Practice question
Question
What is the orbital period of an electron in the \( n = 2 \) orbit of a hydrogen atom if \( v_1 = 2.2
\times 10^6 \, \text{m/s} \) and \( r_1 = 5.3 \times 10^{-11} \, \text{m} \)?
Explanation
**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
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.