What is the significance of inverse meters in astronomy calculations?

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    Inverse Meter
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Discussion Overview

The discussion centers around the use of inverse meters in astronomy calculations, specifically in relation to Rydberg's constant and its application in wavelength calculations. Participants explore the implications of using inverse meters in equations related to spectral lines.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the meaning of inverse meters in the context of Rydberg's constant, questioning if it translates to 1/109737361.6 m.
  • Another participant confirms that inverse meters simply means per meter.
  • A participant seeks clarification on how to apply the equation 1/lambda = R (1/nf^2 - 1/ni^2) with Rydberg's constant, asking if the process differs from plugging in other numerical values.
  • A later reply suggests that after multiplying R by the fraction, one should reciprocate the result to obtain the wavelength.

Areas of Agreement / Disagreement

There is no explicit consensus on the application of the equation or the interpretation of the results, as participants provide different insights and methods without resolving the initial confusion.

Contextual Notes

Participants do not clarify the assumptions behind their interpretations of inverse meters or the specific steps involved in the calculations, leaving some mathematical steps unresolved.

Who May Find This Useful

Individuals interested in astronomy calculations, particularly those dealing with spectral lines and Rydberg's constant, may find this discussion relevant.

velvetymoogle
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Hi,
I'm doing some astronomy work and one problem involves using Rydberg's constant. However, it is inverse meters and I'm confused. Does that mean that it is 1/109737361.6 m? Thanks.
 
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Yes - it just means per metre.
 
Last edited:
mgb_phys said:
Yes - it just means per metre.

Okay so if I have the 1/lambda = R (1/nf^2 - 1/ni^2) equation, how would I go about plugging it in? Is it any different than plugging in any other numbers?
 
after you multiply R by the fraction,
you probably want to reciprocate the result (1/x)
to obtain the wavelength.
 

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