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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?

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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.

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