Atomic Structure

Physical Chemistry Weightage: 2-3 Questions (8-12 Marks) JEE Unit 2

“Welcome back! These notes assume you have already read the NCERT chapter and know the basic definitions, so they skip the textbook story and keep only what JEE Main actually asks. Atomic Structure is one of the most reliable scoring chapters: it usually gives two to three questions per paper, and most of them are short calculations or counting problems that can be done in under a minute if you know the pattern. The favourites are Bohr-model energy and spectral lines, quantum-number counting, de Broglie and uncertainty numericals, photoelectric effect and electronic configurations. Learn the scaling laws and the counting shortcuts below and most of this chapter turns into arithmetic.”

— SCORECHEM ACADEMIC TEAM

1. Light and the Photoelectric Effect

c = νλ, wave number = 1/λ, E = hν = hc/λ. Shortcut: E (eV) = 1240 / λ (nm). Energy is proportional to frequency and wave number and inversely proportional to wavelength.
Ordering photons: convert everything to one quantity. Example: 400 nm photon, 10¹⁶ s⁻¹ photon and 10⁴ cm⁻¹ photon. Energies: 4.97 × 10⁻¹⁹ J, 6.63 × 10⁻¹⁸ J and 1.99 × 10⁻¹⁹ J; so B > A > C.
Photoelectric effect: hv = W₀ + KE_max, KE_max = h(ν − ν₀) = eV_s. Given two metals with work-function ratio and equal photon energy, set up (E − W_A)/(E − W_B) = the given KE ratio and solve for W.

Fig. 1: KE max vs frequency in the photoelectric effect, and what controls what.

⚠️ JEE Trap: Intensity vs frequency. Higher intensity gives MORE photoelectrons (larger current) but the same KE. Only higher frequency raises KE. Below the threshold frequency nothing is emitted, however bright the light. Also, the slope of KE vs ν is h for every metal; only the intercept (−W₀) changes with the metal.

2. Bohr Model Toolkit

Valid ONLY for one-electron species: H, He⁺, Li²⁺, Be³⁺. Remember these as scaling laws, not as formulas to rederive.

Fig. 3: Bohr model scaling laws (single-electron species only).

Energy: E_n = −13.6 Z²/n² eV = −2.18 × 10⁻¹⁸ Z²/n² J. Here R_H = 2.18 × 10⁻¹⁸ J is an ENERGY (Rydberg energy), not the wave-number Rydberg constant 1.097 × 10⁷ m⁻¹.
KE = −E, PE = 2E. For hydrogen first excited state (E = −3.4 eV): KE = +3.4 eV.
Ionisation energy (from n) = 13.6 Z²/n² eV.
Radius / speed: r = 0.529 n²/Z Å, v = 2.18 × 10⁶ Z/n m/s. For the first orbit of H, v = 2.18 × 10⁶ m/s = 21.8 × 10⁵ m/s.
Energy to excite a mole: ΔE per atom × N_A. Excitation 1 → 2 needs 0.75 R_H per atom.
De Broglie link: 2πr = nλ, so λ = 2πr/n. In the 4th orbit of H, r = 16a₀ and λ = 8πa₀.
Frequency of the de Broglie wave in orbit = v/λ = mv²/h = 2KE/h.

⚠️ JEE Trap: Sign and Z. Energies of bound states are NEGATIVE. Options that show +2.18 × 10⁻¹⁸ J for a stationary state are always wrong. Check Z²/n²: third orbit of Li²⁺ gives 9/9 = 1, so E = −2.18 × 10⁻¹⁸ J exactly.

3. Hydrogen Spectrum

Fig. 2: Hydrogen energy levels and spectral series (arrows: 1st, 2nd, 3rd line).

Energy of a line: ΔE = 13.6 Z² (1/n₁² − 1/n₂²) eV; wave number = R Z² (1/n₁² − 1/n₂²).
Series: Lyman (n₁=1, UV), Balmer (2, visible), Paschen (3, near IR), Brackett (4, IR), Pfund (5, far IR). Only Balmer is in the visible region.
Ordering lines: compute the fraction (1/n₁² − 1/n₂²) for each. Example: Paschen 1st (3→4): 7/144 = 0.049; Brackett 4th (4→8): 3/64 = 0.047; Paschen 3rd (3→6): 1/12 = 0.083; Balmer 2nd (2→4): 3/16 = 0.19. Ascending: Brackett 4th < Paschen 1st < Paschen 3rd < Balmer 2nd.
Number of lines: from n to lower levels in a bulk sample: n(n−1)/2 to the ground state; between n₁ and n₂: Δn(Δn+1)/2 with Δn = n₂ − n₁. A SINGLE atom emits at most n − 1 lines.
Ratio of Lyman 1st to Balmer 1st: 10.2 eV / 1.89 eV = 5.4.
Different species, same line: He⁺ (Z = 2) transition n₁→n₂ equals the hydrogen line with levels n₁/2 and n₂/2. So the 2nd Balmer line of He⁺ (4→2) has the same energy as the 1st Lyman line of H (2→1): 10.2 eV.
Given the sum and difference of levels (n₁ + n₂ = 4, n₂ − n₁ = 2): n₂ = 3, n₁ = 1.

⚠️ JEE Trap: Which line is "second"? The second line of the Balmer series is 4→2, not 3→2. The "series limit" is n₂ = ∞. In a wavelength question for Li²⁺, do not forget the Z² = 9: 1/λ = 1.097 × 10⁷ × 9 × (8/9) m⁻¹ for 3→1.

4. de Broglie and Uncertainty

λ = h/mv = h/√(2m·KE) = h/√(2mqV). For an electron, λ = 12.27/√V Å.
Same accelerating voltage, different particles: λ ∝ 1/√(mq). If charges are equal, λ₁/λ₂ = √(m₂/m₁); for masses 1 and 4 amu the ratio is 2. If the charges differ, include q.
Heisenberg: Δx · mΔv ≥ h/4π. For an electron confined to a nucleus (Δx = 10⁻¹⁵ m): Δv = 6.626 × 10⁻³⁴ / (4 × 3.14 × 9.1 × 10⁻³¹ × 10⁻¹⁵) ≈ 5.8 × 10¹⁰ m/s. That exceeds the speed of light, which is why electrons cannot exist inside the nucleus.
Macroscopic objects have negligible uncertainty; the principle matters only for microscopic particles.

⚠️ JEE Trap: h/4π vs h/2π. The uncertainty relation uses h/4π, while Bohr's angular momentum uses nh/2π. Also convert Δx to metres and mass to kg before dividing, and use the exact constants the question gives (6.626 × 10⁻³⁴, π = 3.14) when the answer is a nearest integer.

5. Quantum Numbers and Orbitals

Number	Values	Tells you	Counts
n	1, 2, 3 ...	shell, size, energy	n subshells, n² orbitals, 2n² electrons
l	0 to n−1	shape (s, p, d, f)	2l + 1 orbitals in a subshell
m_l	−l to +l	orientation	2l + 1 values
m_s	+½ or −½	spin	2 per orbital

Counting shortcuts (JEE loves these):

Orbitals with a given n and m_l = m: l can run from |m| to n − 1, so the number of orbitals is n − |m|. For n = 4, m_l = 0: 4 orbitals (s, p, d, f).
Orbitals with given n and |m_l| = 1: m_l = +1 and −1 each give n − 1 orbitals, so 2(n − 1). For n = 4 that is 6, and the number of electrons with m_s = −½ is also 6.
Given n = 5, m_l = −1: l = 1, 2, 3, 4 gives 4 orbitals, so a maximum of 8 electrons. With all four numbers (n, l, m_l, m_s) fixed, exactly 1 electron.
Orbital angular momentum = √(l(l+1)) ℏ (zero for s). Spin-only moment = √(n(n+2)) B.M. with n unpaired electrons.

Fig. 4: Nodes: radial = n - l - 1, angular = l, total = n - 1.

⚠️ JEE Trap: Orbit vs orbital and l vs m_l. An orbital is a probability region with a maximum of 2 electrons, not a path; s orbitals have no angular nodes but still have radial nodes from 2s onwards. And "m_l = 0" does not mean "s orbital only": every subshell has an orbital with m_l = 0.

6. Electronic Configuration

Fig. 5: (n + l) rule. Same n + l: the lower n is lower in energy and fills first.

Aufbau: fill in increasing (n + l); ties go to the lower n. Pauli: no two electrons with all four quantum numbers equal. Hund: singly occupy degenerate orbitals with parallel spins first.
Energy comparison for a multi-electron atom: (n + l) decides; equal (n + l) means the lower n is lower in energy. Example (2024): 3p (4) < 4s (4) < 3d (5) < 4p (5) < 4d (6). For hydrogen-like species, energy depends on n only.
Exceptions to know: Cr [Ar] 3d⁵ 4s¹, Cu [Ar] 3d¹⁰ 4s¹, Mo [Kr] 4d⁵ 5s¹, Ag [Kr] 4d¹⁰ 5s¹, Pd [Kr] 4d¹⁰, Pt [Xe] 4f¹⁴ 5d⁹ 6s¹, Au [Xe] 4f¹⁴ 5d¹⁰ 6s¹. Lanthanum and gadolinium take 5d¹; actinium and curium take 6d¹.
f-block check: Es (Z = 99) is [Rn] 5f¹¹ 7s² (regular, no 6d electron).
Cations of transition metals: remove the ns electrons first, then (n−1)d (see Unit 11).
Dalton's theory (asked in 2024): the INCORRECT statements are "atoms are divisible" and "all atoms of an element have different properties". The theory says atoms are indivisible, atoms of an element are identical in mass, compounds form in fixed ratios and reactions reorganise atoms. (Modern view: atoms are divisible; isotopes differ in mass.)

⚠️ JEE Trap: The 4s/3d order. 4s fills BEFORE 3d, but 4s electrons are removed FIRST on ionisation. Fe²⁺ is [Ar] 3d⁶, not [Ar] 3d⁴ 4s².

7. How JEE Frames Questions

If you see	Do this
"Arrange spectral lines by energy"	Compute (1/n₁² − 1/n₂²) for each; ignore the constant
Photon energies in different units	Convert all to J (or eV) with E = hc/λ = hν = hc(wave number)
One-electron ion, transition wavelength	1/λ = R Z² (1/n₁² − 1/n₂²); watch cm vs m
"Number of electrons with n, l, m_l, m_s"	Count orbitals with the given restrictions, then multiply by 1 or 2
Two metals with a ratio of work functions	Set (E − W_A)/(E − W_B) = KE ratio; solve
Velocity, radius or frequency in Bohr orbit	Use Z/n and n²/Z scaling from the first orbit
"Incorrect statement" list	Check sign, unit and definition of each; one wrong number is enough
Uncertainty or de Broglie NAT	Convert to SI, use the given constants, round only at the end

8. Quick Sheet and Checklist

Point	Formula
Photon	E = hc/λ = 1240/λ eV (nm)
Photoelectric	hv = W₀ + KE_max; KE = eV_s
Bohr energy	−13.6 Z²/n² eV; KE = −E; PE = 2E
Bohr radius / speed	0.529 n²/Z Å; 2.18 × 10⁶ Z/n m/s
Line count	Δn(Δn+1)/2 (bulk); n − 1 (single atom)
de Broglie	λ = h/mv = h/√(2mqV); 2πr = nλ
Uncertainty	Δx · mΔv ≥ h/4π
Nodes	radial n−l−1; angular l; total n−1
Orbitals for n and m_l=m	n −
Spin-only moment	√(n(n+2)) B.M.

Before the exam, check you can:

Order any set of hydrogen lines by energy without a calculator.
Compute energy, radius, speed and wavelength for He⁺ and Li²⁺ using scaling.
Count electrons or orbitals for any set of quantum numbers in under 30 seconds.
Do a de Broglie and an uncertainty numerical to the nearest integer.
Write the configuration of Cr, Cu, Pd, Pt, Au and of the f-block exceptions.