## solving 1A #15

Kaylee Nezwek_2G
Posts: 30
Joined: Wed Sep 30, 2020 10:02 pm

### solving 1A #15

The question asks "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 attempted to solve by first converting nanometers to meters, getting 1.026*10^-7 m. I believe this falls in the visible light range, making it apart of the Balmer series. However, im not sure what to do from here... I think you would use the Rydberg equation but kinda confused on what numbers to plug in where. Anyone know if this is correct so far and what to do next?

Melody Haratian 3K
Posts: 54
Joined: Wed Sep 30, 2020 9:48 pm
Been upvoted: 1 time

### Re: solving 1A #15

Hello! The line will fall into the UV radiation spectrum making it a part of the Lyman series.The wavelengths of the Balmer series are much larger ( around 400-700 nm). What you should do next is find the frequency of atomic hydrogen by plugging in the wavelength into the c = λν. After finding the frequency, you can use the Rydberg equation.
You’ll be solving for n2 in the equation, and you know your n1=1 since the atomic hydrogen is part of the Lyman series. Rearrange the Rydberg equation to have it be:
1/n22 = 1/n12 -ν/R

We know that R = 3.29 x 1015, n1=1 and you can find the frequency from the speed of light equation. From plugging in those values, you’ll find the value of n2, giving you the final energy level of the electron.

Kaylee Nezwek_2G
Posts: 30
Joined: Wed Sep 30, 2020 10:02 pm

### Re: solving 1A #15

Thank you so much!!