1A.15 Practice Problem

[Diego Salgues 2K]

"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 get we convert the wavelength to a frequency and then set up the Rydberg Equation. However, there are two variables we need to solve for...n1 and n2. How do we find one shell's n value? Is there something I'm not seeing?

Thank you,
D

[Anna Dai-Liu 3B]

Since the emitted light is in the UV region, that tells us that it's part of the Lyman series in the H emission spectrum. All of the lines in the Lyman series involve transitions from a higher energy level to the level n = 1, meaning that one of your energy levels must be 1. Once you find the energy level of n = 1, you can use that to calculate the other level (since the energy of the emitted light corresponds to the amount of energy lost).

[Nathan Sigel 2k]

I was confused on this too. It turns out the solution is pretty simple: The way we know the n1 value is by determining if the wavelength is in the Balmer or Lyman series for hydrogen. For UV light we say n1 = 1. For visible light, we say n1=2. I hope this helps.

[Qinyan Feng 1H]

I was confused why for UV light, we say n1 = 1? When I was doing this question, I just simply found two integers that can satisfy the calculated value of (1/n1 - 1/n2), so I think I missed some points.

[Travis Wang 2G]

I think to fully solve the problem, you do need to work backwards from the Rydberg equation and start punching in guesses into a calculator. When we solve for 1/(n1)^2-1/(n2)^2 using the known wavelength, we know that the difference of the n-squareds needs to be around 0.88. The only step left is to find the n1 and n2 that subtract to 0.88 and they happen to be reasonable numbers. As already noted, because we are in the UV region for the emissions of a hydrogen atom (Lyman series), n1 is automatically assigned a number, in this case, 1.