## 6th edition 1.15

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

CaminaB_1D
Posts: 63
Joined: Fri Sep 28, 2018 12:16 am

### 6th edition 1.15

I am having trouble figuring this out:

1.15 says "In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line."

I understand we are given the wavelength, but I am not sure where to start or what equation to use. I calculated the change in energy needed to emit that wavelength, not sure if that is in the right direction.

Chem_Mod
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### Re: 6th edition 1.15

That is a good direction, now you can use the Rydberg formula, and the energy you calculated will be E.

ryanhon2H
Posts: 60
Joined: Fri Sep 28, 2018 12:28 am

### Re: 6th edition 1.15

The way I did it was using the Rydberg equation, which is

1/wavelength (in meters) = R(1/n12 - 1/n22)

The wavelength is given, and R is the Rydberg constant. Since this is in the ultraviolet spectrum of hydrogen, it is the Lyman series, where n1 is 1. Now all you need to do is plug in and solve for n2.