(a) determine its energy in eV (b) determin its wavelength in nm Consider the sp
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Question
(a) determine its energy in eV
(b) determin its wavelength in nm
Consider the spectral line of shortest wavelength corresponding to a transition shown in the figure.
(c) find its photon energy in eV
(d) find its wavelength in nm
(e) what is the shortest possible wavelength in the Balmer series?
The Salmer 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 in eV (b) determin its wavelength in nm Consider the spectral line of shortest wavelength corresponding to a transition shown in the figure. (c) find its photon energy in eV (d) find its wavelength in nm (e) what is the shortest possible wavelength in the Balmer series?Explanation / Answer
energy corresponding to longest wavelength is -3.401+0.378 = -3.023 eV
E = 3.023*1.6*10^-19 J = 4.8 *10^-19
wavelength = hc/E = 6.626*10^-34*3*10^8/4.8*10^-19
= 414 nm
1. energy of shortest wavlength = -3.401 - 1.512 = 1.889 eV = 3.022*10^-19 j
2. wavelength = hc/E = 6.626* 10^-34*3*10^8/ 3.022*10^-19 = 6.57 nm
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