GREPhysics.NET: GR9677 #9

GR9677 #9

Problem

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Atomic $\Rightarrow$ }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.)

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. Reply to this comment
senatez 2006-11-01 10:51:41	Yes, it should be $[1/(1/1^2-1/2^2]/[1/(1/2^2-1/3^2)] = 5/27$ Reply to this comment

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.

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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!!!

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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.

Reply to this comment

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

Reply to this comment

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.

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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 + 1

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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.

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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.

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