NCERT Plus One (Class 11) Chemistry • Chapter 2

Structure of Atom

Subatomic Discoveries, Rutherford & Bohr Models, Hydrogen Spectra, de Broglie, Heisenberg, Quantum Numbers & Orbitals

📄 Official NCERT Textbook PDF

Topic 2.1 – 2.2: Subatomic Particles, Thomson, Rutherford Gold Foil, Isotopes & Isobars

Imagine firing thousands of fast bullets straight through a sheet of tissue paper—only to watch one out of every 20,000 bullets ricochet straight back at your nose! Ernest Rutherford said it was almost as incredible as firing a 15-inch artillery shell at tissue paper and having it bounce back. That shocking moment in 1911 shattered the solid-ball atom and uncovered the tiny, super-dense atomic nucleus.

1. Discovery of Electron, Proton & Neutron

  • Cathode Rays → Electron (J.J. Thomson, 1897): In a discharge tube at very low pressure & high voltage, rays move from cathode (−) to anode (+), travel in straight lines, deflect like negative particles, and are identical for ALL gases/electrodes. Thomson measured specific charge: e/me = 1.758820 × 1011 C kg-1.
  • Millikan's Oil Drop Experiment (1909): Showed electric charge is quantized (q = ne), giving electron charge e = −1.602176 × 10-19 C and mass me = 9.1094 × 10-31 kg.
  • Canal Rays → Proton (1919): Positively charged gaseous ions whose e/m depends on the gas. Lightest ion came from hydrogen (proton, +1.602 × 10-19 C).
  • Neutron (James Chadwick, 1932): Bombarding thin beryllium with α-particles emitted neutral particles slightly heavier than protons.
ParticleSymbolAbsolute Charge (C)Relative ChargeMass (kg)Approx Mass (u)
Electrone−1.602176 × 10-19−19.109382 × 10-310 (0.00054)
Protonp+1.602176 × 10-19+11.672622 × 10-271 (1.00727)
Neutronn001.674927 × 10-271 (1.00867)

2. Thomson vs Rutherford Atomic Models

  • Thomson's Plum-Pudding / Watermelon Model (1898): Uniform positive sphere (radius ≈ 10-10 m) with electrons embedded like seeds; explained electrical neutrality but failed scattering tests.
  • Rutherford's α-Scattering (Gold Foil ~100 nm + ZnS Screen): (i) Most α-particles passed undeflected → atom is mostly empty space. (ii) Small fraction deflected by small angles → concentrated positive core. (iii) Very few (~1 in 20,000) bounced back (~180°) → positive charge & mass pack inside a minuscule nucleus (rnucleus ≈ 10-15 m vs ratom ≈ 10-10 m — like a cricket ball inside a 5 km radius stadium!).
  • Drawbacks of Rutherford Model: According to Maxwell's EM theory, an accelerating orbiting electron should continuously emit radiation and spiral into the nucleus within 10-8 s (fails atomic stability); also says nothing about electronic distribution or energies.
Atomic Composition & Nuclides (AZX):
Atomic Number (Z) = Protons = Electrons (in neutral atom)  |  Mass Number (A) = Protons (Z) + Neutrons (n)
Isotopes (same Z, different A): 11H, 21D, 31T & 3517Cl, 3717Cl (identical chemical properties!).
Isobars (same A, different Z): 146C and 147N.
NCERT Problems 2.1 & 2.2:
• In neutral 8035Br: Protons = 35, Electrons = 35, Neutrons = 80 − 35 = 45.
• A species has 18 electrons, 16 protons, 16 neutrons: Z = 16 (Sulphur), A = 16 + 16 = 32, charge = 16 − 18 = −2 → 3216S2-.
Always write Mass Number (A) as the top-left superscript and Atomic Number (Z) as the bottom-left subscript (e.g., 7935Br—never flip them!). Also remember: isotopes show the same chemical properties because chemical reactions depend on electrons (controlled by Z), NOT neutrons!

Topic 2.3.1 – 2.3.2a: Electromagnetic Waves, Spectrum, Blackbody Radiation & Planck's Quantum Theory

Watch an iron rod glowing inside a blacksmith's furnace: first it turns dull red, then bright orange-red, then white, and at blistering temperatures turns bluish-white! Classical wave physics predicted every hot body should blind us with infinite ultraviolet light (the "ultraviolet catastrophe"). In 1900, Max Planck cracked the puzzle by revealing that light energy isn't a smooth ramp—it comes in discrete stair-steps called quanta!

1. Wave Nature of Electromagnetic Radiation (Maxwell, 1870)

  • Oscillating electric and magnetic fields are mutually perpendicular and perpendicular to wave propagation; require no medium and travel in vacuum at c = 3.0 × 108 m s-1.
  • Electromagnetic Spectrum (increasing frequency ν, decreasing wavelength λ):
    Radio (~106 Hz) < Microwave (~1010 Hz) < Infrared (~1013 Hz) < Visible (4.0×1014 to 7.5×1014 Hz; Red 750 nm → Violet 400 nm) < UV (~1016 Hz) < X-rays < γ-rays.
Wave Equations:   c = ν λ   |   Wavenumber ν̄ = 1 / λ (SI unit: m-1; common: cm-1)
Planck's Quantum Relation (1900):   E = hν = hc / λ  (where h = 6.626 × 10-34 J s)
Allowed oscillator energies are quantized like staircase steps: E = 0, hν, 2hν, 3hν, ..., nhν.

2. Why Classical Physics Failed & Blackbody Radiation

Wave theory explains diffraction and interference, but fails to explain: (i) blackbody radiation, (ii) photoelectric effect, (iii) heat capacity of solids at low T, and (iv) atomic line spectra. An ideal blackbody absorbs and emits all frequencies uniformly (approximated by a hollow cavity with a pinhole). As temperature rises, its intensity peak shifts toward shorter wavelengths (red → white → blue)—explained accurately only by Planck's energy quanta!

NCERT Problems 2.5 & 2.6:
• Problem 2.5: Yellow light λ = 5800 Å = 5.8 × 10-7 m = 5.8 × 10-5 cm.
ν̄ = 1 / (5.8 × 10-5 cm) = 1.724 × 104 cm-1  |  ν = c / λ = (3.0 × 108) / (5.8 × 10-7) = 5.17 × 1014 s-1.
• Problem 2.6: Energy of 1 mole of photons at ν = 5 × 1014 Hz:
E1 = 6.626 × 10-34 × 5 × 1014 = 3.313 × 10-19 J → For 1 mol = 3.313 × 10-19 × 6.022 × 1023 = 199.51 kJ mol-1.
Shorter wavelength (violet, 400 nm) means HIGHER frequency and HIGHER photon energy than red (750 nm)! Also watch units: always convert nm (10-9 m) or Å (10-10 m) to metres before calculating frequency, and multiply by Avogadro's number (6.022 × 1023) when asked for energy of one mole of photons!

Topic 2.3.2b: The Photoelectric Effect & Einstein's Photon Explanation

Shine a blindingly bright red spotlight onto clean potassium metal for hours—not a single electron budges! Now flick on even a faint, dim yellow-green torch—BAM! Electrons leap off the metal instantaneously! Classical waves couldn't explain why weak high-frequency light beats intense red light. In 1905, Albert Einstein realized light hits metal not as a gentle ocean tide, but as individual particle bullets called photons!

1. Experimental Facts (H. Hertz, 1887)

When light strikes clean alkali metals (K, Rb, Cs) inside a vacuum tube:

  1. Zero Time Lag: Electrons eject the instant light hits the surface.
  2. Threshold Frequency (ν0): Below a metal's characteristic minimum frequency ν0 (e.g., for Potassium ν0 = 5.0 × 1014 Hz), NO electrons eject regardless of brightness (red light 4.3–4.6 × 1014 Hz fails; weak yellow 5.1 × 1014 Hz succeeds!).
  3. Role of Frequency vs Intensity: Kinetic energy of ejected electrons depends ONLY on frequency (increases linearly with ν > ν0). Increasing intensity (brightness) increases ONLY the number of ejected photoelectrons (photocurrent).
Einstein's Photoelectric Equation (1905 — Nobel Prize 1921):
hν = hν0 + ½ mev2   ⇒   KEmax = h(ν − ν0) = hc(1/λ − 1/λ0)
Minimum escape energy W0 = hν0 is the Work Function (Cs: 1.9 eV, K: 2.25 eV, Na: 2.3 eV, Mg: 3.7 eV, Ag: 4.3 eV, Cu: 4.8 eV).

2. Dual Nature of Electromagnetic Radiation

Light exhibits wave-like behaviour (interference, diffraction) while propagating, and particle-like behaviour (photons in blackbody & photoelectric effect) when interacting with matter.

NCERT Problem 2.9: Threshold frequency ν0 = 7.0 × 1014 s-1, incident ν = 1.0 × 1015 s-1.
KE = h(ν − ν0) = (6.626 × 10-34 J s)(10.0 × 1014 − 7.0 × 1014) s-1 = (6.626 × 10-34)(3.0 × 1014) = 1.988 × 10-19 J!
Top MCQ trap: What happens to photoelectron Kinetic Energy if you double the light's brightness (intensity)? Answer: KE stays UNCHANGED! Only the number of electrons emitted per second doubles. Also watch the wavelength condition: ejection needs ν ≥ ν0, which means incident wavelength must be LESS than or equal to threshold wavelength (λ ≤ λ0)!

Topic 2.3.3 – 2.4: Atomic Line Spectra, Bohr's Model of Hydrogen & Rydberg Series

Every element writes its own signature in neon-bright lines—an atomic barcode! In 1868, astronomers pointed a prism at sunlight and spotted mysterious yellow lines belonging to no Earthly element—they named the new cosmic element Helium (from Greek Helios, Sun) nearly thirty years before finding it on Earth! In 1913, Niels Bohr cracked how Hydrogen prints its crisp spectral barcode.

1. Continuous vs Line Spectra & Hydrogen Series

White light spreads into a seamless rainbow (continuous spectrum). Excited gaseous atoms emit sharp bright lines on a dark background (line emission spectrum), while cool gases absorb those exact lines from white light (absorption spectrum—photographic negative of emission).

Hydrogen SeriesLower Level (n1)Upper Level (n2)EM Spectral Region
Lyman12, 3, 4, ...Ultraviolet (UV)
Balmer23, 4, 5, ...Visible (only visible series!)
Paschen34, 5, 6, ...Infrared (IR)
Brackett45, 6, 7, ...Infrared (IR)
Pfund56, 7, 8, ...Infrared (IR)
Rydberg Formula:   ν̄ = 1/λ = 109,677 × (1/n12 − 1/n22) cm-1  (or 1.09677 × 107 m-1)
Bohr Quantization:   Angular Momentum mevr = nh / (2π)  (n = 1, 2, 3, ...)
Orbit Radii (H & 1-e- ions He+, Li2+):   rn = 52.9 n2 / Z pm  (Bohr radius a0 = 52.9 pm)
Orbit Energy:   En = −2.18 × 10-18 (Z2 / n2) J   |   ΔE = 2.18 × 10-18 (1/ni2 − 1/nf2) J

2. Why is En Negative & Bohr's Limitations

A free electron infinitely far away (n = ∞) is assigned zero energy (E∞ = 0). As the electron is attracted closer to the nucleus, energy is released and drops below zero—making n = 1 the most negative (most stable) ground state.
Limitations of Bohr's Model: Fails for multi-electron atoms (even He), fails to explain closely spaced fine-structure doublets, magnetic splitting (Zeeman effect), electric splitting (Stark effect), and chemical bond formation.

NCERT Problems 2.10 & 2.11:
• Problem 2.10 (n = 5 → n = 2 in H): ΔE = 2.18×10-18(1/25 − 1/4) = −4.58 × 10-19 J (emitted) → ν = 4.58×10-19 / 6.626×10-34 = 6.91 × 1014 Hz → λ = c/ν = 434 nm (Balmer violet-blue).
• Problem 2.11 (He+ first orbit, Z=2, n=1): E1 = −2.18×10-18 × (22/12) = −8.72 × 10-18 J; r1 = 0.0529 × (12/2) = 0.02645 nm (26.45 pm).
Bohr's equations work ONLY for one-electron species (H, He+, Li2+, Be3+)—NEVER apply them to neutral He or Li! Also remember: Balmer is the ONLY series in the Visible region (Lyman is UV; Paschen/Brackett/Pfund are IR). Maximum emission lines when dropping from level n to ground state = n(n − 1) / 2 (e.g., n = 6 gives 15 lines).

Topic 2.5: Dual Behaviour of Matter (de Broglie) & Heisenberg's Uncertainty Principle

Picture trying to measure the speed of a tiny invisible mosquito in pitch darkness by bouncing heavy basketballs off it! The moment a basketball hits the mosquito and tells you where it was, it violently blasts the mosquito off at an unknown speed! Werner Heisenberg proved electrons face that exact cosmic trade-off: pinpointing an electron's position automatically blurs its velocity!

1. de Broglie's Matter Waves (1924)

Louis de Broglie proposed that just as light has dual wave-particle nature, moving matter also possesses a wavelength inversely proportional to momentum (confirmed experimentally via electron diffraction, used in electron microscopes magnifying ~15 million times).

de Broglie Relation:   λ = h / (mv) = h / p
Heisenberg Uncertainty Principle (1927):   Δx · Δpx ≥ h / (4π)  ⇒  Δx · Δvx ≥ h / (4πm)

2. Why Macroscopic Objects Don't Show Wave or Uncertainty Effects

  • Because Planck's constant h = 6.626 × 10-34 J s is unimaginably tiny, ordinary objects (large mass m) have wavelengths ~10-34 m and uncertainties ~10-33 m—completely undetectable!
  • For a microscopic electron (m = 9.11 × 10-31 kg), both λ and Δx·Δv are huge on the atomic scale—making Bohr's sharp circular paths/trajectories physically impossible!
NCERT Problems 2.13, 2.15 & 2.16:
• Problem 2.13 (de Broglie): Electron (m = 9.1 × 10-31 kg) with KE = 3.0 × 10-25 J → v = √(2·KE/m) = 812 m s-1 → λ = h/(mv) = 896.7 nm.
• Problem 2.15 (Heisenberg, Electron): Locate electron within Δx = 0.1 Å = 10-11 m → Δv = h / (4πmΔx) = 5.79 × 106 m s-1 (huge!).
• Problem 2.16 (Heisenberg, Golf ball): 40 g (0.040 kg) at 45 m/s (±2% → Δv = 0.9 m/s) → Δx = 1.46 × 10-33 m (negligible!).
Watch TWO classic exam errors: (1) Heisenberg's denominator is 4π (NOT 2π: Δx · Δp ≥ h / 4π)! (2) In both de Broglie and Heisenberg formulas, ALWAYS convert mass into kilograms (kg)—never leave grams in the denominator, because 1 J = 1 kg m2 s-2!

Topic 2.6 – 2.6.1: Quantum Mechanical Model, Schrödinger Equation & The 4 Quantum Numbers

Sending a letter to an electron doesn't work with circular race-track coordinates—you need a 4-line Quantum Postal Address! Think of it just like mailing a parcel in Kerala: State (n) tells how big the shell is, District (l) tells the orbital's 3D shape, Street (ml) picks its spatial direction, and Resident (ms) tells which way the electron spins!

1. Schrödinger Equation & Meaning of ψ vs |ψ|2

In 1926, Erwin Schrödinger wrote Ĥψ = Eψ incorporating wave-particle duality. The wave function ψ itself has no physical meaning, but |ψ|2 (Max Born) equals the probability density of finding the electron at a point. Regions enclosing ~90% probability are called Atomic Orbitals.

Quantum NumberSymbolAllowed ValuesPhysical Meaning & Key Formula
1. Principaln1, 2, 3, 4 ... (K, L, M, N)Shell size & main energy; n subshells, n2 orbitals, max 2n2 e-
2. Azimuthall0 to (n − 1)Subshell & 3D shape: l = 0(s), 1(p), 2(d), 3(f), 4(g)
3. Magneticml−l, ..., 0, ..., +lSpatial orientation; (2l + 1) orbitals per subshell
4. Spinms+½ (↑) or −½ (↓)Uhlenbeck & Goudsmit (1925): intrinsic electron spin orientation
Orbit vs Orbital: An Orbit (Bohr) is a flat 2D circular track holding up to 2n2 e- (violates Heisenberg). An Orbital (Quantum) is a 3D probability region around the nucleus holding at most 2 electrons with opposite spins.
NCERT Problems 2.17 & 2.18:
• n = 3 Shell: l = 0 (one 3s), l = 1 (three 3p), l = 2 (five 3d) → Total orbitals = 1 + 3 + 5 = 9 = n2 (max 18 e-).
• Notations: (n=2, l=1) → 2p; (n=4, l=0) → 4s; (n=5, l=3) → 5f; (n=3, l=2) → 3d.
Highest-frequency exam question: Identify the impossible orbital or quantum set! Remember TWO strict bounds: (1) l must be strictly LESS than n (so 1p, 2d, 3f do NOT exist; lowest shell for g-orbital with l=4 is n = 5!). (2) |ml| can never exceed l (for l=1, ml = −2 is impossible). Also note: ms did NOT come from Schrödinger's equation!

Topic 2.6.2 – 2.6.3: Orbital Shapes (s, p, d), Radial & Angular Nodes, and (n + l) Energies

Why do textbooks draw an orbital at 90% probability instead of 100%? Because an electron's probability mist never hits zero at any finite distance—to catch 100%, you'd have to draw a sphere as big as the entire universe! Even inside that misty cloud, silent zero-probability sheets called nodes carve spheres into onions (2s), dumbbells (p), and four-leaf clovers (d)!

1. Shapes of Atomic Orbitals

  • s-orbitals (l = 0): Spherically symmetric around the nucleus. Size grows as 4s > 3s > 2s > 1s. While 1s peaks at the nucleus and falls smoothly, 2s drops to zero at a spherical radial node before rising to a second smaller peak!
  • p-orbitals (l = 1): Dumbbell-shaped with two opposite lobes separated by a nodal plane through the nucleus. Three mutually perpendicular orbitals: px, py, pz (degenerate in energy).
  • d-orbitals (l = 2, min n = 3): Five equivalent-energy orbitals: four cloverleaf shapes (dxy, dyz, dxz, dx²-y²) plus the unique dz² (z-axis dumbbell surrounded by an equatorial doughnut/torus ring!).
Node Formulas (Where |ψ|2 = 0):
• Radial (Spherical) Nodes = n − l − 1   |   Angular Nodes (Nodal Planes) = l   |   Total Nodes = n − 1
Examples: 1s → 0 nodes; 2s → 1 radial; 2p → 1 angular; 3p → 1 radial + 1 angular = 2 total; 3d → 0 radial + 2 angular = 2 total.

2. Energies of Orbitals: Hydrogen vs Multi-Electron Atoms

  • Hydrogen (1-electron): Energy depends ONLY on n: 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f (orbitals of same n are degenerate).
  • Multi-Electron Atoms (Shielding & Zeff): Inner electrons screen nuclear charge, and penetration follows s > p > d > f—so subshells split (s < p < d < f) and follow the (n + l) Rule:
    1. Lower (n + l) value → Lower energy (e.g., 4s has 4+0=4 < 3d which has 3+2=5, so 4s fills before 3d!).
    2. Equal (n + l) → Lower n has lower energy (e.g., 3d vs 4p: both n+l=5, so 3d < 4p).
Don't mix up Radial Nodes (n − l − 1) with Total Nodes (n − 1)! For a 3p orbital, radial nodes = 3 − 1 − 1 = 1, angular nodes = 1, total = 2. Also, 3s, 3p, 3d have the EXACT SAME energy in Hydrogen, whereas in multi-electron atoms 3s < 3p < 3d!

Topic 2.6.4 – 2.6.6: Electronic Configuration: Aufbau, Pauli, Hund & Cr / Cu Stability

Boarding electrons into an atom works just like passengers taking seats on an intercity express bus! First, they grab the lowest front seats (Aufbau). Every double-bench fits only two people facing opposite directions (Pauli). And when three empty benches sit side-by-side (px, py, pz), three strangers ALWAYS sit one per bench before anyone shares a seat (Hund)! Yet Chromium (24) and Copper (29) make a clever last-second swap for extra peace!

1. Three Golden Rules of Electronic Configuration

  1. Aufbau Principle (German: building up): Orbitals fill in increasing energy order: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p...
  2. Pauli Exclusion Principle (1926): No two electrons in an atom can have the exact same set of all four quantum numbers → an orbital holds max 2 electrons with opposite spins (↑↓).
  3. Hund's Rule of Maximum Multiplicity: Pairing in degenerate orbitals (p, d, f) does NOT occur until each orbital is singly occupied with parallel spins (pairing begins at 4th p, 6th d, and 8th f electron; e.g., N: 1s2 2s2 2px1 2py1 2pz1).

2. Exceptional Configurations: Chromium (Z = 24) & Copper (Z = 29)

Chromium (Z = 24):  Expected [Ar] 3d4 4s2  →  Actual: [Ar] 3d5 4s1 (Half-filled d5, 6 unpaired e-!)
Copper (Z = 29):     Expected [Ar] 3d9 4s2  →  Actual: [Ar] 3d10 4s1 (Completely filled d10, 1 unpaired e-!)

Why are half-filled (p3, d5, f7) & fully filled (p6, d10, f14) subshells extra stable?

  • (i) Symmetrical Distribution: Symmetry leads to stability; electrons in the same subshell shield one another weakly and feel stronger nuclear attraction.
  • (ii) Large Exchange Energy: Parallel-spin electrons in degenerate orbitals swap positions releasing exchange energy. Shifting one 4s electron to 3d jumps Cr from 6 exchanges (in d4: 3+2+1) to 10 exchanges (in d5: 4+3+2+1)!
Two super-common traps: (1) Chromium has 6 unpaired electrons (five in 3d5 + one in 4s1), NOT 4! (2) When forming transition-metal cations like Fe2+ or Cu+, electrons are ALWAYS removed from the outer 4s orbital FIRST before 3d (Fe2+ = [Ar] 3d6, NOT [Ar] 3d4 4s2)!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated Favorite) • 2 Marks): State Heisenberg's Uncertainty Principle and give its mathematical expression. Why is it insignificant for macroscopic bodies?

Statement: It is impossible to determine simultaneously both the exact position and the exact momentum (or velocity) of a microscopic moving particle such as an electron.

Mathematical Form: Δx · Δpx ≥ h / (4π), or Δx · m Δvx ≥ h / (4π).

Why negligible for macroscopic bodies: Because Planck's constant h (6.626 × 10-34 J s) is extremely small and macroscopic mass m is very large, the product Δx · Δv ≈ 10-33 is too tiny to be observable.

Q2 (Kerala HSE SAY / Model (Most Asked) • 2 Marks): Write the ground-state electronic configurations of Chromium (Z = 24) and Copper (Z = 29). Give two reasons for their exceptional stability.

Cr (Z = 24): 1s2 2s2 2p6 3s2 3p6 3d5 4s1 (or [Ar] 3d5 4s1)
Cu (Z = 29): 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (or [Ar] 3d10 4s1)

Reasons for extra stability of half-filled (d5) & fully filled (d10) subshells: (1) Symmetrical distribution of electrons resulting in lower shielding and stronger nuclear pull, and (2) Maximum exchange energy released by parallel-spin electrons exchanging positions across degenerate 3d orbitals.

Q3 (Kerala HSE March • 3 Marks): Explain the significance of the four quantum numbers (n, l, ml, ms). Which of them does NOT arise from the Schrödinger wave equation?

1. Principal (n): Shell, size and main energy of the orbital.
2. Azimuthal (l): Subshell and 3D shape of the orbital (0=s, 1=p, 2=d, 3=f).
3. Magnetic (ml): Spatial orientation of the orbital (2l + 1 values from −l to +l).
4. Spin (ms): Orientation of intrinsic electron spin (+½ or −½).

The spin quantum number (ms) does NOT come from Schrödinger's equation (proposed by Uhlenbeck & Goudsmit, 1925).

Q4 (Kerala HSE Improvement / NCERT Ex 2.30 • 2 Marks): Identify which of the following orbitals are NOT possible and justify: 1p, 2s, 2d, 3f.

For a given principal quantum number n, allowed values of l are 0 to (n − 1).

  • 1p (n=1, l=1): Not possible (since l cannot equal n).
  • 2s (n=2, l=0): Possible.
  • 2d (n=2, l=2): Not possible (max l for n=2 is 1).
  • 3f (n=3, l=3): Not possible (max l for n=3 is 2).

Q5 (Kerala HSE March (Standard Numerical) • 3 Marks): The threshold frequency ν0 for a metal is 7.0 × 1014 s-1. Calculate the kinetic energy of an electron emitted when radiation of frequency ν = 1.0 × 1015 s-1 hits the metal.

By Einstein's photoelectric equation: KE = h(ν − ν0)

KE = (6.626 × 10-34 J s) × (1.0 × 1015 − 7.0 × 1014) s-1
= (6.626 × 10-34) × (3.0 × 1014) = 1.988 × 10-19 J.

Q6 (Kerala HSE Model • 2 Marks): Find the number of radial nodes, angular nodes, and total nodes in (i) 2s and (ii) 4p orbitals.

Formulas: Radial nodes = n − l − 1; Angular nodes = l; Total nodes = n − 1.

(i) 2s (n=2, l=0): Radial = 2 − 0 − 1 = 1; Angular = 0; Total = 1.
(ii) 4p (n=4, l=1): Radial = 4 − 1 − 1 = 2; Angular = 1; Total = 3.

Q7 (Kerala HSE SAY • 3 Marks): What are the main limitations of Bohr's model of the atom? Calculate the radius and energy of the first orbit of He+ ion.

Limitations: (1) Fails to explain spectra of multi-electron atoms (like He), (2) Fails to explain fine-structure doublets, (3) Cannot explain Zeeman effect (magnetic splitting) or Stark effect (electric splitting), (4) Contradicts Heisenberg's uncertainty principle & de Broglie duality, and ignores chemical bonding.

For He+ (Z = 2, n = 1):
r1 = 52.9 × (12 / 2) = 26.45 pm (0.02645 nm).
E1 = −2.18 × 10-18 × (22 / 12) = −8.72 × 10-18 J.

Q8 (Kerala HSE March • 2 Marks): State Pauli's Exclusion Principle and Hund's Rule of Maximum Multiplicity with suitable examples.

Pauli Exclusion Principle: No two electrons in an atom can have the same set of all four quantum numbers (an orbital holds max 2 electrons with opposite spins: ↑↓). E.g., He (1s2) has n=1, l=0, ml=0, ms = +½ and −½.

Hund's Rule: Pairing of electrons in degenerate orbitals of the same subshell does not take place until each orbital is singly occupied with parallel spins. E.g., Nitrogen (Z=7): 1s2 2s2 2px1 2py1 2pz1.

Q9 (Kerala HSE SAY • 1 Marks): What is the lowest value of principal quantum number 'n' that allows g-orbitals to exist?

For a g-orbital, l = 4. Since the minimum n satisfies l ≤ n − 1, the lowest value is n = 5.

Q10 (Kerala HSE March • 4 Marks): Name the five series of lines in the Hydrogen atomic spectrum with their spectral regions. Calculate the wavelength of the photon emitted during transition from n = 5 to n = 2 in a Hydrogen atom.

Series & Regions: Lyman (n1=1, UV), Balmer (n1=2, Visible), Paschen (n1=3, IR), Brackett (n1=4, IR), Pfund (n1=5, IR).

Transition ni = 5 → nf = 2:
ΔE = 2.18 × 10-18 (1/52 − 1/22) = −4.58 × 10-19 J (emitted).
ν = 4.58 × 10-19 / 6.626 × 10-34 = 6.91 × 1014 Hz.
λ = c / ν = (3.0 × 108) / (6.91 × 1014) = 4.34 × 10-7 m = 434 nm (Visible Balmer line).

Q11 (Kerala HSE / NCERT Exemplar • 2 Marks): State de Broglie's dual-behaviour equation. Calculate the de Broglie wavelength of a 0.1 kg ball moving at 10 m s-1.

Equation: λ = h / (mv), where h = 6.626 × 10-34 J s.
λ = (6.626 × 10-34) / (0.1 × 10) = 6.626 × 10-34 m (far too small to observe wave behaviour).

Q12 (Kerala HSE Improvement • 1 Marks): Identify the orbitals represented by: (a) n = 4, l = 2 and (b) n = 3, l = 0. How many electrons in n = 4 can have ms = −½?

(a) n = 4, l = 2 → 4d; (b) n = 3, l = 0 → 3s.
Shell n = 4 has 42 = 16 orbitals (32 electrons total); exactly half, i.e., 16 electrons, can have ms = −½.