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**.