Hydrogen emission spectrum calculation

In summary, the conversation discusses the methods used to find the unknown value of n in an equation. The first method involved using f = v/λ to find the frequency, then E = hf to find the energy and rearranging the equation E = (-13.6eV/n^2) - (-13.6eV/n^2) to solve for n. However, this method gave the same value of n as the original entry level. The second method involved using 1/λ = R(1/4 - 1/n^2) to solve for n, but also resulted in a value of 5. The conversation also mentions the possibility of using different values for n1 and n2 in the equations.
  • #1
benca
19
0
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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  • #2
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
oops, yes I've worked it out now. thanks
 
  • #4
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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Related to Hydrogen emission spectrum calculation

1. What is the hydrogen emission spectrum and why is it important to calculate?

The hydrogen emission spectrum is a pattern of colored lines that appear when hydrogen gas is heated and the electrons within the atoms move to different energy levels. It is important to calculate because it can provide information about the energy levels and structure of the hydrogen atom, which is fundamental to understanding atomic and molecular processes.

2. How is the hydrogen emission spectrum calculated?

The hydrogen emission spectrum can be calculated using the Rydberg formula, which is an equation that relates the wavelengths of the spectral lines to the energy levels of the hydrogen atom. The formula takes into account the Rydberg constant, which is a fundamental constant in physics, and the energy levels of the atom.

3. What factors affect the calculation of the hydrogen emission spectrum?

The calculation of the hydrogen emission spectrum can be affected by several factors, including the temperature and pressure of the hydrogen gas, the presence of other elements or impurities, and the accuracy of the measurements taken.

4. How is the hydrogen emission spectrum used in scientific research?

The hydrogen emission spectrum is used in various fields of scientific research, including astronomy, chemistry, and physics. It can provide information about the composition and temperature of stars, the structure and behavior of molecules, and the laws of quantum mechanics.

5. Are there any real-life applications of the hydrogen emission spectrum?

Yes, the hydrogen emission spectrum has several real-life applications, such as in the development of new technologies for energy production and storage, as well as in the field of spectroscopy, which is used in various industries for chemical analysis and quality control.

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