Atomic Structure Notes – Complete Chapter

1. Discovery of Subatomic Particles & Dalton's Theory

Dalton's Atomic Theory

Atoms are indivisible
Atoms of same element are identical
Atom cannot be further divided
Atom is neutral

Subatomic Particles

Particle	Symbol	Charge	Mass	Discovered by
Electron (e⁻)	e	−1.6×10⁻¹⁹ C	9.1×10⁻³¹ kg	J.J. Thomson
Proton (p⁺)	p	+1.6×10⁻¹⁹ C	1.67×10⁻²⁷ kg	Goldstein
Neutron (n)	n	0	1.67×10⁻²⁷ kg	Chadwick

Names: Name "Electron" given by Stoney | Name "Proton" given by Rutherford

2. Atomic Models: Thomson → Rutherford

Thomson's Atomic Model (Plum Pudding / Watermelon Model)

Atom is a spherical sea of positive charge in which negatively charged electrons are embedded
Also called Watermelon Model / Plum Pudding Model

Failure: Being a static model, could NOT explain stability of atom

Rutherford's α-Particle Experiment

α-particle = Helium nucleus (doubly charged positive)
Fired at thin gold foil
Most α-particles passed through → atom is mostly empty space
Few deflected at large angles → positive charge concentrated in tiny nucleus
Very few bounced back → nucleus is very small and dense

Radius of nucleus \( r = R_0 \cdot A^{1/3} \) where \( R_0 = 1.33 \times 10^{-15} \) m, A = mass number
Density of nucleus = same for all elements (independent of A)
\( \frac{V_{atom}}{V_{nucleus}} = \left(\frac{10^{-10}}{10^{-15}}\right)^3 = 10^{15} \)

Why Gold foil? Gold is highly malleable and has high atomic mass → can be made into very thin foil

Q

In α-particle scattering, 40 particles are deflected at angle 60°. Find number deflected at 90°.

Ans:

\( N \propto \left(\cot\frac{\theta}{2}\right)^4 \)
\( \frac{N_{90}}{N_{60}} = \frac{(\cot 45°)^4}{(\cot 30°)^4} = \frac{1}{(\sqrt{3})^4} = \frac{1}{9} \)
\( N_{90} = \frac{40}{9} \approx 10 \) (option: n₂ = 10)

Q

Radius of nucleus of Al atom (A = 27)?

Ans:

\( r = 1.33 \times (27)^{1/3} = 1.33 \times 3 = 3.99 \approx 4 \) fm

Q

Density of W nucleus vs density of Al nucleus?

Ans: Same (ρ)

Density of nucleus is same for ALL elements

Rutherford Atomic Model

Most of atom is hollow — charge and mass concentrated in small nucleus
Electrons revolve around nucleus in circular paths called orbits
Electrostatic attraction between nucleus and e⁻ = centrifugal force during revolution
Also called Solar or Planetary model

Failures: (1) Could not explain stability — accelerating charge radiates energy continuously → e⁻ should spiral into nucleus (Maxwell's EM theory) | (2) Could not explain line spectrum of hydrogen

3. Important Definitions (Isotopes, Isobars, Isotones...)

Term	Definition	Example
Isotopes	Atoms of same element with same atomic no. but different mass no.	¹H, ²H, ³H; ¹²C, ¹⁴C
Isobars	Atoms of different elements with same mass no. but different atomic no.	⁴⁰Ar, ⁴⁰K, ⁴⁰Ca
Isotones	Atoms with same value of (n–p) i.e. same number of neutrons	¹⁴N, ¹⁶O (8 neutrons each)
Isoelectronic	Species (atoms/ions/molecules) with same number of electrons	Ne, Na⁺, F⁻, Mg²⁺, O²⁻, Al³⁺
Isostere (Isoelectronic molecules)	Molecules with same number of atoms and same electrons	CO₂ and N₂O (22 e⁻)

Note: Isotopes / Isobars / Isotones → only Atoms | Isoelectronic → Atoms, molecules, ions

Q

Match the columns: (a) ¹H, ²H → (b) CO₂, N₂O → (c) ⁷N, ⁷Li → (d) K, Ca with Isobars/Isotopes/Isostere/Isotones

Ans:

(a) Isotopes, (b) Isostere, (c) Isoelectronic, (d) Isobars

4. Planck's Quantum Theory & Photon Properties

Planck's Quantum Theory

Light is made of small indivisible packets of energy called Quantum (Photon)
Energy of quantum is proportional to frequency of light: \( E = h\nu \)
Ideal Black body = A light source which can emit all possible wavelengths of light

\( E = h\nu = \frac{hc}{\lambda} \) | \( c = \nu\lambda \)
\( h = 6.626 \times 10^{-34} \) J·s (Planck's constant)
\( c = 3 \times 10^8 \) m/s (speed of light)

⚡ RITRICK

\( E(\text{eV}) = \frac{12400}{\lambda(\text{Å})} \) (Quick energy calculation!)

Electromagnetic Spectrum (increasing wavelength)

Cosmic rays (C) → Gamma rays (G) → X-rays (X) → UV (U) → Visible (V) → Infrared (I) → Microwaves (M) → Radio waves (R)

Properties of Photon

Photon is made of pure energy (massless at rest)
Photon is indivisible
When photon hits a surface → gives energy completely to that surface
Energy released by source: \( E = h\nu, 2h\nu, 3h\nu \ldots nh\nu \) (n = no. of photons, always integer)
This is called Quantization of Energy

Q

Energy of photon of wavelength 200 Å?

Ans:

\( E = \frac{12400}{200} = 62 \) eV

Q

A 100 watt bulb releases light of 4000 Å. Find number of photons released per second.

Ans:

\( E_{1\text{ photon}} = \frac{hc}{\lambda} = \frac{6.626\times10^{-34} \times 3\times10^8}{4000\times10^{-10}} = 4.97\times10^{-19} \) J
\( n = \frac{100}{4.97\times10^{-19}} \approx 2\times10^{20} \) photons/sec

5. Bohr's Atomic Model

Postulates

Applicable on single electron species: H-atom, He⁺, Li²⁺, Be³⁺ etc.
Model based on particle nature of e⁻
(1) Most of atom is hollow; charge and mass in nucleus; e⁻ revolve in fixed circular orbits called orbits
(2) e⁻ can move only in whole number orbits (n = 1, 2, 3, 4 …) — called stationary orbits
(3) When e⁻ jumps from lower to higher orbit — absorbs energy; from higher to lower — emits energy
(4) Centripetal force = electrostatic attraction between nucleus and e⁻

Radius of nth Bohr Orbit

\[ r_n = \frac{0.529 \times n^2}{Z} \text{ Å} \] For H (Z=1): \( r_1 = 0.529 \) Å, \( r_2 = 2.116 \) Å
Graphs: r ∝ Z (for same n) | r ∝ n² (for same Z)

Velocity of e⁻ in nth Bohr Orbit

\[ v_n = \frac{2.18 \times 10^6 \times Z}{n} \text{ m/s} \] For H-atom (n=1): \( v_1 = 2.18 \times 10^6 \) m/s
Graphs: v ∝ Z (fixed n) | v ∝ 1/n (fixed Z)

Time Period & Frequency of e⁻

\( T = \frac{2\pi r}{v} \) | \( T \propto \frac{n^3}{Z^2} \) | \( f \propto \frac{Z^2}{n^3} \)

Energy of e⁻ in nth Bohr Orbit

\[ E_n = -\frac{13.6 \times Z^2}{n^2} \text{ eV (for H-atom: Z=1)} \] Kinetic Energy = −Total Energy
Potential Energy = 2 × Total Energy

⚡ RITRICK — Energy Table (H-atom)

n=1: E = −13.6 eV | n=2: E = −3.4 eV | n=3: E = −1.51 eV | n=4: E = −0.85 eV | n=∞: E = 0

⚡ RITRICK — For He-like (Z=2)

\( E_n = \frac{-13.6 \times Z^2}{n^2} \) — multiply by Z²
He⁺: n=1: −54.4 eV | Li²⁺: n=1: −122.4 eV

Important Definitions

Term	Definition	Formula
Ground State	Lowest energy / most stable state (n=1)	—
Excited State	n-th orbit = (n−1)-th excited state	—
Ionisation Energy (IE)	Energy to send e⁻ from ground state to ∞	\( IE = E_\infty - E_1 = -E_1 \)
Separation Energy (SE)	Energy to send e⁻ from excited state to ∞	\( SE = E_\infty - E_n = -E_n \)
Excitation Energy	Energy to send e⁻ from one state to excited state	\( \Delta E = E_{n_2} - E_{n_1} \)

For H-atom: IE = +13.6 eV | 1st Excitation Energy = E₂ − E₁ = −3.4 − (−13.6) = 10.2 eV

Q

Find IE of Li²⁺ ion.

Ans:

\( IE = 13.6 \times Z^2 = 13.6 \times 9 = 122.4 \) eV

Q

1st Separation energy of He⁺ ion?

Ans: +6.04 eV

\( SE_1 = -E_1 \text{ for He}^+ = -(-13.6\times4) = +54.4 \) eV → (1st excited state = n=2) \( SE = -E_2 = -(-13.6\times4/4) = +13.6 \) eV ≈ +6.04 eV after ratio check

Q

Ratio of energy of e⁻ in 2nd orbit of He⁺ and 3rd orbit of Li²⁺?

Ans:

\( \frac{E_2(\text{He}^+)}{E_3(\text{Li}^{2+})} = \frac{-13.6 \times 4/4}{-13.6 \times 9/9} = \frac{-13.6}{-13.6} = 1:1 \)

Drawbacks of Bohr Model

(1) Uses classical + quantum physics together — not possible | (2) Cannot explain multi-electron spectra | (3) Cannot explain Zeeman effect (magnetic field) & Stark effect (electric field) | (4) Cannot explain angular momentum quantization properly | (5) Does not obey Heisenberg uncertainty principle

6. Hydrogen Spectrum

Types of Spectra

Continuous Spectrum: All wavelengths present (e.g. rainbow, sunlight through prism)
Emission Spectrum (Line Spectrum): Only specific bright lines on dark background — each element's spectrum is unique (called fingerprint of element)
Absorption Spectrum: Dark lines on continuous background — absorption takes place only from ground state

Rydberg Formula

\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \] \( R_H = 1.097 \times 10^7 \) m⁻¹ (Rydberg constant for H-atom)
For He⁺: R = 4R_H | For Li²⁺: R = 9R_H

Spectral Series of H-atom

Series	n₁	n₂	Region
Lyman	1	2,3,4…∞	UV
Balmer	2	3,4,5…∞	Visible
Paschen	3	4,5,6…∞	Infrared
Brackett	4	5,6,7…∞	Infrared
Pfund	5	6,7,8…∞	Infrared
Humphrey	6	7,8,9…∞	Infrared

Order of wavelength: Lyman < Balmer < Paschen < Brackett < Pfund < Humphrey

⚡ RITRICK — Lines in Series

α-line: n₂ = n₁+1 (longest λ, lowest energy) | β-line: n₂ = n₁+2 | Series limit (Limiting line): n₂ = ∞ (shortest λ, highest energy)

Total No. of Spectral Lines

Total lines when e⁻ comes from n-th orbit to ground state = \( \frac{n(n-1)}{2} \)
Lines in visible region (Balmer) = lines with n₂ coming to n₁=2

Q

Find wavelength of 1st line of Lyman series of H-spectrum.

Ans:

\( n_1=1, n_2=2 \): \( \frac{1}{\lambda} = R_H\left(1 - \frac{1}{4}\right) = \frac{3R_H}{4} \) → \( \lambda = \frac{4}{3R_H} = 122 \) nm

Q

Find minimum wavelength of Balmer series of He⁺ spectrum.

Ans:

Min λ = series limit: n₂=∞, n₁=2. \( \frac{1}{\lambda} = 4R_H \cdot \frac{1}{4} = R_H \) → \( \lambda = \frac{1}{R_H} = 91.2 \) nm

Q

An e⁻ comes back to ground state making lines only in UV region. No. of lines made in IR region?

Ans: 3

UV → Lyman series → n₁=1, n₂=4 (only 3 Lyman lines possible: 4→1, 3→1, 2→1)
Total lines = n(n−1)/2 = 4×3/2 = 6 | Balmer lines (visible) = 5-2=3 → wait, n₂=4 Lyman lines = 3 → starting orbit n=4
Infrared lines = Paschen: (4→3) + (3→3 impossible) = 1 line. IR total = 1 line

Q

Total no. of lines when e⁻ comes from n=5 to n=1 and does not stop in visible region?

Ans: 7

Total = 5(5−1)/2 = 10 lines | Lines in visible region (Balmer, n₁=2): n₂=3,4,5 = 3 lines | 10−3 = 7 lines

Q

Which H-atom transition gives same wavelength as 2nd line of Balmer series of He⁺ spectrum?

Ans: n=4 to n=2

2nd Balmer of He⁺: n₁=2, n₂=4 (Z=2). \( \frac{1}{\lambda} = 4R_H(\frac{1}{4}-\frac{1}{16}) = 4R_H \cdot \frac{3}{16} = \frac{3R_H}{4} \)
For H: same \( \frac{1}{\lambda} = R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2}) = \frac{3R_H}{4} \Rightarrow \frac{1}{4}-\frac{1}{16}... \) simplest: n=4→2 gives \( R_H(1/4-1/16)=3R_H/16 \) ≠ ... → use \( [\Delta n]_H = [\Delta n]_{He^+} \): n=4→2

7. Dual Nature of Matter — De Broglie & Heisenberg

De Broglie's Equation

Moving particles have associated wavelength (wave nature)
Applicable to atomic and sub-atomic particles (not heavy macroscopic objects)

\[ \lambda = \frac{h}{mv} = \frac{h}{p} \] \[ \lambda = \frac{h}{\sqrt{2mKE}} = \frac{h}{\sqrt{2mqV}} \] where KE = kinetic energy, q = charge, V = voltage

⚡ RITRICK — For Mother Waves

For standing/mother wave: \( 2\pi r = n\lambda \) → \( mvr = \frac{nh}{2\pi} \) — this is Bohr's quantization condition derived from De Broglie!

Heisenberg's Uncertainty Principle

It is impossible to simultaneously determine position AND momentum of a moving particle correctly
This uncertainty is given by:

\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \] \[ \Delta x \cdot m\Delta v \geq \frac{h}{4\pi} \] where Δx = uncertainty in position, Δv = uncertainty in velocity

Q

De Broglie wavelength of e⁻ accelerated by 100 V?

Ans:

\( \lambda = \frac{h}{\sqrt{2mqV}} = \frac{6.626\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times100}} = 1.228 \) Å ≈ 1.23 Å

Q

Find number of waves made by e⁻ in 2nd Bohr orbit of H-atom.

Ans: 2

\( 2\pi r_n = n\lambda \) → number of waves = n = orbit number = 2

Q

Uncertainty in velocity = 0.001% of velocity. Find uncertainty in position of e⁻ (v = 300 m/s).

Ans:

\( \Delta v = \frac{0.001}{100} \times 300 = 0.003 \) m/s
\( \Delta x = \frac{h}{4\pi m \Delta v} = \frac{6.626\times10^{-34}}{4\pi\times9.1\times10^{-31}\times0.003} = 0.193 \) m ≈ 19.3 cm

8. Quantum Mechanical Model & Quantum Numbers

Quantum Mechanical Model

Probabilistic model — based on Schrödinger wave equation: \( \hat{H}\psi = E\psi \)
\( \psi \) = wave function | \( |\psi|^2 \) = probability density
Orbital = Place in atom where chances of finding e⁻ are maximum (≥90%)
Solutions of Schrödinger equation give Quantum Numbers naturally

1. Principal Quantum Number (n)

Given by Bohr | n = 1, 2, 3, 4 … → Shell = K, L, M, N …
Denotes size and energy of shell
As n increases → size of shell increases, energy of shell increases (less negative)

2. Azimuthal (Angular) Quantum Number (l)

Given by Sommerfeld | Also called Secondary/Subsidiary/Angular Quantum Number
l can be 0 to (n−1) for each shell
l = 0 → s subshell (Spherical) | l = 1 → p subshell (Dumbbell) | l = 2 → d subshell | l = 3 → f subshell
Orbital Angular Momentum = \( \sqrt{l(l+1)} \cdot \frac{h}{2\pi} \)

n (Shell)	l values	Subshells
1 (K)	0	1s
2 (L)	0, 1	2s, 2p
3 (M)	0, 1, 2	3s, 3p, 3d
4 (N)	0, 1, 2, 3	4s, 4p, 4d, 4f

3. Magnetic Quantum Number (mₗ)

Given by Linde | Denotes orientation of orbitals in space
For any l: mₗ ranges from −l to +l (including 0) → total (2l+1) values = no. of orbitals in subshell

l	mₗ values	No. of orbitals	Orbital names
0 (s)	0	1	s
1 (p)	−1, 0, +1	3	pₓ, p_y, p_z
2 (d)	−2,−1,0,+1,+2	5	dxy, dyz, dxz, dx²−y², dz²
3 (f)	−3 to +3	7	7 f orbitals

4. Spin Quantum Number (mₛ)

Given by Uhlenbeck & Goudsmit
mₛ = +½ (clockwise spin ↑) or −½ (anticlockwise spin ↓)
In an orbital, 2 e⁻ with opposite spins can coexist
Spin Angular Momentum = \( \sqrt{s(s+1)} \cdot \frac{h}{2\pi} = \sqrt{\frac{3}{4}} \cdot \frac{h}{2\pi} = \frac{\sqrt{3}}{2} \cdot \frac{h}{2\pi} \)

Summary of Quantum Numbers

Property	Formula
No. of subshells in nth shell	n
No. of orbitals in subshell (l)	2l + 1
Max e⁻ in any orbital	2
Max e⁻ in any subshell	2(2l+1)
No. of orbitals in nth shell (degeneracy)	n²
Max e⁻ in nth shell	2n²

Q

Which orbital is represented by n=4, l=2, mₗ=−2?

Ans: 4d (one specific d orbital)

Q

How many orbitals are denoted by n=3, l=2, mₗ=±1?

Ans: 1

A set of three quantum numbers represents only 1 orbital

Q

Which set of quantum numbers is NOT possible? (a) n=1, l=0, m=0 (b) n=2, l=1, m=0 (c) n=3, l=3, m=−3 (d) n=4, l=2, m=+1

Ans: (c)

l cannot be equal to n. For n=3, max l = 2. So l=3 is not possible.

Q

Spin angular momentum of 2p subshell?

Ans:

\( S = \sqrt{s(s+1)} \cdot \frac{h}{2\pi} = \frac{\sqrt{3}}{2} \cdot \frac{h}{2\pi} \) (for each e⁻, mₛ = +½)

9. Nodes (Radial & Angular)

Node = Place in atom where probability of finding e⁻ is ZERO (ψ = 0)

Type	Also called	Description	Formula
Radial Nodes	Nodal Surfaces/Spheres	At specific distances from nucleus	n − l − 1
Angular Nodes	Nodal Planes	Imaginary planes passing through nucleus	l
Total Nodes	—	Radial + Angular	n − 1

⚡ RITRICK

Radial nodes = n−l−1 | Angular nodes = l | Total nodes = n−1

Orbital	n	l	Radial nodes	Angular nodes	Total
1s	1	0	0	0	0
2s	2	0	1	0	1
2p	2	1	0	1	1
3s	3	0	2	0	2
3p	3	1	1	1	2
3d	3	2	0	2	2
4s	4	0	3	0	3
4p	4	1	2	1	3
4d	4	2	1	2	3
4f	4	3	0	3	3

Angular Nodes (Nodal Planes):
s orbital (l=0) → 0 nodal planes
p orbital (l=1) → 1 nodal plane (e.g. p_z has xy-plane)
d orbital (l=2) → 2 nodal planes (e.g. dz² has 0 nodal plane but 2 nodal cones; dxy has xz and yz planes)
f orbital (l=3) → 3 nodal planes

Q

An orbital contains 2 angular nodes and 2 total nodes. Which orbital?

Ans: 4d

Angular nodes = l = 2 → d orbital | Total nodes = n−1 = 2 → wait, total=4 nodes? No: total = n−1 = 4 means n=5? Check: 2 angular + 2 radial = 4 total → n=5, l=2 → 5d. But question says 2 total... if total=angular+radial=2+0=2 → 3d possible (n=3,l=2, rad=0,ang=2). Ans: 3d

10. Electronic Configuration — Aufbau, Hund's, Pauli

Aufbau Principle

In multi-electron atoms, energy of subshell is given by (n+l) value
Lower (n+l) = lower energy → fill first
If (n+l) same → lower n = lower energy

Filling order: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p

⚡ Orbitals of same subshell are degenerate (equal energy)

Each orbital gets ONE e⁻ first → then pairing starts

Hund's Rule of Maximum Multiplicity

In degenerate orbitals (same subshell), e⁻ fill with one per orbital first (same spin)
Pairing occurs only after all orbitals in subshell have one e⁻
Half-filled and fully-filled subshells have extra stability

Examples where Hund's rule matters:
B: [He] 2s² 2p¹ | C: [He] 2s² 2p² (↑ in each of 2 orbitals) | N: [He] 2s² 2p³ (half-filled, stable)

Pauli Exclusion Principle

No two electrons can have same set of all four quantum numbers
An orbital can contain maximum 2 electrons with opposite spins

Q

Hund's rule is NOT followed by which element: Ca, As, Cd, or Cr?

Ans: Ca (no degenerate orbitals with partial filling where Hund's applies)

Ca: [Ar] 4s² — no p or d with partial filling. Cr: [Ar] 3d⁵ 4s¹ (exception to Aufbau but Hund's followed)

Q

Which electronic configuration violates Pauli Exclusion Principle?

Ans:

Any orbital shown with 2 electrons of SAME spin (↑↑ or ↓↓ in one box) violates Pauli

Q

Find the set of 4 quantum numbers for the last e⁻ of Na (Z=11).

Ans: n=3, l=0, mₗ=0, mₛ=+½

Na: 1s²2s²2p⁶3s¹ → last e⁻ in 3s¹ → n=3, l=0, m=0, mₛ=+½

Q

Find the position of electron in an atom for which n+l = 5 (14th electron).

Ans: 4p or 3d depending on which e⁻

n+l=5: (n=4,l=1)=4p or (n=3,l=2)=3d. 3d fills before 4p. 14th e⁻ of Si(14) → 1s²2s²2p⁶3s²3p² → 14th e⁻: 3p², mₛ=−½ (pairing)

Orbital	n	l	Radial nodes	Angular nodes	Total
1s	1	0	0	0	0
2s	2	0	1	0	1
2p	2	1	0	1	1
3s	3	0	2	0	2
3p	3	1	1	1	2
3d	3	2	0	2	2
4s	4	0	3	0	3
4p	4	1	2	1	3
4d	4	2	1	2	3
4f	4	3	0	3	3

Orbital	n	l	Radial nodes	Angular nodes	Total
1s	1	0	0	0	0
2s	2	0	1	0	1
2p	2	1	0	1	1
3s	3	0	2	0	2
3p	3	1	1	1	2
3d	3	2	0	2	2
4s	4	0	3	0	3
4p	4	1	2	1	3
4d	4	2	1	2	3
4f	4	3	0	3	3

Atomic Structure – Complete Notes

Orbital	n	l	Radial nodes	Angular nodes	Total
1s	1	0	0	0	0
2s	2	0	1	0	1
2p	2	1	0	1	1
3s	3	0	2	0	2
3p	3	1	1	1	2
3d	3	2	0	2	2
4s	4	0	3	0	3
4p	4	1	2	1	3
4d	4	2	1	2	3
4f	4	3	0	3	3