Why Can You Only Observe the Balmer Series Lines in Hydrogen Spectroscopy?

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Discussion Overview

The discussion revolves around the observation of the Balmer series lines in hydrogen spectroscopy, particularly why only these lines are visible in a classroom experiment involving a hydrogen lamp and a diffraction grating. The scope includes concepts from quantum physics and spectroscopy.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the Balmer series is visible because it falls within the visible wavelength region, while other series like Lyman and Paschen are in the ultraviolet and infrared ranges, respectively.
  • One participant mentions the use of the Rydberg formula to calculate the wavelengths of the Balmer series lines, suggesting that the visible lines correspond to specific principal quantum numbers (n=3, 4, 5, 6, etc.).
  • Another participant emphasizes the importance of holding one quantum number constant (m=2) to define the Balmer series while varying the other (n) to obtain the visible wavelengths.
  • There is a calculation presented for the wavelengths corresponding to the Balmer series, indicating that the results align with the visible lines observed in the experiment.

Areas of Agreement / Disagreement

Participants generally agree that the Balmer series is the only visible series in the context of the experiment, but there is no consensus on the completeness or accuracy of the calculations presented, as some participants express uncertainty about the methodology.

Contextual Notes

Some limitations include potential dependencies on the power supply used in the experiment and the need for further verification of the calculations related to the wavelengths of the Balmer series.

UrbanXrisis
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I'm learning about quantum physics and did an experiement in my class. It's a spectroscopy activity where you place a hydrogen lamp in front of a defraction grating. We must verify that only the Balmer lines are visible but why do I only observe the Balmer series lines and not others?
 
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Because the Balmer series is in the visible wavelength region...
 
UrbanXrisis said:
I'm learning about quantum physics and did an experiement in my class. It's a spectroscopy activity where you place a hydrogen lamp in front of a defraction grating. We must verify that only the Balmer lines are visible but why do I only observe the Balmer series lines and not others?

Only the Bamer lines are in the visible range for your eyes. The Lyman series, for example, are in the ultraviolet range, while the Paschen series are in the infrared. You can verify this by looking at the wavelength typical for each series. All these lines are probably there (depending on the power supply attached to your discharge tubes), but you just can't see them.

Zz.
 
Sorry to bore you. Like I said, I'm totally new at this. The experiment asks me to verify that only the Balmer lines are visible with the equation:

1/lambda=R(1/n^2)-(1/n^2)
 
UrbanXrisis said:
Sorry to bore you. Like I said, I'm totally new at this. The experiment asks me to verify that only the Balmer lines are visible with the equation:

1/lambda=R(1/n^2)-(1/n^2)

So then take the n values (principle quantum number) for just the balmer series. I think they are n=[3,4,5,6...]. But you simply go back and look up in the textbook (or web) that they are visible wavelengths that result from putting these numbers in the above Balmer formula...right? All other series (like Paschen) will calculate to be outside your vision. Now the tough part is arranging your equation properly. It says above there are two n's, but its easier to think of them separately as m and n where m<n. We will hold m=2 as constant for the whole series (that's what defines the Balmer series) and increment the n to get all the lines. So the first will be m=2 and n=3.

1. R[1/m^2 - 1/n^2] = R[1/4 - 1/9] = 1.09723 10E-3[0.13888] = 1/(6562.08 Angstroms)

and that's correct. You are using the Rydberg constant above. Then,

2. R[1/4 - 1/16] = 1/(4860.7 A)

Is that right? I'll bet all the lines you saw in your lab correspond to everything you calculate with this method. Now keep going. I don't know how many you calculate for all the visible lines, but as I said you can look that up
 

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