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**.