Rydberg formula
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Rydberg formula
How do you solve this problem: "In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. 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'm not really sure where to start at, thank you!
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Re: Rydberg formula
I think you would start by using the formula "(hc)/wavelength = change in energy" to find the change in the energy since the wavelength is given to you and h and c are constants. Then you would use Rydberg's formula.
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Re: Rydberg formula
If you didn’t know the Rydberg formula is in
En = -hR/n^2 or frequency= -R( 1/n1^2 - 1/n2^2)
En = -hR/n^2 or frequency= -R( 1/n1^2 - 1/n2^2)
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Re: Rydberg formula
The Rydberg formula is just a simplified version of the formula, ΔE=[(-hR)/(N22)-(-hR)/(N12)]=hν. Cancel out and substitute some terms and you should end up with c/(Rλ)=1/n12-1/n22, and it should be pretty clear what the two values of n are.
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Re: Rydberg formula
Do we need to solve problems with the Rydberg formula? I remember that there is one equation we are not supposed to use in practice problems.
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Re: Rydberg formula
I am also confused about the Rydberg formula and which one we are supposed to use. Also - is the Rydberg formula only for Hydrogen?
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Re: Rydberg formula
I'm also confused. I thought Lavelle said we didn't have to know how to solve questions with Schroedinger's equation. Is it rydberg or schroedinger?
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Re: Rydberg formula
Rydberg formula can be used for other elements I believe, giving us a spectral line for any element. My question is, are those spectral lines of other elements than hydrogen visible to the human eye?
Re: Rydberg formula
It's also important to note that N1 is going to be 1 because the element used is hydrogen. That should clear up any confusion as to missing variables.
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