## Lyman Series

Rayna Irving 3H
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Joined: Wed Sep 30, 2020 9:33 pm

### Lyman Series

If the Lyman Series is ever mentioned in a problem should we just assume E(final) is 1?

Giselle Granda 2
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### Re: Lyman Series

Yes, the ground state for the Lyman series is always 1, so your n1 or nfinal is equal to 1.

David Chibukhchian 1H
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### Re: Lyman Series

Yeah, we should assume that the final energy state in the Lyman series is n=1. This is because the jump from n=1 to other energy levels requires a lot of energy, specifically from light in the UV region (which is where the Lyman series is found). Since the Balmer series is visible light and thus can't really provide enough energy to bring the electron up from n=1, we should make this assumption about the Lyman series.

sophie esherick 2K
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### Re: Lyman Series

Yes like the other peers mentioned we should assume that the final energy state or your n1 in the Rydberg Eqn. Lyman series is n=1. This is because when you working with in the Lyman series, you are talking about the UV region. This area is higher energy than other series, for example n=2 which is the Balmer series and visible light region.

Anna Yakura 3E
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Joined: Wed Sep 30, 2020 9:46 pm

### Re: Lyman Series

From what some of the TAs said, just assume n=1 is somewhere in the equation. I think whether its nfinal or ninitial depends on whether its an emission or absorbtion. If emission, since it is going back to the lower state, n=1 would be the final. If absorbtion, because the electron is being excited up to an energy level, n=1 would be the initial. Hope this helps!

Alara Aygen 2E
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### Re: Lyman Series

Hi! If it is Lyman series, n=1 should be in the equation. However, don't assume it only has to be the final state. If the e- is absorbing energy than n=1 is probably the initial state. Similarly, if e- is emitting energy than n=1 is probably the final state. Hope this helps!

Alara Aygen 2E
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### Re: Lyman Series

Hi! If it is Lyman series, n=1 should be in the equation. However, don't assume it only has to be the final state. If the e- is absorbing energy than n=1 is probably the initial state. Similarly, if e- is emitting energy than n=1 is probably the final state. Hope this helps!

Katie Le 2K
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### Re: Lyman Series

Yes Lyman ends with n=1. Balmer ends with n=2

Haochen He 3L
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### Re: Lyman Series

Yes, if it is Lyman series n1 should be 1

Jared Limqueco 1H
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### Re: Lyman Series

Lyman series is always n=1 but not always n(final)=1. It's UV light too

Daniela_Martinez_3B
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### Re: Lyman Series

Yes! Remember that anything in the UV region involves n=1, anything in the visible region involves n=2, and anything in the infrared region involves n=3.

Jalaia Jackson 2J
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Joined: Wed Sep 30, 2020 9:45 pm

### Re: Lyman Series

This is how I remembered it: The Lyman series corresponds to UV light and the electrons rest at the energy level n=1. While the Balmer series corresponds to visible light and the electron's rest at the energy level n=2.

Posts: 42
Joined: Wed Sep 30, 2020 9:31 pm

### Re: Lyman Series

You would only assume this if the it was an emission spectrum problem. If it was an absorption problem, you would assume that n=1 is the initial.

Immi Lee - 2F
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Joined: Wed Sep 30, 2020 9:58 pm

### Re: Lyman Series

I would say you cannot necessarily assume that E(final) = 1, because depending on the question they may be asking based on the absorption or emission spectrum, but definitely one of the energy levels is n=1!

sabrina ghalambor 1E
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### Re: Lyman Series

yes it should end in n=1! Lyman corresponds to UV light so it has a higher energy level

Andre Fabian 1G
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### Re: Lyman Series

Yes, the Lyman series always ends in n=1 (ground state), while the Balmer series always ends in n=2.

Hope this helps!
Andre Fabian

Nathan Chu 1J
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Joined: Wed Sep 30, 2020 9:48 pm

### Re: Lyman Series

Yes, the Lyman series involves an an electron moving down to the n=1 energy level or up from the n=1 level. However, you cannot assume that the value is for n(final), as the problem could be describing emission or absorption.

Julianna_flores3L
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### Re: Lyman Series

The Lyman series is the UV so n1=1 while the baller series is in visible light so n1=2.

Lauren Strickland 2a
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Joined: Wed Sep 30, 2020 9:47 pm

### Re: Lyman Series

Yes for the Lyman series n1=1

Gian Boco 3G
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Joined: Wed Sep 30, 2020 9:36 pm

### Re: Lyman Series

Yes, but for future references, keep in mind that this is for Hydrogen atom only.

Ralph Zhang 2L
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### Re: Lyman Series

If you click on the hints or solutions(if you already did it) sections of sapling it should tell you about it.

AlbertGu_3G
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### Re: Lyman Series

Adding on to other answers, the Balmer series would be n=2 and the Paschen series would be n=3. Hope this helps!

Grace_Remphrey_2G
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Joined: Wed Sep 30, 2020 9:54 pm

### Re: Lyman Series

If the problem mentions the Lyman Series, there should be an n=1 somewhere in the equation. The Lyman series is in the ultraviolet while the Balmer series is in the visible and the Paschen, Brackett, Pfund, and Humphreys series are in the infrared.
Hope this helped!

Aria Movassaghi 2k
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### Re: Lyman Series

Since Lyman Series is talking about Uv region, you should assume n1=1

305572629
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### Re: Lyman Series

Always remember: Lyman series N1 is 1 and Balmer series N1 is 2.

Ethan Goode 3H
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### Re: Lyman Series

Yes, lyman series has much more energy. As the gap from n2 to n1 is the largest gap of energy, lyman will always end up at n1 since it has the most energy.

America Ramirez 3H
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### Re: Lyman Series

Yes, Lyman =1 and Balmer =2.

Jacob Schwarz
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### Re: Lyman Series

Yes, if they Lyman series is mentioned, you should assume n=1 as the final state.

Michelle Nguyen 2C
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### Re: Lyman Series

Yes, with the Lyman series, just assume that the final stage is n=1!

Karina Grover 1K
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### Re: Lyman Series

In Lyman series, the final principal quantum number/energy level is always n = 1.

DPatel_2L
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Joined: Wed Sep 30, 2020 9:41 pm

### Re: Lyman Series

Yes for that series n1=1