# Question #5dc82

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The first thing that you need to do here is to calculate the **wavelength** of the photon emitted when your electron undergoes a

The wavelength of the photon will then help you determine its **energy**.

So, you know that when an electron in a hydrogen atom makes a transition from an **initial energy level** **final energy level** **wavelength** is equal to

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

Here

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

You can rearrange the above equation, which is called the **Rydberg equation**, as

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

which gets you

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

In your case, you have

#lamda = 1/(1.097 * 10^7color(white)(.)"m"^(-1)) * (5^2 * 2^2)/(5^2 - 2^2)#

#lamda = 4.34 * 10^(-7)color(white)(.)"m"#

Expressed in *nanometers*--recall that

#lamda = color(darkgreen)(ul(color(black)("434 nm")))#

Now, in order to find the **energy** of the photon, you need to use a variation of the **Planck - Einstein relation**.

#E = h * c/(lamda)#

Here

#E# is theenergyof the photon#h# isPlanck's constant, equal to#6.626 * 10^(-34)# #"J s"# #c# is thespeed of lightin a vacuum, usually given as#3 * 10^8# #"m s"^(-1)#

Plug in your value to find

#E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * (3 * 10^8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(4.34 * 10^(-7) color(red)(cancel(color(black)("m"))))#

#E = color(darkgreen)(ul(color(black)(4.58 * 10^(-19)color(white)(.)"J")))#

I'll leave both answers rounded to three **sig figs**.

It's worth mentioning that the **Balmer series**, is located in the visible portion of the EM spectrum.