The first thing the you need to do below is to usage the frequency the the emitted photon to calculation its wavelength.

As you know, frequency and also wavelength have actually an inverse relationship explained 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 offered 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 relationship between the wavelength the the emitted photon and also the principal quantum numbers that the orbitals indigenous which and also to which the change is gift made is provided 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 that is lower in energy#n_2# represents the principal quantum number the the orbital that is higher in energy

Now, you understand that her electron emits one electron, so appropriate from the start, you understand that the principal quantum number of the orbital that is higher in power will it is in #n_2 = 5#.

In other words, the principal quantum number of the final orbital in this shift must be # because a photon is being emitted, no absorbed.

So, rearrange 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) suggests 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 deserve to say that your electron is undergoing a #n=5 -> n= 2# transition. This shift is situated in the visible part of the EM spectrum and is component of the Balmer series.