A hydrogen atom makes a downward transition from the n = 20 state to the n = 5 state. Find the wavelength of the emitted photon. The lowest level energy state of hydrogen is -13.6 eV. (h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J, c = 3.00 × 10⁸ m/s) A) 2.43 μm B) 1.46 μm C) 1.94 μm D) 2.92 μm

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Quizplus · Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The energy of a hydrogen atom in the n = 20 state is -13.6 eV / (20)^2 = -0.034 eV. The energy of a hydrogen atom in the n = 5 state is -13.6 eV / (5)^2 = -0.544 eV. The energy of the emitted photon is therefore 0.544 eV - 0.034 eV = 0.51 eV. The wavelength of the emitted photon is therefore λ = hc/E = (6.626 × 10^-34 J s)(3.00 × 10^8 m/s) / (0.51 eV * 1.60 × 10^-19 J/eV) = 2.43 μm.

A Hydrogen Atom Makes a Downward Transition from the N