Practice question
Question
In the Bohr model, what is the ratio of the kinetic energy of an electron in the \( n = 2 \) state to
that in the \( n = 1 \) state?
Explanation
**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. K = (e²/8πepsilon₀ r) , r_n ∝ n² . K_n ∝ (1/n²) . Ratio = (K₂/K₁) = (1/2²/1/1²) = (1/4) . 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 0.25, consistent with Bohr model and nuclear binding energy systematics.
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.