The Balmer series for the hydrogen atom corresponds to electronic transitions th

Question

The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number n = 2 as shown in the figure below.

Consider the photon of longest wavelength corresponding to a transition shown in the figure. (a) Determine its energy. eV (b) Determine its wavelength. nm Consider the spectral line of shortest wavelength corresponding to a transition shown in the figure. (c) Find its photon energy. eV (d) Find its wavelength. nm (e) What is the shortest possible wavelength in the Balmer series? nm

Explanation / Answer

a) biggest wavelength means smallest energy

so -1.512+3.401=1.889 eV

b) lambda = 1240/1.889=656.4 nm

c) now biggest energy

so 3.401 -0.378=3.023

d) wavelength = 1240/3.023=410.2 nm

e) wavelength = 1240/3.401=364.6 nm

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