Chapter 12: Atoms | Class 12 Physics | AAI ATC Exam Prep

📌 Chapter Overview

🔬 Thomson's Plum Pudding Model

💥 Rutherford's Scattering Experiment

🪐 Nuclear Model of Atom

🌈 Atomic Spectra

⚛️ Bohr's Model & Energy Levels

🌊 de Broglie's Explanation

Alpha-Particle Scattering & Rutherford's Nuclear Model

In 1911, Geiger and Marsden, under Rutherford's guidance, directed a beam of 5.5 MeV α-particles (from ²¹⁴₈₃Bi source) at a thin gold foil of thickness 2.1 × 10⁻⁷ m. Scattered particles were detected using a ZnS screen and microscope.

Key Observations

Most α-particles passed straight through. Only ~0.14% scattered by more than 1°, and about 1 in 8000 deflected by more than 90° (backward scattering).

      💡 Backward scattering required a large repulsive force — possible only if the entire positive charge and most mass were concentrated in a tiny, dense nucleus.
    

Rutherford's Nuclear Model

Nucleus size: 10⁻¹⁵ m to 10⁻¹⁴ m. Atom size: 10⁻¹⁰ m. The atom is mostly empty space — electrons orbit the nucleus like planets around the sun.

Coulomb Repulsive Force on α-particle

F = (1/4πε₀) × (2e × Ze) / r²

where Z = 79 for gold, r = distance between α-particle and nucleus.

Impact Parameter (b)

The impact parameter is the perpendicular distance of the α-particle's initial velocity from the nucleus centre. Smaller b → larger scattering angle. Head-on collision (b = 0) → maximum deflection (~180°).

🖼️

ADD IMAGE HERE
Geiger-Marsden Experimental Setup (Fig. 12.1 / 12.2)

🖼️

ADD IMAGE HERE
α-particle trajectory showing impact parameter b and scattering angle θ (Fig. 12.4)

🎯 Practice MCQs — Alpha-Particle Scattering

Q1 In the Geiger-Marsden experiment, the distance of closest approach for a 7.7 MeV α-particle hitting a gold nucleus (Z = 79) is approximately:

A 6 fm

B 30 fm

C 10 fm

D 100 fm

✅ d = 2Ze² / (4πε₀ K)
d = 2 × 79 × (9×10⁹) × (1.6×10⁻¹⁹)² / (1.2×10⁻¹²)
d = 3.84×10⁻¹⁶ × 79 ≈ 3.0 × 10⁻¹⁴ m = 30 fm

Q2 An α-particle of kinetic energy 5 MeV is aimed at a silver nucleus (Z = 47). The distance of closest approach is: (1 MeV = 1.6×10⁻¹³ J)

A 13.6 fm

B 30 fm

C 27.1 fm

D 48 fm

✅ d = 2Ze² / 4πε₀K = 2 × 47 × (9×10⁹) × (1.6×10⁻¹⁹)² / (5×1.6×10⁻¹³)
d = (2 × 47 × 9×10⁹ × 2.56×10⁻³⁸) / (8×10⁻¹³)
d ≈ 2.71 × 10⁻¹⁴ m = 27.1 fm

Q3 In the Rutherford scattering experiment using gold (Z = 79), the ratio of the size of the atom to the size of the nucleus is approximately:

A 10

B 1000

C 10,000 to 1,00,000

D 100

✅ Atom size ≈ 10⁻¹⁰ m; Nucleus size ≈ 10⁻¹⁵ to 10⁻¹⁴ m
Ratio = 10⁻¹⁰ / 10⁻¹⁵ = 10⁵ (or 10⁴ to 10⁵)

Electron Orbits — Energy in Classical Model

In Rutherford's model, the electrostatic force of attraction (Fₑ) between the electron and nucleus provides the centripetal force (F꜀). For a stable orbit in hydrogen:

Fₑ = F꜀ → e² / (4πε₀r²) = mv²/r

Orbit Radius vs Velocity

r = e² / (4πε₀mv²)

Kinetic & Potential Energy

K = ½mv² = e²/(8πε₀r) | U = −e²/(4πε₀r)

Total Energy of Electron

E = K + U = −e²/(8πε₀r)

⚠️ The total energy is negative → the electron is bound to the nucleus. A positive E would mean the electron escapes (unbound).

Problem with Classical Model

A revolving (accelerated) charged particle must radiate energy (classical EM theory). The electron would spiral into the nucleus in ~10⁻⁸ s — the atom would collapse! Also, it would emit a continuous spectrum, not the observed discrete lines.

🖼️

ADD IMAGE HERE
Electron spiralling into nucleus due to energy radiation (Fig. 12.6)

🎯 Practice MCQs — Electron Orbits

Q4 The total energy of an electron in the ground state of hydrogen atom is −13.6 eV. Its kinetic energy is:

A −13.6 eV

B +13.6 eV

C −27.2 eV

D +6.8 eV

✅ K = −E (total) = −(−13.6) = +13.6 eV
(Kinetic energy = −½ × potential energy = −E for Coulomb orbits)

Q5 The orbital radius of an electron in hydrogen atom when total energy is −13.6 eV is:

A 1.06 × 10⁻¹⁰ m

B 5.3 × 10⁻¹¹ m

C 2.12 × 10⁻¹⁰ m

D 0.265 × 10⁻¹⁰ m

✅ E = −e²/(8πε₀r) → r = −e²/(8πε₀E)
r = (9×10⁹ × (1.6×10⁻¹⁹)²) / (2 × 2.2×10⁻¹⁸) = 5.3 × 10⁻¹¹ m (Bohr radius a₀)

Q6 The velocity of an electron in the ground state of hydrogen atom (r = 5.3 × 10⁻¹¹ m) is approximately:

A 3 × 10⁸ m/s

B 2.2 × 10⁸ m/s

C 2.2 × 10⁶ m/s

D 4.4 × 10⁶ m/s

✅ v = e / √(4πε₀mr) = 1.6×10⁻¹⁹ / √(4π × 8.85×10⁻¹² × 9.1×10⁻³¹ × 5.3×10⁻¹¹)
v ≈ 2.2 × 10⁶ m/s (~0.7% of speed of light)

Atomic Spectra

Each element emits a characteristic spectrum when electrically excited. For rarefied gases, only discrete wavelengths are emitted — appearing as bright lines on dark background: emission line spectrum.

Absorption Spectrum

When white light passes through a gas, dark lines appear in the continuous spectrum at the same wavelengths the gas emits. This is the absorption spectrum — unique "fingerprint" of each element.

Hydrogen Spectrum — Balmer Series (1885)

Johann Balmer gave an empirical formula for wavelengths in the visible region of hydrogen spectrum. The spectrum has series of lines: Lyman (UV), Balmer (visible), Paschen (IR), etc.

      🌈 The discrete wavelengths in hydrogen spectrum hint at a deep connection between atomic structure and energy quantization — which Bohr later explained.
    

🖼️

ADD IMAGE HERE
Hydrogen emission spectrum showing Lyman, Balmer, Paschen series (Fig. 12.5)

🎯 Practice MCQs — Atomic Spectra

Q7 A difference of 2.3 eV separates two energy levels. The frequency of radiation emitted during transition is:

A 3.68 × 10¹⁴ Hz

B 5.55 × 10¹⁴ Hz

C 2.3 × 10¹⁴ Hz

D 1.1 × 10¹⁵ Hz

✅ hν = ΔE = 2.3 eV = 2.3 × 1.6×10⁻¹⁹ = 3.68×10⁻¹⁹ J
ν = 3.68×10⁻¹⁹ / 6.63×10⁻³⁴ = 5.55 × 10¹⁴ Hz

Q8 The initial frequency of light emitted by an electron revolving around a proton at radius 5.3 × 10⁻¹¹ m with velocity 2.2 × 10⁶ m/s is:

A 3.3 × 10¹⁵ Hz

B 6.6 × 10¹⁵ Hz

C 1.32 × 10¹⁵ Hz

D 6.6 × 10¹⁴ Hz

✅ ν = v / (2πr) = (2.2×10⁶) / (2π × 5.3×10⁻¹¹)
ν = 2.2×10⁶ / 3.33×10⁻¹⁰ ≈ 6.6 × 10¹⁵ Hz

Q9 A hydrogen atom absorbs a photon and the electron jumps from n=1 to n=4. The energy of the absorbed photon is: (Ground state energy = −13.6 eV)

A 10.2 eV

B 12.09 eV

C 12.75 eV

D 13.6 eV

✅ E₁ = −13.6 eV; E₄ = −13.6/16 = −0.85 eV
ΔE = E₄ − E₁ = −0.85 − (−13.6) = 12.75 eV

Bohr's Model of the Hydrogen Atom

Niels Bohr (1885–1962) modified Rutherford's model using quantum ideas (1913). He introduced three postulates:

Postulate 1 — Stationary Orbits

Electrons revolve in certain stable orbits (stationary states) without radiating energy. Each state has a definite total energy.

Postulate 2 — Quantisation of Angular Momentum

L = nh/2π , where n = 1, 2, 3, ... (principal quantum number)

Only those orbits are allowed where the angular momentum is an integral multiple of h/2π.

Postulate 3 — Photon Emission/Absorption

hν = Eᵢ − E_f (emission) | E_f = Eᵢ + hν (absorption)

Radius of nth Orbit

rₙ = (n²/m)(h/2π)² × (4πε₀/e²) ∝ n²

For n=1 (ground state): r₁ = a₀ = 0.529 Å = 5.3 × 10⁻¹¹ m (Bohr radius)

Energy of nth Orbit

Eₙ = −me⁴ / (8n²ε₀²h²) = −13.6/n² eV

      📌 Key values: E₁ = −13.6 eV (ground state) | E₂ = −3.40 eV | E₃ = −1.51 eV | E₄ = −0.85 eV | E∞ = 0 (ionised)
    

Ionisation Energy

Minimum energy to remove the electron from the ground state = 13.6 eV

🎯 Radii: rₙ ∝ n² → r₂ = 4a₀, r₃ = 9a₀. Energies: Eₙ ∝ 1/n² → higher n means less negative, more energy.

🖼️

ADD IMAGE HERE
Energy level diagram of hydrogen atom showing n=1 to n=∞ (Fig. 12.7)

🎯 Practice MCQs — Bohr's Model

Q10 The radius of the electron orbit in hydrogen atom for n = 3 is: (Bohr radius a₀ = 5.3 × 10⁻¹¹ m)

A 5.3 × 10⁻¹¹ m

B 21.2 × 10⁻¹¹ m

C 47.7 × 10⁻¹¹ m

D 15.9 × 10⁻¹¹ m

✅ rₙ = n² × a₀ = 9 × 5.3 × 10⁻¹¹ = 47.7 × 10⁻¹¹ m

Q11 Energy required to excite hydrogen atom from n=2 to n=3 is:

A 10.2 eV

B 1.89 eV

C 12.09 eV

D 3.4 eV

✅ E₂ = −3.40 eV; E₃ = −1.51 eV
ΔE = E₃ − E₂ = −1.51 − (−3.40) = 1.89 eV

Q12 In a hydrogen atom, the speed of the electron in n = 2 orbit is: (v₁ = 2.2 × 10⁶ m/s)

A 4.4 × 10⁶ m/s

B 1.1 × 10⁶ m/s

C 2.2 × 10⁶ m/s

D 0.55 × 10⁶ m/s

✅ vₙ = v₁/n (from Bohr model: mvr = nh/2π and Coulomb force)
v₂ = 2.2×10⁶/2 = 1.1 × 10⁶ m/s

Line Spectra of Hydrogen Atom

When an electron transitions from orbit nᵢ (higher) to n_f (lower), a photon is emitted:

hν_if = Eₙᵢ − Eₙ_f = 13.6(1/n_f² − 1/nᵢ²) eV

Spectral Series of Hydrogen

      🔵 Lyman Series: n_f = 1, nᵢ = 2,3,4,… → UV region

      🟢 Balmer Series: n_f = 2, nᵢ = 3,4,5,… → Visible region

      🔴 Paschen Series: n_f = 3, nᵢ = 4,5,6,… → Near IR

      🟠 Brackett Series: n_f = 4 → IR region

      🟡 Pfund Series: n_f = 5 → Far IR

🎯 Bohr won the Nobel Prize in Physics in 1922 for his hydrogen atom model.

Wavelength Formula (Rydberg)

1/λ = R_H (1/n_f² − 1/nᵢ²) where R_H = 1.097 × 10⁷ m⁻¹

🎯 Practice MCQs — Line Spectra

Q13 A hydrogen atom in ground state absorbs a photon of 12.5 eV energy. To which orbit does the electron jump?

A n = 2

B n = 3

C n = 3 (closest possible)

D n = 4

✅ E₁=−13.6; E₂=−3.4; E₃=−1.51; E₄=−0.85 eV
E₁+12.5 = −13.6+12.5 = −1.1 eV → closest to E₃ = −1.51 eV
The atom can absorb exactly 12.09 eV (n=1→3) or 10.2 eV (n=1→2). For 12.5 eV beam, n=3 transitions will dominate (some KE to electron).

Q14 The wavelength of photon emitted when hydrogen electron jumps from n=4 to n=2 (Balmer series) is approximately: (R_H = 1.097 × 10⁷ m⁻¹)

A 656 nm

B 486 nm

C 410 nm

D 122 nm

✅ 1/λ = 1.097×10⁷ × (1/4 − 1/16) = 1.097×10⁷ × (3/16)
1/λ = 2.057×10⁶ m⁻¹ → λ ≈ 486 nm (Hβ line, blue-green)

Q15 The quantum number characterising Earth's revolution around the Sun (r = 1.5 × 10¹¹ m, v = 3 × 10⁴ m/s, M = 6 × 10²⁴ kg) using Bohr's condition is:

A 10⁵⁰

B 2.57 × 10⁷⁴

C 6 × 10⁶⁸

D 1.5 × 10⁸⁰

✅ L = Mvr = nh/2π → n = 2πMvr/h
n = (2π × 6×10²⁴ × 3×10⁴ × 1.5×10¹¹) / (6.63×10⁻³⁴)
n ≈ 2.57 × 10⁷⁴ — an astronomically large quantum number!

de Broglie's Explanation & Limitations of Bohr's Model

de Broglie's Explanation (1923)

Louis de Broglie explained Bohr's quantisation using the wave nature of electrons. An electron in a circular orbit must form a standing wave. For this, the circumference of the orbit must be an integral multiple of the de Broglie wavelength:

2πrₙ = nλ = n(h/mvₙ) → mvₙrₙ = nh/2π

This directly gives Bohr's angular momentum quantisation condition! The quantised orbits correspond to resonant standing waves — only specific radii "fit" whole wavelengths.

🖼️

ADD IMAGE HERE
Standing wave on circular orbit for n=4 (Fig. 12.8)

Limitations of Bohr's Model

❌ Cannot be applied to multi-electron atoms (e.g., Helium) — does not account for electron-electron repulsion.

❌ Cannot explain relative intensities of spectral lines — why some transitions are more probable than others.

❌ Inconsistent with Heisenberg's Uncertainty Principle — assumes well-defined orbit (position + momentum simultaneously).

🎯 Practice MCQs — de Broglie & Model Limitations

Q16 According to de Broglie's explanation, for a stable orbit the circumference of the orbit (2πr) must equal:

A h/mv

B λ/n

C nλ (integral multiples of de Broglie wavelength)

D n²λ

✅ For standing wave: 2πrₙ = nλ, where λ = h/mvₙ
This gives: mvₙrₙ = nh/2π — Bohr's 2nd postulate!

Q17 For n=4 orbit in hydrogen atom, how many de Broglie wavelengths fit in the circumference of the orbit?

A 1

B 2

C 3

D 4

✅ 2πrₙ = nλ → for n = 4, exactly 4 wavelengths fit in the orbit circumference. (As shown in Fig. 12.8 of NCERT)

Q18 Bohr's model gives correct results for:

A All multi-electron atoms

B Only helium atom

C Hydrogenic atoms (single-electron systems)

D Only hydrogen atom, not He⁺ or Li²⁺

✅ Bohr's model works for hydrogenic atoms — H, He⁺, Li²⁺, Be³⁺, etc. (one electron, any nuclear charge Z). It fails for multi-electron atoms.

🚀 Master AAI ATC Physics with Aviate Learnings!

Watch video lectures, join live classes, and submit your doubts directly to Kanishk Sir

▶ Subscribe on YouTube 📋 Submit Your Query

⚛️ Atoms