Atomic Spectra

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Rose_Malki_3G
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Atomic Spectra

Postby Rose_Malki_3G » Fri Oct 16, 2020 3:48 pm

For the equation V = -R((1/n^2)-(1/n^2))
How do we know which n value (the final or initial n value) to put first in the equation?

Vivian Chang 3L
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Re: Atomic Spectra

Postby Vivian Chang 3L » Fri Oct 16, 2020 3:55 pm

Hi!

You can manipulate the equation you mentioned to get v(frequency) = RH (1/n2^2 - 1/n1^2), where RH is Rydberg's constant, 3.29 x 10^15 s^/1.

In this equation, n1 is always the higher energy level and n2 is the lower energy level.

Anh Trinh 1J
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Re: Atomic Spectra

Postby Anh Trinh 1J » Fri Oct 16, 2020 3:59 pm

The first n in the equation would be the final energy level. Dr. Lavelle said in today’s lecture to think of it as the change in energy = final energy - initial energy. Usually in a question it would describe the transition that the electron makes. Ex: “an electron makes a transition from the fourth to the second principal quantum level,” the fourth energy level would be the initial and the second energy level would be the final.

LeanneBagood_2F
Posts: 96
Joined: Wed Sep 30, 2020 9:32 pm

Re: Atomic Spectra

Postby LeanneBagood_2F » Fri Oct 16, 2020 9:52 pm

Rose_Malki_3L wrote:For the equation V = -R((1/n^2)-(1/n^2))
How do we know which n value (the final or initial n value) to put first in the equation?

Some of our classmates already responded with this but the difference of something is always going to be found by calculating final - initial.
Because of this, the first n in the equation should be n-final whereas the second should be n-initial. Here's an image of Rydberg's equation that I got from Google if my explanation does not make sense! Image

little disclaimer: i got this image from a quick Google search of "rydberg's equation" so it does not exactly align with the equation that was originally being asked about but it gives the gist of it

AndrewNguyen_2H
Posts: 95
Joined: Wed Sep 30, 2020 9:59 pm

Re: Atomic Spectra

Postby AndrewNguyen_2H » Sat Oct 17, 2020 9:19 pm

Vivian Chang 2H wrote:Hi!

You can manipulate the equation you mentioned to get v(frequency) = RH (1/n2^2 - 1/n1^2), where RH is Rydberg's constant, 3.29 x 10^15 s^/1.

In this equation, n1 is always the higher energy level and n2 is the lower energy level.


Isn't n1 the initial??

Justin Nguyen 3E
Posts: 97
Joined: Wed Sep 30, 2020 9:35 pm

Re: Atomic Spectra

Postby Justin Nguyen 3E » Sat Oct 17, 2020 10:47 pm

I think n1 is the initial and also the higher energy level because the electrons emit energy when an excited electron (higher energy level) drops back down to its ground state.


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