Structure in JEE Main 2027?
Typically 2–3 questions, mostly numerical (Bohr radius/energy, de Broglie, photoelectric).
Should I memorise spectrum series wavelengths?
Memorise the series names + their final n + spectrum region. Specific wavelengths are derivable from Rydberg’s formula in the exam.
Which is the highest-yield formula here?
Rydberg’s: 1/λ = RH·Z²(1/n₁² − 1/n₂²). Combined with En = −13.6Z²/n² eV, these two cover 70% of JEE Atomic Structure numerics.
Are anomalous configs of Cr, Cu important?
Yes — Cr [Ar]3d⁵4s¹ and Cu [Ar]3d¹⁰4s¹ are recurrent JEE asks for “exception to Aufbau”.