Hydrogen emission spectrum calculation

  • Thread starter benca
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  • #1
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Homework Statement:
The hydrogen spectrum contains a blue line with a wavelength of 434 nm. Photons of blue light are emitted when hydrogen's electron drops from the fifth energy level to a lower energy level What is the lower energy level?
Relevant Equations:
E = (-13.6eV/n^2) - (-13.6eV/n^2)
1/λ = R(1/4 - 1/n^2)
E = hf
h = Planck's constant
I've first method I tried was using f = v/λ to find the frequency, then E = hf to find the energy and then using E = (-13.6eV/n^2) - (-13.6eV/n^2) to rearrange and solve for the unknown n. However I got 5, the same as the original entry level.

I also tried using 1/λ = R(1/4 - 1/n^2) to solve for n and got 5 again. I'm not really sure what I'm doing wrong, but I think it has something to do with the way I used E = hf to get the energy and used that value for E in E = (-13.6eV/n^2) - (-13.6eV/n^2). Any hints?
 

Answers and Replies

  • #2
mjc123
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What is n in your equation for E? As it stands E ≡ 0. There should be two n values, one for the upper and one for the lower level. Try putting n1 = 5 and calculating n2.
Your equation for 1/λ should be analogous. Where does the term 1/4 come from? Is it assuming n2 = 2? Perhaps you then get n1=5 because the right answer for n2 happens to be 2.
 
  • #3
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oops, yes I've worked it out now. thanks
 
  • #4
mjc123
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Check: the visible lines of H (the Balmer series) are due to the transitions that end on n=2. Perhaps your 1/λ equation was specifically for this series, rather than generally for all H lines.
 

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