Chemistry I / Atomic Spectra and the Quantum Atom
Practice question · Numerical answer

An electron falls from n=3n = 3 to n=2n = 2 in a hydrogen atom. Using En=−2.18×10−18/n2E_n = -2.18 \times 10^{-18}/n^2 J, compute the emitted photon energy in units of 10−1910^{-19} J, to two decimal places.

Hints
  1. Emitted energy is the gap between the two levels.
  2. Compute 2.18e-18 x (1/4 - 1/9).
Show the answer

3.03 (answers within ±0.06 count)

Why

ΔE=2.18×10−18(1/22−1/32)=2.18×10−18×0.1389=3.03×10−19\Delta E = 2.18 \times 10^{-18}(1/2^2 - 1/3^2) = 2.18 \times 10^{-18} \times 0.1389 = 3.03 \times 10^{-19} J, the red H-alpha line at 656 nm.

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