Quantum mechanics

  • #1
Hello,
I don't know if this is the right place to post..
this is a self study question from Levine, it is told that is important but i have no clue in solving it ..

The Balmer series corresponds to a set of emissions involving the electron in a hydrogen atom relaxing from a high energy state into the n = 2 level. The emission peaks in this series corresponding to transitions from n = 6, 5, 4, and 3 to n = 2 all appear in the visible region of the electromagnetic spectrum. Prove that these are the only four transitions between energy levels in the hydrogen atom that lead to emissions in the visible part of the spectrum.
 

Answers and Replies

  • #2
1,187
5
The Lyman series (UV emission) involve similar transitions, with the exceptions of the resting state being n = 1 and the inclusion of the transition from n = 2 to n = 1. The same thing applies to the Paschen series (infrared), with the exception that the resting state is n = 3. Thus, the Balmer series is bounded between the two. If you show that there are no other transitions that lie between the Paschen and Lyman series, then the listed Balmer series is the only one observed as visible light. In addition, you might want to calculate the transition between n = 7 and n = 2 to see if it is visible or not.
 
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  • #3
I am not sure if i am getting it right ..
since n must be an integer ... n=2 should be the only state between 1 and 3 ??
I understand in theory .. there is only balmer between lyman n'=1 and Paschen n'=3 ..
but I have no idea how to proved by calculations ...
 
  • #4
1,187
5
Transitions smaller than the Paschen series will produce longer wavelength radiation and transitions larger than the Lyman series will produce smaller wavelength radiation. Thus, if you show that all transitions with n = 1 as the resting state and n = 3 as the resting state are not visible emissions, then the transitions that lie between the two are ones with n = 2 as the resting state. If you show that the transition from n = 7 to n = 2 is not visible radiation, then the Balmer series has the only visible emissions.
 
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  • #5
o i got it !
so all I got to do is just to find the difference in energy between n= 7 and n = 2 and figure out the frequency of the wave out by E=hv. Then compare our frequency with the frequecy of visible light region ?? :)
 
  • #6
1,187
5
Yup, but before that you should show that transitions with n = 1 and n = 3 as resting states are not visible emissions.
 
  • #7
Great! I got it !
Thanks !
 

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