Chemistry- Rydberg Equation

Reply Thu 6 May, 2010 11:20 am
I am having trouble with this whole thing please help if you have the time!


Use the Rydberg formula for hydrogen...

1.) Rearrange to resemble the equation of a line: y=mx+b
2.) What is y? What is m? What is x?
3.) What is the best value for n(sub f)
4.) What is R(sub H)
5.) For this series, what is the limiting wavelength (the shortest wavelength that could possibly be observed)??
6.) Use Table one below..... Complete the analysis of the deuterium emission lines (i.e. answer questions 3-5 above).
7.) Compare the value of R(sub D) with the value of R(sub H) that you found earlier. How does this ratio compare with the theoretical ratio predicted by Bohr's Theory?

Table 1:

Line number, n.) Wavelength, nm
1.) 656.1
2.) 486.0
3.) 433.9
4.) 410.0
5.) 396.9
6.) 388.8
7.) 383.4
8.) 379.7
9.) 377.0
10.) 375.0
11.) 373.4
12.) 372.1
13.) 371.0

Even if you can help with some of the question it would be greatly appreciated!

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High Seas
Reply Thu 6 May, 2010 07:48 pm
Wils, try entering likely numbers in the first 3 spaces here >
> atomic number, quantum number of initial state, quantum number of final state,
and then plot the results you get from the online calculations. You'll get a feeling for it fairly quickly. Example, leave atomic number and final state as they come up on the first screen but enter final state as 5. See dark blue photon come up in the calculation (further down on same page), it's 433.9nm wavelength.
High Seas
Reply Thu 6 May, 2010 08:00 pm
@High Seas,
To get a red photon, enter final state 3 (leaving initial state at 2 and atomic number at 1), that's a wavelength of 656.1 nm.
High Seas
Reply Fri 7 May, 2010 02:24 am
@High Seas,
The red photon (from 3=final, 2=initial, 1=atomic number) belongs to the Balmer series, which has these visible spectral lines for hydrogen.
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