NCERT Solutions for Class 12 Physics Chapter 12 Atoms
NCERT Solutions for Class 12 Physics Chapter 12 Atoms provides detailed coverage of Rutherford's alpha-particle scattering experiment, Bohr's model of the hydrogen atom, spectral series of hydrogen, and de Broglie's explanation of Bohr's second postulate.
Overview of Topics & Subtopics (Click to Jump)
1. Alpha-Particle Scattering Experiment & Rutherford's Atomic Model
Short Note:
• Geiger-Marsden Experiment: Alpha particles (He²⁺) directed at a thin gold foil. Observations showed most α-particles pass un-deviated, while a tiny fraction (~1 in 8000) bounce back by > 90°.
• Key Inferences: Entire positive charge and almost all mass of the atom are concentrated in a tiny central region called the nucleus (size ~ 10⁻¹⁵ m to 10⁻¹⁴ m), while atom size is ~ 10⁻¹⁰ m.
• Distance of Closest Approach (r₀): Estimated size of nucleus: r₀ = (1 / 4πε₀) × (2 Z e² / K), where K is the initial kinetic energy of the α-particle.
• Impact Parameter (b): Perpendicular distance of the velocity vector of α-particle from the central nucleus line. Small b leads to large deflection angle θ.
• Limitations of Rutherford Model: Could not explain the electrical stability of an orbiting accelerating electron radiating continuous energy, nor the discrete line spectrum of atoms.
2. Bohr's Model of the Hydrogen Atom
Short Note:
• Bohr's Postulates:
1. Stationary Orbits: Electrons revolve around the nucleus only in certain non-radiating, stable circular orbits.
2. Quantization of Angular Momentum: Orbital angular momentum is an integral multiple of h / (2π): L = m v r = (n h) / (2π), where n = 1, 2, 3... is the principal quantum number.
3. Frequency Condition: Energy is emitted or absorbed as a photon only during transition between orbits: h ν = E₂ - E₁.
• Radius of n-th Orbit: r_n = (n² h² ε₀) / (π m e²) = n² × r₁ (Bohr radius a₀ = r₁ ≈ 0.53 Å = 0.053 nm).
• Velocity of Electron in n-th Orbit: v_n = e² / (2 ε₀ n h) ≈ c / (137 n).
• Energy of Electron in n-th Orbit:
- Kinetic Energy: K = -E_total = 13.6 / n² eV
- Potential Energy: U = 2 E_total = -27.2 / n² eV
- Total Energy: E_n = -13.6 / n² eV (Negative sign confirms electron is bound to nucleus).
3. Line Spectra of Hydrogen & De Broglie's Explanation
Short Note:
• Rydberg Formula for Hydrogen Wavelength: 1 / λ = R × (1 / n₁² - 1 / n₂²), where R is Rydberg constant (R ≈ 1.097 × 10⁷ m⁻¹).
• Spectral Series Summary:
- Lyman Series: n₁ = 1, n₂ = 2, 3, 4... (Ultraviolet Region)
- Balmer Series: n₁ = 2, n₂ = 3, 4, 5... (Visible Region)
- Paschen Series: stream n₁ = 3, n₂ = 4, 5, 6... (Near Infrared Region)
- Brackett Series: n₁ = 4, n₂ = 5, 6, 7... (Infrared Region)
- Pfund Series: n₁ = 5, n₂ = 6, 7, 8... (Far Infrared Region)
• De Broglie's Explanation of Quantization: Standing wave condition for an electron wave along a circular orbit requires circumference to equal integer number of wavelengths: 2π r = n λ = n (h / m v) => m v r = (n h) / (2π).
Intext & Conceptual Questions
• Bohr's Postulate: Bohr postulated that electrons in an atom revolve only in those non-radiating stationary orbits for which the total orbital angular momentum (L) is an integral multiple of h / (2π):
L = m v r = (n h) / (2π), where n = 1, 2, 3...
• De Broglie's Explanation: De Broglie proposed that an electron revolving in a circular orbit behaves as a matter wave. For a stable stationary orbit, the electron wave must form a constructive standing wave around the circular path, meaning the total perimeter of the orbit must equal an integral number of de Broglie wavelengths:
2π r = n λ
Substituting de Broglie wavelength λ = h / (m v) into the equation:
2π r = n (h / m v) => m v r = (n h) / (2π)
Thus, de Broglie's wave hypothesis provided a theoretical derivation for Bohr's empirical quantization rule.
| Feature | Lyman Series | Balmer Series |
|---|---|---|
| Lower Orbit (n₁) | n₁ = 1 (Ground State) | n₁ = 2 (First Excited State) |
| Upper Orbits (n₂) | n₂ = 2, 3, 4, 5... | n₂ = 3, 4, 5, 6... |
| Spectral Region | Ultraviolet (UV) Region | Visible Light Region |
| Energy Transition | Higher energy transitions (larger frequency) | Moderate energy transitions |
Textbook Chapter End Exercise Questions
For Balmer series, the lower quantum level is n₁ = 2.
Rydberg Formula: 1 / λ = R × (1 / 2² - 1 / n₂²) = R × (1 / 4 - 1 / n₂²)
Part (a): Longest Wavelength (λ_max)
Longest wavelength corresponds to the minimum energy photon transition, which occurs when n₂ = 3:
1 / λ_max = R × (1 / 4 - 1 / 9) = R × (5 / 36)
λ_max = 36 / (5 × R) = 36 / (5 × 1.097 × 10⁷) ≈ 656.3 nm
Part (b): Shortest Wavelength (λ_min)
Shortest wavelength corresponds to the maximum energy photon transition (series limit), which occurs when n₂ = ∞:
1 / λ_min = R × (1 / 4 - 0) = R / 4
λ_min = 4 / R = 4 / (1.097 × 10⁷) ≈ 364.6 nm
• Longest wavelength = 656.3 nm (H-alpha line)
• Shortest wavelength = 364.6 nm (Balmer limit)
Given for Hydrogen (Z = 1, n = 1):
• Charge of electron (e) = 1.6 × 10⁻¹⁹ C
• Planck's constant (h) = 6.63 × 10⁻³⁴ J s
• Permittivity of free space (ε₀) = 8.854 × 10⁻¹² F/m
Step 1: Calculate Electron Velocity (v₁)
v₁ = e² / (2 ε₀ h)
v₁ = (1.6 × 10⁻¹⁹)² / (2 × 8.854 × 10⁻¹² × 6.63 × 10⁻³⁴)
v₁ ≈ 2.18 × 10⁶ m/s (roughly 1/137 of light speed c)
Step 2: Calculate Radius of Ground Orbit (r₁)
r₁ = 0.53 Å = 0.53 × 10⁻¹⁰ m
Step 3: Calculate Orbital Period (T)
T = (2π r₁) / v₁ = (2 × 3.1416 × 0.53 × 10⁻¹⁰) / (2.18 × 10⁶) ≈ 1.53 × 10⁻¹⁶ s
• Speed of electron in n = 1 level = 2.18 × 10⁶ m/s
• Orbital time period = 1.53 × 10⁻¹⁶ s
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