## Problem 1.15 Textbook

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Gisselle Sainz 2F
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### Problem 1.15 Textbook

For this problem would we consider the ultraviolet's energy level of n=1 (Lyman's series) as the final energy level (n, final) or as the initial energy level (n, initial)?

Chem_Mod
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### Re: Problem 1.15 Textbook

To emit a photon, the electron in the H-atom loses energy. So it falls to a lower energy. nfinal = 1 for Lyman series.

Amanda 1A
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### Re: Problem 1.15 Textbook

in the solutions manual for the textbook, it says the transition is n1=1 to n2=3. If 1 is considered n(final) since it's an emission of energy, do we have to say the transition is from n1=3 to n1=1 in order to get the answer correct? or is it still correct to answer it like the solution manual?

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### Re: Problem 1.15 Textbook

Conservation of energy when emitting a photon:
Etotal,initial = Etotal,final
Eninitial = Ephoton + Enfinal
-Rh/ninitial2 = hv + -Rh/nfinal2

isolate hv and divide by h
v = -R(1/ninitial2 - 1/nfinal2 ) = R(1/nfinal2 - 1/ninitial2 )

Jack Dias
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### Re: Problem 1.15 Textbook

Yes, you will want to phrase it as "Initial=3, final=1". This is because the energy is emitted, which only happens when the energy level drops, in this case, 3 to 1. The energy of course is released from the atom and then observed. Make sure you have the energy as negative for the atom (because it loses energy) and positive for energy emitted (the surrounding system/area gains energy).

Best,
Jack Dias

Kara Justeson 1B
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### Re: Problem 1.15 Textbook

for reference the problem says:

in the uv spectrum of atomic hydrogen, a line is observed at 102.6nm. 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 understand how you find n=3 as being one of the energy levels if you know n=1 is also one, but how do you know that n=1 is one of the energy levels?

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### Re: Problem 1.15 Textbook

UV implies Lyman series, meaning nfinal= 1