## Balmer and Lyman Series [ENDORSED]

Paola_RubioRod_1F
Posts: 18
Joined: Fri Jan 08, 2016 3:00 am

### Balmer and Lyman Series

This was a homework question for chapter 1. What is common to the lines within a series that makes grouping them together logical? (Balmer Series, Lyman Series, Paschen Series)

Jenna_Hakel_2A
Posts: 34
Joined: Fri Jul 15, 2016 3:00 am

### Re: Balmer and Lyman Series  [ENDORSED]

We talked about this in my discussion yesterday and my TA told us that for each series grouped together, the energy levels where the electrons start is always the same. For example, for the Lyman series, the electrons always either start at or return to (depending on if the light is absorbed or emitted) the energy level n=1. For the Balmer series, the electrons always start at or return to n=2. The other series in that question are the same with n=3 and n=4, respectively. So their principal quantum levels are always the same for each series!

Joseph Nguyen 3L
Posts: 24
Joined: Fri Jul 22, 2016 3:00 am

### Re: Balmer and Lyman Series

This would be important in other problems where they give you the energy difference or only the wavelength and ask you to find the n positions of both energy levels. For example, if they only gave you a wavelength of light of 500nm, you would know at least it is visible light, and therefore a Balmer series. You would then know, depending on whether energy is absorbed or emitted, that the base energy level is n=2. Now you can isolate the other n and solve for it to be able to solve for it.

104822659
Posts: 23
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Balmer and Lyman Series

Hi, I'm having trouble understanding what principle quantum level is and is there an equation where the n can be computed from?

Katherine_Zhuo_3B
Posts: 23
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Balmer and Lyman Series

The principal quantum number is an integer that describes the energy level of the electron. As n increases, the number of electronic shells increases and the electron is further from the nucleus and therefore has a higher energy. To compute n, we can utilize Rydberg's equation given the energy: En=-hR/n^2.