solving 1A #15

[Kaylee Nezwek_2G]

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 1B]

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]

Thank you so much!!