The Balmer series for the hydrogen atom corresponds to electronic transitions th
ID: 1376479 • Letter: T
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
E = hc/(lambda)
so to have longest wavelength, E must be smallest
E = 3.401 - 1.512 = 1.889 eV
E = hc/(lambda)
lambda = 6.63*10^-34 * 3*10^8/(1.6*10^-19 * 1.889) = 6.58*10^-7 m = 658 nm
Now,
E = hc/(lambda)
so to have shortest wavelength, E must be largest
E = 3.401 - .378 = 3.023 eV
E = hc/(lambda)
lambda = 6.63*10^-34 * 3*10^8/(1.6*10^-19 * 3.023) = 4.11*10^-7 m = 411 nm
Shortest possible wavelength will be where energy chnage is largest, so 411nm is answer
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