#orbital period

Hydrogen Atom Properties - Radius, Speed and Energy

What is the orbital period of an electron in the \( n = 3 \) orbit if \( v_1 = 2.2 \times 10^6 \, \text{m/s} \) and \( r

**Hydrogen atom radius** r_n = n² a₀, a₀=5.3×10⁻¹¹ m first Bohr radius, r₂=4a₀=2.12×10⁻¹⁰ m, ratio r₄/r₂ =16/4=4, r₃=9a₀, circumference 2πr_n =2π n² a₀, for n=3 circumference=2π×9×5.3×10⁻¹¹=3×10⁻⁹ m. Orbital period T =2πr/v, v_n = v₁/n, v₁=2.2×10⁶ m/s, T₂=2πr₂/v₂, v₂=1.1×10⁶ m/s, T₂≈1.21×10⁻¹⁵ s. v₃ = (2.2 × 10⁶/3) ≈ 7.33 × 10⁵ m/s . r₃ = 9 × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m . T = (2π r₃/v₃) = (2 × 3.14 × 4.77 × 10⁻¹⁰/7.33 × 10⁵) ≈ 4.09 × 10⁻¹⁵ s . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Hydrogen Atom Properties - Radius, Speed and Energy

What is the orbital period of an electron in the first orbit of a hydrogen atom if its speed is \( 2.2 \times 10^6 \, \t

**Angular momentum** L_n = n h/2π, h=6.6×10⁻³⁴ J·s, L₂=2×6.6×10⁻³⁴/2π=2.11×10⁻³⁴ J·s, speed v_n = e²/(2 ε₀ h) ×1/n ≈2.2×10⁶/n m/s, kinetic energy ½ m v² =13.6/n² eV, illustrating Bohr model predictions for hydrogen-like atoms. T = (2π r/v) = (2 × 3.14 × 5.3 × 10⁻¹¹/2.2 × 10⁶) ≈ 1.51 × 10⁻¹⁶ s . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 1.51 × 10⁻¹⁶ s, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Hydrogen Atom Properties - Radius, Speed and Energy

What is the orbital period of an electron in the \( n = 2 \) orbit of a hydrogen atom if \( v_1 = 2.2 \times 10^6 \, \te

**Angular momentum** L_n = n h/2π, h=6.6×10⁻³⁴ J·s, L₂=2×6.6×10⁻³⁴/2π=2.11×10⁻³⁴ J·s, speed v_n = e²/(2 ε₀ h) ×1/n ≈2.2×10⁶/n m/s, kinetic energy ½ m v² =13.6/n² eV, illustrating Bohr model predictions for hydrogen-like atoms. v₂ = (v₁/2) = 1.1 × 10⁶ m/s , r₂ = 4 × 5.3 × 10⁻¹¹ = 2.12 × 10⁻¹⁰ m . T = (2π r₂/v₂) = (2 × 3.14 × 2.12 × 10⁻¹⁰/1.1 × 10⁶) ≈ 1.21 × 10⁻¹⁵ s . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 1.21

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Universal Law of Gravitation

A planet orbits the Sun with a period of 5 years. If Earth’s orbital radius is 1.5×1011m, what is its semi-major axis?

Kepler’s third law: Tp2TE2 = ap3aE3. TE = 1year, Tp = 5years, aE = 1.5×1011m. 5212 = ap3(1.5×1011)3. 25 = ap33.375×1033. ap3 = 25×3.375×1033 = 8.4375×1034. ap = (8.4375×1034)1/3≈4.39×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.4 × 10¹¹ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.