The hydrogen atom in the ground state is excited by means of monochromatic radiations of wavelength #lambda# #Å#. The resulting spectrum consist of maximum 15 different lines. What is the value of #lambda#? #R_H = "109737 cm"^(-1)#?

1 Answer
Truong-Son N.
May 15, 2018

#lambda = 937.306# #Å#

The 15 emission lines result from five groups of possible relaxations, consisting of #5,4,3,2,# and #1# transition(s) respectively; this means it started from at most #n = 6# and ended at anywhere from #n = 1# to #n = 5#:

#6->5, 6->4, 6->3, 6->2, 6->1##" "" "("5 transitions")#
#5->4, 5->3, 5->2, 5->1##" "" "" "" "color(white)(/)("4 transitions")#
#4->3, 4->2, 4->1##" "" "" "" "" "" "" "color(white)(.)("3 transitions")#
#3->2, 3->1##" "" "" "" "" "" "" "" "" "color(white)(....)("2 transitions")#
#2->1##" "" "" "" "" "" "" "" "" "" "" "" "" "("1 transition")#

Inspect those combinations to see that there are no repeats.

An ensemble of hydrogen atoms really experiences all of these transitions in some distribution if the excited electron makes it to #n = 6# from #n = 1# by absorbing incoming light.

First we begin with the Rydberg equation for hydrogen atom:

#DeltaE = -R_H(1/n_f^2 - 1/n_i^2)#

Then the electron must have absorbed energy to go from #n = 1# to #n = 6# so #DeltaE > 0#:

#DeltaE = -"109737 cm"^(-1) cdot (1/6^2 - 1/1^2)#

#= "106689 cm"^(-1)#

This form of the energy is already in reciprocal wavelength units, so we need only to convert to angstroms.

#1/"106689 cm"^(-1) = 9.37306 xx 10^(-6) "cm"#

#color(blue)(lambda) = 9.37306 xx 10^(-6) cancel"cm" xx cancel"1 m"/(100 cancel"cm") xx (1color(white)(.)Å)/(10^(-10) cancel"m")#

#=# #color(blue)("937.306 Å")#

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