Practice question
Question
A hydrogen atom emits a photon when an electron drops from \( n = 5 \) to \( n = 2 \). What is the
energy of the photon? (Use \( E_n = -\frac{13.6}{n^2} \, \text{eV} \))
Explanation
**Quantization basis** de Broglie standing wave requires constructive interference, integer wavelengths in orbit, otherwise destructive, so only certain radii allowed r_n = n² a₀, a₀=0.53 Å, angular momentum L = n h/2π, de Broglie explains why orbits are stationary - electron wave closed on itself. E₅ = -(13.6/5²) = -0.544 eV , E₂ = -3.4 eV . Δ E = -0.544 - (-3.4) = 2.856 eV ≈ 2.86 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 2.86 eV, 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.