Help with Homework 1A.15

[Ellis Song 4I]

1A.15 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.

Reading this question, I have no idea where to begin. Is the question asking at which energy level the initial and final energy level are? How would you do this and how do you know which equations to use?

[Ada Chung 1C]

Here is how I would approach the problem and kind of my thought process
1. Because the problem tells you that this is in the UV spectrum you would be able to know that n1=1 (from the Lyman series)
2. Since you are given the wavelength you can use wavelength x frequency=speed of light to find the frequency. Once you have found frequency you can then find and solve for n1 by plugging in the frequency you go into Rydberg equation which is v=R((1/1^2)-(1/n2^2)) now solve for n2

[Alex Chen 2L]

The line wavelength is your starting point. Using that, you should be able to find the frequency of that line of light, so you can use the equation E=hv to find the energy that the electron emitted as a photon. Furthermore, remember that because UV light has the highest energy of light, the electron has to make the largest jump in energy levels, meaning the final energy level will be n = 1. After that, you can use the emperical equation En = -hR/n^2 to determine the energy level of n = 1 and add that to the energy of the line of light (from E=hv) in order to solve for the intial energy level n through the emperical equation again.

[Hannah Lee 2F]

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.

First, you would find the change in energy (ΔE). The equations we know are: c = λv and E = hv. Since the wavelength is given, you could find E via E = hc / λ.
Therefore: E = (6.626 x 10^-34 J s)(2.998 x 10⁸ m/s) / (102.6 x 10^-9 m) = 1.936 x 10^-18 J
It's important to realize that energy is being emitted, which means the e- is transitioning from a higher to lower energy level, so ΔE < 0. (The energy of electrons can be either positive or negative, but the energy of photons emitted/absorbed must always be positive.) Therefore, ΔE = -1.936 x 10-¹⁸ J.

The problem indicates that the line is observed in the UV spectrum, so it deals with the Lyman series, which ends at n = 1 (n_final = 1).
E_final = -hR / n_final² = -(6.626 x 10^-34 J s)(3.28984 x 10¹⁵ Hz) / 1² = -2.180 x 10^-18 J

We can then plug in the values we found into the equation ΔE = E_final - E_initial. So:
-1.936 x 10^-18 J = -2.180 x 10^-18 J - E_initial
E_initial = -2.180 x 10^-18 J - (-1.936 x 10^-18 J)
E_initial = -2.438 x 10^-19 J = -hR / n_initial²
n_initial² = -hR / -2.438 x 10^-19 J = - (6.626 x 10^-34 J s)(3.28984 x 10¹⁵ Hz) / (-2.438 x 10^-19 J)
n_initial = 3, so the e- transitions down from the energy level n = 3 to n = 1.