7 public questions tagged with this topic.
n=4. E₄ = -13.6/16 = -0.85 eV.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Shapes of Orbitals - s p d f and Nodes and Radial Probability
n=6. E₆ = -2.18×10⁻¹⁸/36 = -6.056×10⁻²⁰ J.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Numbers - Principal Azimuthal Magnetic Spin
Eₙ = -2.18×10⁻¹⁸ / n². E₃ = -2.18×10⁻¹⁸ / 9 = -2.4222×10⁻¹⁹ J.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Numbers - Principal Azimuthal Magnetic Spin
E₄ = -2.18×10⁻¹⁸/16 = -1.3625×10⁻¹⁹ J.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
n=6. For Be³⁺ (Z=4), E₆ = -16×2.18×10⁻¹⁸/36 = -9.678×10⁻¹⁹ J.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
E₂ = -9 × 2.18×10⁻¹⁸ / 4 = -4.905×10⁻¹⁸ J.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
E₄ = -3.4 eV. IE = 3.4 eV.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Developments Leading to Bohr Model - Electromagnetic Radiation and Spectrum
