11 public questions tagged with this topic.
Transitions: 5→2, 5→3→2, 5→4→2, 4→2, 3→2. Total = 5.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Shapes of Orbitals - s p d f and Nodes and Radial Probability
Transitions: 7→4, 7→5→4, 7→6→4, 6→4, 5→4. Total = 5.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Shapes of Orbitals - s p d f and Nodes and Radial Probability
Number of lines = (9-6)(9-6+1)/2 = 6.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Shapes of Orbitals - s p d f and Nodes and Radial Probability
Brackett: n₁=4, n₂=6. λ = 2625 nm.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Shapes of Orbitals - s p d f and Nodes and Radial Probability
Number of lines = (n₂-n₁)(n₂-n₁+1)/2 = 6.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Shapes of Orbitals - s p d f and Nodes and Radial Probability
Number of lines = (8-4)(8-4+1)/2 = 10; actual distinct transitions = 7.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Numbers - Principal Azimuthal Magnetic Spin
Transitions: 4→2, 4→3→2, 3→2. Total = 3.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
Number of lines = (7-3)(7-3+1)/2 = 10.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Developments Leading to Bohr Model - Electromagnetic Radiation and Spectrum
Lyman: n₁=1, third line n₂=4. ν̄ = RH (1 - 1/16) = 1.028×10⁷ m⁻¹.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Developments Leading to Bohr Model - Electromagnetic Radiation and Spectrum
Balmer: n₁=2, n₂=6. ν̄ = 2.437×10⁶ m⁻¹.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Atomic Models - Thomson Rutherford and Bohr Model
Number of lines = (8-5)(8-5+1)/2 = 6.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Atomic Models - Thomson Rutherford and Bohr Model
