GR9677 #9



Alternate Solutions 
hamood 20070407 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.   senatez 20061101 10:51:41  Yes, it should be  

Comments 
sina2 20130921 09:37:53  We should consider:
or
So the longest wavelength belongs to lowest possible energy.
5/27 is correct.   nontradish 20120406 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!!!
  hamood 20070407 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.   senatez 20061101 10:51:41  Yes, it should be
cartonn30gel 20110403 21:45:14 
5/27 is correct

  sblusk 20061024 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.   kevglynn 20061022 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 + 1   daschaich 20051107 23:20:46  Actually, the longest wavelength results when and the shortest wavelength when . The result given in A is exact.  

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