The first point that you must perform below is to use the frequency of the emitted photon to calculate its wavelength.

As you understand, frequency and also wavelength have an inverse relationship defined by the equation

#color(blue)(ul(color(black)(nu * lamda = c)))#

Here

#nu# is the frequency of the photon#c# is the rate of light in a vacuum, commonly given as #3 * 10^8# #"m s"^(-1)#

Plug in your value to find

#lamda = (3 * 10^8color(white)(.)"m" color(red)(cancel(color(black)("s"^(-1)))))/(6.90 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))) = 4.348 * 10^(-7)# #"m"#

Now, the connection in between the wavesize of the emitted photon and the major quantum numbers of the orbitals from which and also to which the shift is being made is offered by the Rydberg equation

#1/(lamda_"e") = R * (1/n_1^2 - 1/n_2^2)#

Here

#lamda_"e"# is the wavelength of the emitted photon (in a vacuum)#R# is the Rydberg constant, equal to #1.097 * 10^(7)# #"m"^(-1)##n_1# represents the major quantum number of the orbital that is lower in energy#n_2# represents the principal quantum number of the orbital that is better in energy

Now, you know that your electron emits an electron, so appropriate from the start, you understand that the major quantum number of the orbital that is higher in energy will certainly be #n_2 = 5#.

In other words, the principal quantum variety of the final orbital in this shift must be # because a photon is being emitted, not absorbed. So, rearselection the Rydberg equation to isolate #n_1#

#1/(lamda_ "e") = R/n_1^2 - R/n_2^2#

#R/n_1^2 = 1/(lamda_ "e") + R/n_2^2#

#R/n_1^2 = (n_2^2 + lamda_ "e" * R)/(lamda_ "e" * n_2^2) means n_1 = sqrt((R * lamda_ "e" * n_2^2)/(n_2^2 + lamda_ "e" * R))#

Plug in your values to find

#n_1 = sqrt(( 1.097 * color(blue)(cancel(color(black)(10^7))) color(red)(cancel(color(black)("m"^(-1)))) * 4.348 * color(blue)(cancel(color(black)(10^(-7))))color(red)(cancel(color(black)("m"))) * 5^2)/(5^2 + 4.348 * color(blue)(cancel(color(black)(10^(-7))))color(red)(cancel(color(black)("m"))) * 1.097 * color(blue)(cancel(color(black)(10^7)))color(red)(cancel(color(black)("m"^(-1))))))#

#n_1 = 2.001384 ~~ 2#

Therefore, you can say that your electron is undergoing a #n=5 -> n= 2# transition. This shift is located in the visible component of the EM spectrum and also is component of the Balmer series.