VKC Notes · Quantum chemistry · Civil Services P-I · IFoS P-I

1.4 Hydrogen Atom: Schrödinger Equation, Quantum Numbers, Radial Distribution, Nodes and Orbital Shapes

★★★ 2021–2025: 8 Mains parts, 68 marks (Civil Services 4 parts / 30 m · IFoS 4 / 38 m). Civil Services asks the 1s radial distribution almost every year (most probable radius, mean radius, probability ratio); IFoS asks the equation in spherical coordinates, nodes and quantum numbers. Every question is a short derivation or a count: learn the five formulas in the box below and the radial-distribution sketch.

Syllabus lines. Civil Services: “Schrödinger wave equation (time independent); interpretation of wave function, particle in one-dimensional box, quantum numbers, hydrogen atom wave functions; shapes of s, p and d orbitals”. IFoS: “Schrödinger wave equation (time independent). Interpretation of wave function, particle in one-dimensional box, quantum numbers, hydrogen atom wave functions. Shapes of s, p and d orbitals.” Geo-Scientist asks the same ideas inside its quantum chemistry and periodicity questions.

Two-page short notes for revision All VKC notes

Download PDF PLAN

PDF edition, print-ready. It opens with the VKC plan, together with every locked answer, note and test on the site.

CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

1.4.1 Schrödinger equation in spherical polar coordinates ★★

The proton is about 1836 times heavier than the electron, so take it as fixed at the origin (or use the reduced mass μ ≈ m_e). The potential is the Coulomb attraction, which depends on r only, so polar coordinates (Fig. 1.4.1) separate the problem IFoS 2025 Q2(a)(i) 10m.SN

(1.4.1)

Fig. 1.4.1 Spherical polar coordinates with the nucleus at the origin.

Draw it in 60 s: z up, y right, x slanting down-left; join O to P (length r); mark θ from +z to OP and φ in the xy plane from +x to the foot of the perpendicular from P.

Separation: put , multiply by r²/ and collect: one side depends on r only, the other on θ, φ only, so each equals the same constant l(l + 1). This gives three ordinary equations. Conversions: ; volume element .SN

(1.4.2)

Where the quantum numbers come from (acceptability, 1.1): Φ = e^im_lφ must be single-valued, Φ(φ + 2π) = Φ(φ) ⇒ m_l = 0, ±1, ±2…; Θ must stay finite at θ = 0, π ⇒ l = 0, 1, 2… with |m_l| ≤ l; R must stay finite as r → ∞ ⇒ n = 1, 2, 3… with l ≤ n − 1. The term l(l + 1)ħ²/2μr² is the centrifugal barrier: it keeps electrons with l > 0 away from the nucleus.SN

1.4.2 Energy levels and the spectrum ★★★

(1.4.3)

E depends on n only: the n² orbitals of a shell are degenerate in hydrogen (in many-electron atoms penetration splits ns < np < nd). Levels crowd together towards E = 0, the ionisation limit; IE(H) = 13.6 eV = 1312 kJ mol⁻¹.SN
Series: Lyman (n₁ = 1, UV, 121.6 → 91.2 nm), Balmer (n₁ = 2, visible: Hα 656, Hβ 486, Hγ 434 nm), Paschen (n₁ = 3, IR). Selection rule for one-photon transitions: Δl = ±1, Δn any.SN

Fig. 1.4.2 Hydrogen levels drawn to scale: E ∝ −1/n². Lyman ends on n = 1, Balmer on n = 2.

Draw it in 60 s: horizontal lines at −13.6, −3.4, −1.5, −0.85 eV and a dashed line at 0; down arrows to n = 1 (Lyman, UV), to n = 2 (Balmer, visible), to n = 3 (Paschen, IR).

Solved PYQ CSE 2021 Q2(a) 10m: radiation for n = 101 → 100, and why it is hard to observe.

⇒ λ = 4.63 cm, ν = c/λ = 6.48 × 10⁹ Hz (microwave).SN
Hard to see: such a Rydberg atom has radius n²a₀ ≈ 0.53 µm and binding energy 13.6/10⁴ eV = 1.36 meV, far below kT = 25.7 meV at 298 K, so one collision ionises it; the two levels are almost equally populated and spontaneous emission (∝ ν³) is very slow. It is seen only in thin interstellar gas (the H100α radio line).

1.4.3 Quantum numbers ★★

Number	Allowed values	Fixes	Comes from
principal n	1, 2, 3 …	energy (Eq. 1.4.3), size ⟨r⟩	R finite as r → ∞
azimuthal l	0 … n − 1 (s, p, d, f)	\|L\| = √(l(l+1)) ħ; angular nodes	Θ finite at θ = 0, π
magnetic m_l	−l … +l (2l + 1 values)	L_z = m_lħ; orientation	Φ single-valued
spin m_s	±½	S_z = m_sħ	not from Eq. 1.4.1 (relativistic)

Orbitals per shell = n²; electrons per shell = 2n². Pauli: no two electrons of an atom share all four numbers; Hund: degenerate orbitals are singly filled with parallel spins first.SN

Solved PYQ IFoS 2023 Q1(a) 8m: four quantum numbers for the eight electrons of O (1s² 2s² 2p⁴).

e	1	2	3	4	5	6	7	8
(n, l)	1, 0	1, 0	2, 0	2, 0	2, 1	2, 1	2, 1	2, 1
m_l	0	0	0	0	+1	0	−1	+1
m_s	+½	−½	+½	−½	+½	+½	+½	−½

Electrons 5–7 go singly into the three 2p orbitals with parallel spins (Hund); the eighth pairs. Two unpaired electrons: O is paramagnetic. Any equivalent choice of m_l for the pair, or all spins reversed, is equally correct.

Part of VKC Basic

Read all 9 sections of this topic, its short notes and the PDF with VKC Basic

VKC Basic (₹999 for 6 months) opens every solved answer, the notes, the handbook and the daily answer archive. Every Mains test series includes it.

See what is included Already a member? Sign in

Solved PYQs, short notes and books: vkcscience.comTelegram: t.me/upsc_chemistry_vkc