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Atomic}Rydberg Energy

\frac{1}{\lambda}=R\left( \frac{1}{n_f^2} - \frac{1}{n_0^2} \right). Given the information that the Lyman Series is n_f=1, and the Balmer series is n_f=2, one forms the ratio \lambda_L/\lambda_B=0.25 (taking n_i=\infty). This is closest to choice (A). (Recall that ETS wants the answer that best fits.)

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
hamood
2007-04-07 15:24:52
yeah it makes more sense to go for the smallest Energies (meaning longest wavelengthhs); so n= 2 & 1 for Lyman and n = 3 & 2 for Balmer.Alternate Solution - Unverified
senatez
2006-11-01 10:51:41
Yes, it should be [1/(1/1^2-1/2^2]/[1/(1/2^2-1/3^2)] = 5/27Alternate Solution - Unverified
Comments
sina2
2013-09-21 09:37:53
We should consider:
E\,=\,hf or
E\,=\,h\frac {v  }{ \lambda  }
So the longest wavelength belongs to lowest possible energy.
5/27 is correct.
NEC
nontradish
2012-04-06 20:34:20
There is roughly a 4 to 1 ratio from Balmer to the Lyman series as quoted in this paper, "This is possible because of the near four to one ratio in wavelengths of Balmer and Lyman." http://www.ph.unimelb.edu.au/~chantler/opticshome/xrayopt/LamingChantlerNIM.pdf


Thanks for this site Yosun!!!
NEC
hamood
2007-04-07 15:24:52
yeah it makes more sense to go for the smallest Energies (meaning longest wavelengthhs); so n= 2 & 1 for Lyman and n = 3 & 2 for Balmer.Alternate Solution - Unverified
senatez
2006-11-01 10:51:41
Yes, it should be [1/(1/1^2-1/2^2]/[1/(1/2^2-1/3^2)] = 5/27
cartonn30gel
2011-04-03 21:45:14
5/27 is correct
Alternate Solution - Unverified
sblusk
2006-10-24 06:10:41
The ratio of longest wavelengths corresponds to the smallest energy difference. So, one should not use n_i = infinite, but rather

n_i = 2 and 3 for Lyman and Balmer series, respectively. In this case one obtains exactly 5/27.
Typo Alert!
kevglynn
2006-10-22 11:20:24
I just want to go along with what was mentioned in the last post. For lamda to be a maximum, one would want to minimize its inverse. Therefore, n_i approaching infinity is a wrong assumption. Instead, use the smallest possible value of n_i, which would be n_i = n_f + 1Typo Alert!
daschaich
2005-11-07 23:20:46
Actually, the longest wavelength results when n_i = n_f + 1 and the shortest wavelength when n_i = \infty. The result \frac{5}{27} given in A is exact.NEC

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We should consider: E\,=\,hf or E\,=\,h\frac {v  }{ \lambda  } So the longest wavelength belongs to lowest possible energy. 5/27 is correct.

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