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n=3. v = 1.46×10⁶ m s⁻¹. KE ≈ 9.7×10⁻¹⁹ J.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
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
E = hν = 4.9695×10⁻¹⁹ J.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
E ≈ 9.939×10⁻¹⁹ J. KE ≈ 1.994×10⁻¹⁹ J. W₀ ≈ 7.945×10⁻¹⁹ J.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
v = √(2KE/m) ≈ 2.306×10⁶ m s⁻¹.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
λLyman/λBalmer ≈ 0.185.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
Total electrons = 15. Atomic number = 15.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
E = hν ≈ 3.976×10⁻¹⁹ J. KE = E - W₀ ≈ 1.476×10⁻¹⁹ J.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
Atomic number 16: 1s² 2s² 2p⁶ 3s² 3p⁴ (sulfur).
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
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
v₄ = v₁/4 = 5.475×10⁵ m s⁻¹.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
For Li²⁺ (Z=3), E = Z² EH. Ratio = 9.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
