Rydberg Equation and its Fundamental Equation

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Farah Abumeri 3F
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Rydberg Equation and its Fundamental Equation

Postby Farah Abumeri 3F » Thu Oct 22, 2020 10:35 am

Dr. Lavelle prefers that we use E=(-hR)/n^2 rather than the frequency=R(1/(n1)^2 - 1/(n2)^2) because it's a better conceptual application. On the midterm, are we allowed to use the Rydberg Equation rather than the fundamental equation, because I've notice it's much quicker to attain an answer that way? If so, does n1=n final and n2= n initial? I keep mixing them up and I'm becoming a little confused. If someone could explain, that would be awesome!

Crystal Pan 2G
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Re: Rydberg Equation and its Fundamental Equation

Postby Crystal Pan 2G » Thu Oct 22, 2020 10:39 am

Dr. Lavelle wants us to use precisely because it is easy to mix up the n1 and n2. In the equation he gives us, it is E=nfinal- ninitial.

Marisa Gaitan 2D
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Re: Rydberg Equation and its Fundamental Equation

Postby Marisa Gaitan 2D » Thu Oct 22, 2020 10:39 am

I'm sure we would be able to use either equation. My TA said our scratchwork is not going to be graded, so as long as you get the right answer you will get credit. Also, we did a problem on finding n energy levels in our discussion and my TA used the rydberg equation, so I think it's fine to use.

Chem_Mod
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Re: Rydberg Equation and its Fundamental Equation

Postby Chem_Mod » Thu Oct 22, 2020 10:40 am

The Rydberg equation is really just a derivation of E=(-hR)/n^2. Since the midterm is multiple choice and you are not getting points for your work, either is an acceptable way to solve the problem, and you can decide which is easiest for you to use. N1 will equal your final energy level, and N2 will equal your initial energy level, in order for the equation to make sense. For example, if the transition is from n=4 to n=1, you have (1/1^2)-(1/4^2). The opposite way would result in a negative number.

Kayla Law 2D
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Re: Rydberg Equation and its Fundamental Equation

Postby Kayla Law 2D » Thu Oct 22, 2020 10:43 am

I don't think you're required to show work for the midterm since all of the questions are multiple choice, so you can probably use the Rydberg Equation (but make sure you know the E=(-hR)/n^2 equation too!). I believe that n1 is the n final and n2 is the n initial. I kept mixing them up too lol.

Justin Zhang_1A
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Re: Rydberg Equation and its Fundamental Equation

Postby Justin Zhang_1A » Thu Oct 22, 2020 10:44 am

You can use either equation since the test will be multiple choice and your work won't be graded. However, Dr. Lavelle said there could be some questions where E=(-hR)/n^2 would work better.

VincentLe_3A
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Re: Rydberg Equation and its Fundamental Equation

Postby VincentLe_3A » Thu Oct 22, 2020 10:46 am

I think either equation is acceptable to use on the midterm, and especially since the midterm is timed, it would make sense to use the Rydberg equation frequency=R(1/(n1)^2 - 1/(n2)^2) in order to save time. As long as you understand conceptually where that equation comes from, the Rydberg equation essentially comes from the empirical formula E=(-hR)/n^2. It is just important to pay attention to the energy levels. From the formula on the equations sheet, n1 would be the final energy level and n2 would be the starting energy level.

Farah Abumeri 3F
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Re: Rydberg Equation and its Fundamental Equation

Postby Farah Abumeri 3F » Thu Oct 22, 2020 1:48 pm

Great, that makes a lot of sense! Thank you all!

Kimiya Aframian IB
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Re: Rydberg Equation and its Fundamental Equation

Postby Kimiya Aframian IB » Thu Oct 22, 2020 1:49 pm

Farah Abumeri 1B wrote:Dr. Lavelle prefers that we use E=(-hR)/n^2 rather than the frequency=R(1/(n1)^2 - 1/(n2)^2) because it's a better conceptual application. On the midterm, are we allowed to use the Rydberg Equation rather than the fundamental equation, because I've notice it's much quicker to attain an answer that way? If so, does n1=n final and n2= n initial? I keep mixing them up and I'm becoming a little confused. If someone could explain, that would be awesome!

Hi! Yes, Dr. Lavelle said that we could use whichever one that we prefer, and in the term of the one where it is equal to frequency, it is final minus initial.


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