Hydrogen emission spectrum calculation

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SUMMARY

The discussion centers on calculating the hydrogen emission spectrum using the formulas f = v/λ and E = hf, alongside the Rydberg formula 1/λ = R(1/n1^2 - 1/n2^2). The user initially misapplied the energy equation, leading to confusion about the quantum number n. After clarification, it was established that the visible lines of hydrogen, specifically the Balmer series, correspond to transitions ending at n=2, confirming that n1=5 and n2=2 are correct for this context.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically energy quantization.
  • Familiarity with the Rydberg formula for hydrogen emission spectrum calculations.
  • Knowledge of the relationship between wavelength, frequency, and energy.
  • Basic grasp of the Balmer series and its significance in hydrogen spectroscopy.
NEXT STEPS
  • Study the Rydberg formula in detail, focusing on its application to various spectral series.
  • Learn about the Balmer series and its significance in hydrogen emission spectra.
  • Explore the concept of quantum numbers and their role in electron transitions.
  • Investigate the derivation and implications of the energy equation E = -13.6 eV/n^2 for hydrogen.
USEFUL FOR

Students and educators in physics, particularly those focusing on quantum mechanics and atomic spectroscopy, as well as researchers analyzing hydrogen emission spectra.

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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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