Practice question
Question
What is the frequency of a photon emitted when an electron drops from \( n = 3 \) to \( n = 1 \) in a
hydrogen atom? (Use \( h = 6.6 \times 10^{-34} \, \text{J·s} \), 1 eV = \( 1.6 \times 10^{-19} \,
\text{J} \))
Explanation
**Excitation** energy required to go from n=1 to n=3 is 12.09 eV, from ground to n=∞ ionization 13.6 eV, state n=∞ means ionized, electron free with zero energy, highest level reached by electron beam energy determines which levels can be excited, e.g., 11 eV beam from ground can reach n=2 (10.2 eV) but not n=3 (12.09 eV), so max n=2. Δ E = 12.09 eV = 1.9344 × 10⁻¹⁸ J . nu = (Δ E/h) = (1.9344 × 10⁻¹⁸/6.6 × 10⁻³⁴) ≈ 2.93 × 10¹⁵ Hz . 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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