## 1A.15

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Bailey Giovanoli 1L
Posts: 107
Joined: Wed Sep 30, 2020 9:50 pm

### 1A.15

On number 15 in section A of the work book problems, you are asked to find the initial and final energy levels while given only the wavelength of 102.6nm. So, you would use the Rydberg formula, v=R[1/n(final)-1/n(initial)], and then change the v, frequency to wavelength over speed of light. Even though you know the wavelength, R, and the speed of light, there are still to variables you do not know, n(final) and n(initial). The back of the book has the n(initial) as 1. How would I know this? Do I have to know something about the energy?

Steph Du 1H
Posts: 93
Joined: Wed Sep 30, 2020 9:36 pm
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### Re: 1A.15

The question says it is in the ultraviolet spectrum. Thus, we know it is the Lyman series (Lyman series emits ultraviolet light), and the Lyman series refers to electrons that drop to the ground state, n=1. So then you would use n=1 as your final energy level.

Posts: 159
Joined: Wed Sep 30, 2020 9:46 pm
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### Re: 1A.15

Just remember that the Balmer series consists of lines within the visible region and that n1=2 (and n2=3,4,...). While the Lyman series consists of lines within the Ultraviolet region of the spectrum and that n1=1 (and n2=2,3,...)

Hope this helps!

Bailey Giovanoli 1L
Posts: 107
Joined: Wed Sep 30, 2020 9:50 pm

### Re: 1A.15

Ohhh, thank you so much! I knew I was missing a key concept. I get it now.