# Question #db4d1

##### 1 Answer

#### Explanation:

The **Rydberg equation** allows you to calculate the *wavelength*, **in meters**, of the photon emitted when an electron in a hydrogen atom makes a transition from an **initial energy level** **final energy level**

#1/(lamda) = R * (1/n_f^2 - 1/n_i^2)#

Here

#R# is theRydberg constant, equal to#1.097 * 10^(7)# #"m"^(-1)#

To find the wavelength of the emitted photon, rearrange the Rydberg equation to solve for

#lamda = (n_i^2 * n_f^2)/(n_i^2 - n_f^2) * 1/R#

In your case, the electron is going from

#n_i = 6#

to

#n_f = 2#

a transition that is part of the **Balmer series**, so you should expect the wavelength of the photon to correspond to the **visible part** of the EM spectrum.

Plug in your values to find

#lamda = (6^2 * 2^2)/(6^2 - 2^2) * 1/(1.097 * 10^(7)color(white)(.)"m"^(-1))#

#lamda = 4.10 * 10^(-7)color(white)(.)"m"#

To convert this to *nanometers*, use the fact that

#"1 m" = 10^9# #"nm"#

You will end up with

#lamda = 4.10 * 10^(-7)color(red)(cancel(color(black)("m"))) * (10^9color(white)(.)"nm")/(1color(red)(cancel(color(black)("m")))) = color(darkgreen)(ul(color(black)("410 nm")))#

I'll leave the answer rounded to three **sig figs**.

As you can see, this transition is indeed located in the visible part of the EM spectrum *purple*.