What is the wave number of the hydrogen atom excitation from #n = 1# to #n = 3#?

1 Answer
Jul 29, 2017

About #"97533 cm"^(-1)#.


I assume you mean, what is the frequency in #"cm"^(-1)# (i.e. wavenumbers) of the electronic excitation #n = 1 -> 3# in the hydrogen atom...

We consult the Rydberg equation in #"eV"# (because I am too lazy to remember the Rydberg constant in #"m"^(-1)# or what have you):

#DeltaE = -"13.606 eV"(1/n_f^2 - 1/n_i^2)#

where:

  • #n_k# is the principal quantum number for the #k#th state in the hydrogen atom, i.e. initial #i# or final #f#.
  • #-"13.606 eV"# is the approx. ground state energy of hydrogen atom.

Thus, the change in energy for this excitation (which is positive!) is:

#DeltaE = -"13.606 eV"(1/3^2 - 1/1^2)#

#=# #"12.094 eV"#

And now it is a matter of unit conversion into the equivalent frequency in #"cm"^(-1)#. The following constants will be useful:

  • #h = 6.626 xx 10^(-34) "J"cdot"s"#, Planck's constant.
  • #c = 2.998 xx 10^(10) "cm/s"#, the speed of light.
  • #e ~= 1.602 xx 10^(-19) "J/eV"#, numerically equal to the elementary charge.

#color(blue)(Deltanu) = 12.094 cancel"eV" xx (1.602 xx 10^(-19) cancel"J")/(cancel"1 eV")#

#xx 1/((2.998 xx 10^(10) "cm/"cancel"s")(6.626 xx 10^(-34) cancel"J" cdotcancel"s"))#

#~~ color(blue)("97533 cm"^(-1))#