NCERT Plus Two (Class 12) Physics • Chapter 12

Atoms

Q: Define distance of closest approach in Rutherford's alpha scattering experiment and write its formula. What estimate of nuclear size does it provide?

Definition: The minimum distance from the nucleus centre at which an alpha particle fired head-on momentarily comes to rest before rebounding ( θ = 180° ). At turnaround, K = 1 4πε₀ 2Ze² r₀ ⇒ r₀ = 2Ze² 4πε₀K . Since the alpha turns around outside the nucleus, nuclear radius R ≤ r₀ (~10⁻¹⁴ m) .

Q: State the two major drawbacks of Rutherford's nuclear model of the atom.

1. Atomic Instability: By classical electrodynamics, an orbiting (accelerating) electron continuously radiates EM waves, loses energy, and should spiral into the nucleus in ~10⁻¹¹ s . 2. Continuous Spectrum Prediction: Shrinking orbit frequency would produce a continuous spectrum instead of the observed sharp discrete line spectra.

Q: State the three postulates of Bohr's model of the Hydrogen atom.

1. Stationary Orbits: Electrons revolve in specific stable circular orbits without radiating electromagnetic energy; Coulomb force provides centripetal force. 2. Angular Momentum Quantisation: Allowed orbits satisfy L = mvr = nh 2π ( n = 1, 2, 3... ). 3. Frequency Condition: An electron emits or absorbs a photon only when transitioning between stationary states: hν = Eᵢ − E_f .

Q: Using Bohr's postulates, derive expressions for (i) radius of the n-th orbit and (ii) total energy of the electron in a Hydrogen atom.

(i) Radius: Force balance mv²/r = e²/(4πε₀r²) gives mv²r = e²/(4πε₀) . From 2nd postulate, v = nh/(2πmr) . Substituting v : m (n²h² / 4π²m²r²) r = e²/(4πε₀) ⇒ rₙ = n²h²ε₀ πme² = 0.529 n² Å . (ii) Energy: K = e²/(8πε₀rₙ) and U = −e²/(4πε₀rₙ) , so Eₙ = −e²/(8πε₀rₙ) . Substituting rₙ : Eₙ = − me⁴ 8ε₀²h²n² = − 13.6 n² eV .

Q: The total energy of an electron in the first excited state of Hydrogen is −3.4 eV. Calculate its kinetic energy and potential energy in this state.

In a Bohr orbit, K = −E and U = 2E . • Kinetic Energy: K = −(−3.4) = +3.4 eV • Potential Energy: U = 2 × (−3.4) = −6.8 eV

Q: Explain how de Broglie's hypothesis of matter waves justifies Bohr's quantisation condition for angular momentum.

An electron of mass m and speed v behaves as a matter wave of wavelength λ = h/(mv) . For a standing wave on a circular orbit of radius r to avoid destructive cancellation, the circumference must hold an exact integer number of wavelengths: 2πr = nλ = n h mv . Rearranging directly yields Bohr's 2nd postulate: mvr = nh 2π .

Q: Identify the spectral series of Hydrogen atom lying in (a) ultraviolet region and (b) visible region. Write the Rydberg formula for Balmer's Hα line.

(a) Ultraviolet region: Lyman series ( n₁ = 1 ). (b) Visible region: Balmer series ( n₁ = 2 ). For Hα line ( 3 → 2 ): 1 λ = R ( 1 2² − 1 3² ) = 5R 36 ⇒ λ = 656.3 nm .

Q: Calculate the shortest wavelength present in the Paschen series of Hydrogen spectral lines (Given R = 1.097 × 10⁷ m⁻¹).

Paschen series corresponds to n₁ = 3 ; shortest wavelength occurs at series limit n₂ = ∞ : 1 λ min = R ( 1 3² − 0) = R 9 λ min = 9 1.097 × 10⁷ ≈ 8.204 × 10⁻⁷ m = 820.4 nm (Infrared).

Rutherford alpha scattering & closest approach, nuclear atom puzzle, atomic line spectra, Bohr model radii & energies, hydrogen spectral series & de Broglie standing orbit.

📄 Official NCERT Textbook PDF

Topic 12.2.1: Geiger-Marsden Alpha Scattering & Distance of Closest Approach

Imagine kicking a football toward a massive Palakkad coconut farm that looks empty from afar — except one tiny, super-heavy iron pillar hides dead centre! Almost every ball zooms straight through without touching anything, a few graze nearby and swerve sideways, and roughly one in eight thousand slams head-on and bounces straight back at your shoes. That shock in 1911 forced Rutherford to throw out the soft plum-pudding atom and discover the dense atomic nucleus!

1911 GOLD FOIL DATA99.86% pass straight; 0.14% deflect > 1°; ~1 in 8000 rebound > 90°!

IMPACT PARAMETER (b)b = Ze² cot(θ/2)4πε₀ K • Small b → huge θ; b = 0 → θ = 180°

CLOSEST APPROACH (r₀)At head-on stop: r₀ = 14πε₀ 2Ze²K (~30 fm for 7.7 MeV α on Au)

🧠 Memory Trick: Small b = Big Bounce! Direct bullseye (b = 0) flips the alpha 180° backward, trading all kinetic energy K into Coulomb potential energy!

💡 Simple Kid-Check: If an atom were really a soft mushy watermelon (Thomson's plum-pudding model), a fast 5.5 MeV alpha bullet would punch straight through like a cannonball through tissue paper! Bouncing straight back proved almost all positive charge and mass sit packed inside one tiny hard core.

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Experimental Setup: Geiger and Marsden beamed 5.5 MeV alpha particles (He²⁺, charge +2e) from ²¹⁴Bi onto a ultra-thin gold foil (2.1 × 10⁻⁷ m, Z = 79) and counted flashes on a movable ZnS screen.
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Rutherford Scattering Law: Scattered count drops steeply with angle as N(θ) ∝ 1sin⁴(θ/2). Small impact parameter b (perpendicular offset from nucleus centre) gives large deflection angle θ.
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Distance of Closest Approach (r₀): For a head-on alpha (b = 0), speed drops to zero at distance r₀ where K = U = 14πε₀ (2e)(Ze)r₀. Since r₀ ≈ 10⁻¹⁴ m and atom radius is ~10⁻¹⁰ m, the nucleus is roughly 10,000 times smaller than the atom!

🧒 Super-Easy Example: Find the distance of closest approach for a 7.7 MeV alpha particle fired head-on at a Gold nucleus (Z = 79):
K = 7.7 × 10⁶ × 1.6 × 10⁻¹⁹ = 1.232 × 10⁻¹² J.
r₀ = 9 × 10⁹ × 2 × 79 × (1.6 × 10⁻¹⁹)²1.232 × 10⁻¹² ≈ 3.0 × 10⁻¹⁴ m (30 fm).

Watch two classic Kerala DHSE traps: (1) Remember an alpha particle carries charge +2e (NOT +e!) — forgetting the factor of 2 in 2Ze² cuts your answer in half! (2) Closest approach r₀ is only an upper limit on nuclear radius (R ≤ r₀) because Coulomb repulsion turns the alpha back before it touches the nuclear surface!

Topic 12.2.2: Rutherford Nuclear Atom & Classical Stability Puzzle

Imagine a runner circling a Kochi stadium track while tossing away heavy coins every second — soon she slows down, drifts inward, and collapses right onto the grass centre! According to Maxwell's laws of electricity and magnetism, an accelerating electron orbiting a nucleus MUST continuously shake out light energy and spiral straight into the nucleus in one hundred-billionth of a second — yet atoms around us never collapse!

ELECTRON ORBIT ENERGIESK = +e²8πε₀r • U = −e²4πε₀r • E = −e²8πε₀r

GOLDEN RATIO K : U : EK = −E and U = 2E → always 1 : −2 : −1!

2 FATAL FLAWS(1) Radiating electron spirals into nucleus in ~10⁻¹¹ s. (2) Predicts continuous spectrum, NOT sharp lines!

🧠 Memory Trick: For ANY bound electron orbit: Kinetic energy K = +|E|, Total energy E = −|E|, Potential energy U = −2|E|!

💡 Simple Kid-Check: Circling means changing direction every instant, which is centripetal acceleration! Classical physics says every accelerating charge MUST broadcast electromagnetic waves like a tiny radio antenna — losing energy until the orbit crashes!

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Force Balance in H-Atom: Electrostatic attraction provides centripetal force: mv²r = e²4πε₀r². Multiplying both sides by r/2 gives Kinetic Energy K = 12mv² = +e²8πε₀r.
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Why Total Energy Is Negative: Potential energy is U = −e²4πε₀r, so Total Energy E = K + U = −e²8πε₀r. Negative E simply means the electron is bound to the nucleus!
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Why Rutherford Needed Rescue: Classical theory predicts continuous spiral collapse (unstable matter!) AND continuous rainbow frequencies as r shrinks — flatly contradicted by rock-solid atoms and crisp spectral lines.

🧒 Super-Easy Example: The total energy of an electron in hydrogen's ground state is E = −13.6 eV. Instantly write down its Kinetic and Potential energies without any messy constants:
• Kinetic Energy K = −E = +13.6 eV
• Potential Energy U = 2E = −27.2 eV (check: 13.6 − 27.2 = −13.6 eV!)

Classic 1-mark sign trap: In a bound atom, K is ALWAYS positive and E is ALWAYS negative with equal magnitude (K = −E)! Never give a negative value for Kinetic Energy, and remember Rutherford's model failed TWO things: atomic stability and discrete line spectra!

Topic 12.3: Atomic Emission & Absorption Line Spectra Fingerprint

Walk up to a supermarket billing counter in Ernakulam — one beep on the black-and-white barcode tells the computer the exact biscuit brand! Atoms carry their own light barcodes too: heat low-pressure hydrogen gas with high voltage and it glows only a few razor-sharp coloured stripes; shine white light through cool hydrogen gas and those exact same stripes vanish as dark shadow lines!

EMISSION SPECTRUMHot/excited gas atoms jump down (E₂ → E₁) → bright sharp coloured lines on black background.

ABSORPTION SPECTRUMWhite light passes through cool gas → dark shadow lines at the exact same wavelengths!

ELEMENT FINGERPRINTEach element has unique λ values because each atom has unique energy step gaps ΔE = hc/λ.

🧠 Memory Trick: Emission = Bright Stripes on Dark (atom drops & glows); Absorption = Dark Stripes on Rainbow (atom eats matching photons to climb)!

💡 Simple Kid-Check: A glowing tungsten bulb (hot solid) gives a smooth full rainbow because billions of packed atoms bump into each other. An isolated low-pressure gas atom acts like a solo piano key — it can ONLY play its exact allowed notes!

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Why Discrete Lines Appear: An atom cannot hold random energies; it emits a photon ONLY when jumping between fixed levels: hν = hcλ = E_high − E_low.
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Mirror Match Rule: The wavelengths absorbed by cool gas atoms match a subset of the bright lines emitted when that same gas is heated (famous dark Fraunhofer lines in sunlight revealed Helium & Sodium in the Sun's outer atmosphere!).
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Fast Photon Shortcut: Convert wavelength in nm directly into photon energy in eV using E(eV) ≈ 1240λ(nm).

🧒 Super-Easy Example: Hydrogen's bright red emission line sits at λ = 656.3 nm. What energy jump inside the atom created this photon?
ΔE ≈ 1240656.3 ≈ 1.89 eV (exact match for E₃ − E₂ = −1.51 − (−3.40) = 1.89 eV!).

Key difference between emission and absorption in cool gas: At room temperature, almost all hydrogen atoms sit in the ground state (n = 1), so cool hydrogen gas ONLY absorbs photons starting from n = 1 (Lyman UV lines) — you won't see dark Balmer lines unless hydrogen is already hot enough to populate n = 2!

Topic 12.4: Bohr's Three Postulates, Orbit Radii & Energy Levels

Think of climbing the steep steps inside a Guruvayur temple tank — your feet can stand solidly on Step 1, Step 2, or Step 3, never hovering mid-air at Step 1.7! In 1913, Niels Bohr boldly declared that electrons inside an atom live on strict quantum steps too: while walking smoothly along any allowed step, they radiate ZERO energy, and light flashes ONLY when hopping between steps!

BOHR'S 3 POSTULATES(1) Stable non-radiating orbits. (2) L = mvr = nh2π. (3) Photon on jump: hν = Eᵢ − E_f.

RADII & SPEEDS (H, Z=1)rₙ = 0.529 n² Å (Bohr radius a₀ = 0.529 Å) • vₙ = c137 n

ENERGY LEVELSEₙ = −13.6n² eV → −13.6, −3.4, −1.51, −0.85, −0.54, 0 eV

🧠 Memory Trick: Remember Hydrogen's first 4 rungs by heart: −13.6 → −3.4 → −1.51 → −0.85 eV! And scaling powers: r ∝ n²/Z, v ∝ Z/n, E ∝ −Z²/n²!

💡 Simple Kid-Check: Bohr kept Rutherford's nuclear orbit force law, then added ONE bold lock: angular momentum can only come in whole-number packets of h / (2π)! Instantly, allowed radii jump as 1, 4, 9, 16... times a₀, and energies lock to −13.6 / n² eV!

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Orbit Radius (rₙ): Combine mv²r = Ze²4πε₀r² with mvr = nh2π → rₙ = n² h² ε₀π m Z e² = 0.529 n²Z Å. For Hydrogen (n=1, Z=1), Bohr radius a₀ = 0.529 Å.
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Electron Speed (vₙ): Outer electrons slow down! vₙ = Ze²2ε₀ n h ≈ 2.18 × 10⁶ Zn m/s (for n=1 in H, v₁ ≈ c/137).
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Excitation vs Ionisation Energy: Ground state of H is −13.6 eV. Lifting from n=1 to n=2 needs −3.4 − (−13.6) = 10.2 eV (1st Excitation Energy / Potential 10.2 V). Knocking the electron completely free (n → ∞, E = 0) needs +13.6 eV (Ionisation Energy / Potential 13.6 V).

🧒 Super-Easy Example: For the n = 3 orbit in Hydrogen (Z = 1):
• Radius: r₃ = 0.529 × 3² = 0.529 × 9 = 4.76 Å
• Total Energy: E₃ = −13.6 / 3² = −13.6 / 9 ≈ −1.51 eV
• Angular Momentum: L₃ = 3 × h2π.

Watch three classic DHSE traps: (1) As n increases, Eₙ = −13.6/n² becomes LESS negative — meaning energy INCREASES upward while spacing shrinks! (2) First excitation energy of Hydrogen is 10.2 eV (n=1 → 2), NOT 13.6 eV (ionisation)! (3) Bohr's model fails for multi-electron atoms (even neutral Helium!) and cannot explain fine splitting or relative line intensities.

Topic 12.5: Hydrogen Spectral Series: Lyman, Balmer, Paschen, Brackett & Pfund

Picture five rainwater harvesting tanks stacked at Floors 1, 2, 3, 4, and 5 of a Kozhikode apartment! Water tumbling all the way down to Floor 1 makes a thunderous ultraviolet splash (Lyman); drops landing on Floor 2 create bright visible splashes our eyes can actually watch (Balmer); and gentler drops stopping at Floors 3, 4, or 5 send out warm infrared ripples!

RYDBERG FORMULA1λ = R (1n₁² − 1n₂²) where R = 1.097 × 10⁷ m⁻¹ (1/R ≈ 91.1 nm)

5 SERIES LANDINGSLyman (n₁=1, UV) • Balmer (n₁=2, Visible) • Paschen (3), Brackett (4), Pfund (5) → IR

MAX EMISSION LINESFrom excited state n down to 1: N = n(n − 1)2 possible lines!

🧠 Memory Trick: "Little Boys Play Basketball Properly" → Lyman (1, UV), Balmer (2, Visible), Paschen (3, IR), Brackett (4, IR), Pfund (5, Far-IR)!

💡 Simple Kid-Check: Remember ΔE = hc/λ! A tiny jump (to the very next step, n₂ = n₁ + 1) gives the LONGEST wavelength (λ_max), while a drop from the roof (n₂ = ∞, series limit) gives the SHORTEST wavelength (λ_min)!

Series	Lower `n₁`	Upper `n₂`	EM Region	Longest `λ_max` (`n₁+1 → n₁`)	Limit `λ_min` (`∞ → n₁`)
Lyman	`1`	`2, 3, 4...`	Ultraviolet (UV)	`121.6 nm` (`2→1`)	`91.1 nm` (`∞→1`)
Balmer	`2`	`3, 4, 5...`	Visible (Hα–Hδ)	`656.3 nm` (Hα, `3→2`)	`364.6 nm` (near-UV)
Paschen	`3`	`4, 5, 6...`	Near Infrared	`1875 nm` (`4→3`)	`820.4 nm`
Brackett	`4`	`5, 6, 7...`	Infrared	`4051 nm` (`5→4`)	`1458 nm`
Pfund	`5`	`6, 7, 8...`	Far Infrared	`7458 nm` (`6→5`)	`2279 nm`

🧒 Super-Easy Example: Find the shortest and longest wavelengths of the Balmer series (n₁ = 2):
• Shortest (Series Limit, n₂ = ∞): 1/λ_min = R(1/4 − 0) ⇒ λ_min = 4/R = 4 × 91.15 ≈ 364.6 nm
• Longest (Hα line, n₂ = 3): 1/λ_max = R(1/4 − 1/9) = 5R/36 ⇒ λ_max = 36/(5R) ≈ 656.3 nm (Red!).

Two classic Kerala HSE traps: (1) Students constantly flip longest vs shortest wavelength — smallest jump (n₁+1 → n₁) gives longest λ, whereas ∞ → n₁ gives shortest λ! (2) If an electron drops from n = 4 to ground state in a gas sample, total distinct spectral lines = 4(4−1)/2 = 6 lines (3 Lyman + 2 Balmer + 1 Paschen), NOT 3!

Topic 12.6: de Broglie Standing-Wave Proof of Bohr Quantisation

Tie a long rubber skipping rope into a closed ring in your schoolyard and shake it steadily! When an exact whole number of wave crests (1, 2, 3...) fits neatly around the loop, crests reinforce crests and the ring sings cleanly forever. Try stuffing 2.5 waves around that circle and the crest crashes right into a trough — cancelling itself dead! That exact wave-fitting rule is WHY Bohr's orbits are quantised!

STANDING WAVE CONDITIONCircumference fits integer waves: 2πrₙ = n λ (n = 1, 2, 3...)

DE BROGLIE DERIVATIONPlug λ = h/(mv) into 2πr = nλ ⇒ mvr = nh2π!

WAVELENGTH SCALINGλₙ = 2πrₙn = n (2πa₀) = 3.32 n Å (grows linearly with n!)

🧠 Memory Trick: 2 lines win 2 marks! Write 2πr = nλ, substitute λ = h/mv, rearrange to mvr = nh/2π — Bohr's mystery postulate proved!

💡 Simple Kid-Check: Bohr originally guessed mvr = nh/(2π) without knowing why Nature forbade 1.5 or 2.3! Eleven years later, Louis de Broglie treated the electron as a matter wave — only whole-number loops survive without interfering destructively with themselves!

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Closed Circular Loop: For a stationary electron wave on radius rₙ not to cancel itself after each lap, total circumference must equal an integer multiple of wavelength: 2πrₙ = n λ.
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Instant Proof of Bohr's 2nd Postulate: Using de Broglie's wavelength λ = hp = hmvₙ:
2πrₙ = n hmvₙ ⇒ L = m vₙ rₙ = nh2π!
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Why λ Grows in Higher Orbits: Because electron speed drops as 1/n, its de Broglie wavelength λ = h/(mv) grows proportionally to n: the 3rd orbit fits 3 longer waves, not 3 identical first-orbit waves!

🧒 Super-Easy Example: How many de Broglie wavelengths fit around the n = 4 Bohr orbit, and what is the wavelength if r₁ = 0.529 Å?
• Number of full waves = n = 4 wavelengths.
• Since r₄ = 16 × 0.529 = 8.464 Å, λ₄ = 2πr₄ / 4 = 4 × (2π × 0.529) ≈ 13.3 Å!

Watch this subtle MCQ trap: Does the de Broglie wavelength of an electron stay constant across all Bohr orbits? NO! Since 2πrₙ = nλₙ and rₙ = n²a₀, dividing by n gives λₙ = 2πa₀ · n — the wavelength doubles in n=2 and triples in n=3 because the electron moves slower!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): Define distance of closest approach in Rutherford's alpha scattering experiment and write its formula. What estimate of nuclear size does it provide?

Definition: The minimum distance from the nucleus centre at which an alpha particle fired head-on momentarily comes to rest before rebounding (θ = 180°).
At turnaround, K = 14πε₀ 2Ze²r₀ ⇒ r₀ = 2Ze²4πε₀K.
Since the alpha turns around outside the nucleus, nuclear radius R ≤ r₀ (~10⁻¹⁴ m).

Q2 (Kerala HSE Say • 2 Marks): State the two major drawbacks of Rutherford's nuclear model of the atom.

1. Atomic Instability: By classical electrodynamics, an orbiting (accelerating) electron continuously radiates EM waves, loses energy, and should spiral into the nucleus in ~10⁻¹¹ s.
2. Continuous Spectrum Prediction: Shrinking orbit frequency would produce a continuous spectrum instead of the observed sharp discrete line spectra.

Q3 (Kerala HSE March • 3 Marks): State the three postulates of Bohr's model of the Hydrogen atom.

1. Stationary Orbits: Electrons revolve in specific stable circular orbits without radiating electromagnetic energy; Coulomb force provides centripetal force.
2. Angular Momentum Quantisation: Allowed orbits satisfy L = mvr = nh2π (n = 1, 2, 3...).
3. Frequency Condition: An electron emits or absorbs a photon only when transitioning between stationary states: hν = Eᵢ − E_f.

Q4 (Kerala HSE March • 4 Marks): Using Bohr's postulates, derive expressions for (i) radius of the n-th orbit and (ii) total energy of the electron in a Hydrogen atom.

(i) Radius: Force balance mv²/r = e²/(4πε₀r²) gives mv²r = e²/(4πε₀). From 2nd postulate, v = nh/(2πmr).
Substituting v: m (n²h² / 4π²m²r²) r = e²/(4πε₀) ⇒ rₙ = n²h²ε₀πme² = 0.529 n² Å.
(ii) Energy: K = e²/(8πε₀rₙ) and U = −e²/(4πε₀rₙ), so Eₙ = −e²/(8πε₀rₙ).
Substituting rₙ: Eₙ = −me⁴8ε₀²h²n² = −13.6n² eV.

Q5 (Kerala HSE Model • 2 Marks): The total energy of an electron in the first excited state of Hydrogen is −3.4 eV. Calculate its kinetic energy and potential energy in this state.

In a Bohr orbit, K = −E and U = 2E.
• Kinetic Energy: K = −(−3.4) = +3.4 eV
• Potential Energy: U = 2 × (−3.4) = −6.8 eV

Q6 (Kerala HSE March • 3 Marks): Explain how de Broglie's hypothesis of matter waves justifies Bohr's quantisation condition for angular momentum.

An electron of mass m and speed v behaves as a matter wave of wavelength λ = h/(mv).
For a standing wave on a circular orbit of radius r to avoid destructive cancellation, the circumference must hold an exact integer number of wavelengths:
2πr = nλ = n hmv.
Rearranging directly yields Bohr's 2nd postulate: mvr = nh2π.

Q7 (Kerala HSE Say • 2 Marks): Identify the spectral series of Hydrogen atom lying in (a) ultraviolet region and (b) visible region. Write the Rydberg formula for Balmer's Hα line.

(a) Ultraviolet region: Lyman series (n₁ = 1).
(b) Visible region: Balmer series (n₁ = 2).
For Hα line (3 → 2): 1λ = R (12² − 13²) = 5R36 ⇒ λ = 656.3 nm.

Q8 (NCERT / DHSE Numerical • 2 Marks): Calculate the shortest wavelength present in the Paschen series of Hydrogen spectral lines (Given R = 1.097 × 10⁷ m⁻¹).

Paschen series corresponds to n₁ = 3; shortest wavelength occurs at series limit n₂ = ∞:
1λ_min = R (13² − 0) = R9
λ_min = 91.097 × 10⁷ ≈ 8.204 × 10⁻⁷ m = 820.4 nm (Infrared).

Q9 (Kerala HSE Model • 2 Marks): A Hydrogen atom initially in the ground state absorbs a photon that excites it to the n = 4 level. Find (i) the energy of the absorbed photon and (ii) the maximum number of emission lines possible when it returns to ground state.

(i) ΔE = E₄ − E₁ = (−13.6/16) − (−13.6) = −0.85 + 13.6 = 12.75 eV.
(ii) Total possible transitions N = n(n − 1)2 = 4 × 32 = 6 spectral lines.