Practice question
Question
What is the kinetic energy of an electron in the \( n = 4 \) state of a hydrogen atom? (Use \( E_n =
-\frac{13.6}{n^2} \, \text{eV} \))
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. E₄ = -0.85 eV , K = -E₄ = 0.85 eV . 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 0.85 eV, consistent with Bohr model and nuclear binding energy systematics.
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