Practice question
Question
An electron in a hydrogen atom transitions from \( n = 3 \) to \( n = 1 \). What is the energy of the
emitted photon? (Use \( E_n = -\frac{13.6}{n^2} \, \text{eV} \))
Explanation
**Rutherford's nuclear model** atom has small massive positively charged nucleus with electrons orbiting, size ratio atomic to nuclear ~10⁵, nucleus ~10⁻¹⁵ m atom ~10⁻¹⁰ m, most alpha particles with large impact parameter pass undeflected, small fraction >90° scatter from close approach, centripetal force provided by Coulomb attraction k Z e²/r², fails to explain stability because accelerating charge should radiate and collapse. E₃ = -(13.6/3²) = -1.51 eV , E₁ = -13.6 eV . Δ E = E₃ - E₁ = -1.51 - (-13.6) = 12.09 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE
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