# An excited hydrogen atom with an electron in the n = 5 state emits light having a frequency of #6.90 xx 10^14 s^-1#. What is the principal quantum level for the final state in this electronic transition?

##### 1 Answer

#### Explanation:

The first thing that you need to do here is to use the frequency of the emitted photon to calculate its **wavelength**.

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

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

Here

#nu# is the frequency of the photon#c# is the speed of light in a vacuum, usually 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 relationship between the wavelength of the emitted photon and the **principal quantum numbers** of the orbitals from which and to which the transition is being made is given by the **Rydberg equation**

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

Here

#lamda_"e"# is thewavelengthof the emitted photon (in a vacuum)#R# is theRydberg constant, equal to#1.097 * 10^(7)# #"m"^(-1)# #n_1# represents theprincipal quantum numberof the orbital that islower in energy#n_2# represents theprincipal quantum numberof the orbital that ishigher in energy

Now, you know that your electron **emits** an electron, so right from the start, you know that the principal quantum number of the orbital that is *higher* in energy will be

In other words, the principal quantum number of the final orbital in this transition **must** be *emitted*, not absorbed.

So, rearrange the Rydberg equation to isolate

#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) implies 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 **Balmer series**.