Workshop Question

[SophiaJenny3I]

Hi! Could someone please explain this problem, I know it's Lyman series but I'm struggling to figure out how you know if n=1 is initial or final.
Light with a wavelength of 102.557 nm excites a hydrogen atom gas sample. Is the change in energy of a hydrogen atom positive or negative when it absorbs a photon? What is the principal quantum number of the state the electron was excited to? I understand the first part of the question, but the n initial vs n final is what's getting me. Thanks in advance!

[Vivian Chang 3L]

Since it says the light "excites" the hydrogen atom, we know the electron is going from a lower energy level to a higher energy level. Because 102.557 nm is in the UV light range, it is associated with the Lyman series and you know that one of the energy levels must be n=1. This problem involves an electron going from a lower energy level to a higher energy level, so n initial must be 1. This means you need to find n final, which would be the energy level the electron was excited to.

For how to solve this problem, you can first use deltaE=(hc)/lambda to find the change in energy. You should get about 1.94 x 10^-19 J/photon. Then you can plug this into the equation deltaE=(-hR²)/n_final² - (-hR²)/n_initial². You would plug in 1 for n_initial.

For your answer, you should get n_final= 3.

Hope that helps!

[Annie Tong 2G]

If the problem says the atom is "excited", that means it's going to a higher energy level. So the change in energy for the atom is positive because it's absorbed. Because you already know it is part of the Lyman series, the initial energy level is n =1. You can find the frequency from the given wavelength and substitute it into frequency v = R(1/ initial n^2 - 1/final n^2) to find nfinal. n initial would be 1. The answer should be nfinal = 3.

Hope this helped!