Hydrogen emission spectrum calculation

AI Thread Summary
The discussion revolves around calculating the hydrogen emission spectrum using different methods. The initial approach involved using frequency and energy equations, but the participant consistently arrived at the same n value of 5. They later explored the Rydberg formula but received the same result, prompting confusion about the calculations. Clarifications were provided regarding the need for two n values and the specific context of the equations used. Ultimately, the participant realized that their equation was tailored for the Balmer series, which ends at n=2, leading to the correct understanding of the transitions involved.
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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?
 
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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.
 
oops, yes I've worked it out now. thanks
 
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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