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Question

An electron in a hydrogen atom falls from \( n = 4 \) to \( n = 2 \). What is the energy of the emitted
photon? (Use \( E_n = -\frac{13.6}{n^2} \, \text{eV} \))

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Explanation

**Bohr's quantization** angular momentum L = m v r = n h/2π, for n=2 L=2h/2π= h/π=2.11×10⁻³⁴ J·s, for n=5 L=5h/2π, de Broglie λ = h/p, p= m v, for first orbit v=2.2×10⁶ m/s, λ= h/(m v)=6.6×10⁻³⁴/(9.1×10⁻³¹×2.2×10⁶)=3.3×10⁻¹⁰ m, circumference 2πr=3.33×10⁻¹⁰ m, one wavelength fits for n=1. E₄ = -(13.6/4²) = -0.85 eV , E₂ = -3.4 eV . Δ E = E₄ - E₂ = -0.85 - (-3.4) = 2.55 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 2.55 eV, consistent with Bohr

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