## Balmer vs Lyman Series [ENDORSED]

$c=\lambda v$

Rachel N 1I
Posts: 48
Joined: Thu Jul 27, 2017 3:00 am

### Balmer vs Lyman Series

I was reading through one of the examples in the textbook about Balmer and Lyman series and it stated that because n1=2, the wavelength should match one of the lines in the Balmer series. How are you able to tell which region of the spectrum indicates which series to use? Also why is n1=2 Balmer series and n1=1 Lyman series?

mayapartha_1D
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

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

Hey!

The Lyman series corresponds to n=1 because it refers to wavelengths that start at a given energy level and end at 1 (2-1, 1-1, 3-1). The same goes for the Balmer series but with n=2. As for knowing which series to use, the problem will normally state which series is most appropriate!

Hope this helped.

Ashin_Jose_1H
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

### Re: Balmer vs Lyman Series

The Balmer Series corresponds to the visible light section of the spectrum and the Lyman Series corresponds to the Ultraviolet section of the spectrum.

Lyman Series: When an electron travels from any energy level to the energy level of 1 (i.e, n=3 to n=1, n=10 to n=1), the electron emits energy with wavelengths found in the ultraviolet section.

Balmer Series: When an electron travels from any energy level to the energy level of 2 (i.e, n=3 to n=2, n=10 to n=1), the electron emits energy with wavelengths found in the visible light section.

aaron tang 2K
Posts: 49
Joined: Thu Jul 27, 2017 3:01 am

### Re: Balmer vs Lyman Series

Balmer Series (656.3 nm, 486.1 nm, 434.8 nm, 410.2 nm) is connected to visible light section of the spectrum, while Lyman Series (121.6 nm, 102.6 nm, 97.3 nm) is connected to the ultraviolet section of the spectrum.
The Lyman series is indicated by when the electron is excited from the ground state and jumps from n=1 to n=2. The Balmer series is indicated by when the electron jumps from the ground state n=1 to n=3, etc. The energy emitted will be dependent on where the wavelengths are found.