The an initial thing the you need to do here is to usage the frequency that the emitted photon to calculate its wavelength.

As you know, frequency and also wavelength have an inverse relationship described by the equation

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

Here

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

Plug in your worth 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 relationship between the wavelength of the emitted photon and the principal quantum numbers of the orbitals from which and also to i beg your pardon 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 principal quantum number the the orbital the is lower in energy#n_2# to represent the principal quantum number that the orbital that is higher in energy

Now, you know that her electron emits an electron, so best from the start, you recognize that the primary quantum variety of the orbital the is higher in power will be #n_2 = 5#.

In various other words, the primary quantum variety of the last orbital in this transition must it is in # due to the fact that a photon is being emitted, no absorbed.

*

So, rearrange the Rydberg equation to isolation #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) indicates 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 experience a #n=5 -> n= 2# transition. This change is located in the visible part of the EM spectrum and also is part of the Balmer series.