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Vₛ = KE/e = 1.25 V.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
r₂(H)/r₁(Li²⁺) = 4 / (1/3) = 12.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
Charge = -3. Mass number = 31.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
Δx ≥ h/(4π m Δv) ≈ 1.58×10⁻¹³ m.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
Both have same v, so KE ratio = 1.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
Neutrons = 12. ²³₁₁Na also has 12 neutrons.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
W₀ ≈ 4.417×10⁻¹⁹ J. E = 6.626×10⁻¹⁹ J. KE ≈ 2.209×10⁻¹⁹ J.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
Mass number = 56. Charge = +2.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
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
Mass number = 70. Charge = +2.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
Mass number = 42. Charge = +2.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
KEₑ/KEₚ = mₚ/mₑ ≈ 1836.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
