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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
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
v₃ = 2.19×10⁶ m s⁻¹. KE = ½mv² = 2.18×10⁻¹⁸ J.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Atomic Models - Thomson Rutherford and Bohr Model
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