Correct units to derive Ryberg Constant in ev

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Discussion Overview

The discussion revolves around the calculation of the Rydberg constant in electronvolts (eV) for the hydrogen spectrum, focusing on the appropriate units and conversion methods. Participants explore the significance of the Rydberg formula and seek clarification on unit consistency in their calculations.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant expresses difficulty in obtaining the correct units to derive the Rydberg constant of approximately 13.605 eV.
  • Another participant suggests that the Wikipedia page on the Rydberg constant should provide sufficient information for the calculations.
  • Some participants emphasize the need for a consistent set of units, specifically mentioning the definitions of Planck's constant (h) in eV-s or J-s.
  • A participant proposes using MKS units to derive energy in joules and then converting to eV using the conversion factor of 1 eV = 1.60e-19 J.
  • One participant mentions that their calculations yield results that do not align with the expected energy in joules and questions the multiplication by hc as suggested in the Wikipedia article.
  • Another participant provides a calculation for hc and its relation to the Rydberg constant, suggesting that the final conversion step was missing for one of the participants.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to unit conversion and the derivation of the Rydberg constant. Multiple viewpoints and methods are presented, indicating ongoing uncertainty and debate.

Contextual Notes

Some participants note limitations in their calculations, including potential misunderstandings of unit conversions and the application of constants. There is also mention of the need for more precise calculations, but no specific mathematical steps are resolved.

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Putting together a spreadsheet to calculate the hydrogen spectrum, cannot seem to get the right mix of units to get the Rydberg constant of 13.605... eV

also, what is the significance of R (1/n^2-1) converging to R as n gets large?

can anyone put this together or refer me to a source that lists this out? Going through my text and Wikipedia I am way off
 
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dextercioby said:
I don't understand why this page http://en.wikipedia.org/wiki/Rydberg_constant can't help you. It contains everything you need.

Just looking for a what would comprise a consistent set of units for the constants in R, for example h is defined as either eV-s or J-s
 
If you use MKS units (meters, kilograms, seconds, coulombs, and units derived from them) in the calculation, you should get energy in joules (J). Then convert to eV using the conversion factor 1 eV = 1.60e-19 J.

Otherwise you'll have to show us exactly what you've done already, so we can spot your mistake. Please do this in the Advanced Physics subforum of the "Homework & Coursework Help" forum, not here; this forum is for discussion, not help on specific assignments.
 
jtbell said:
If you use MKS units (meters, kilograms, seconds, coulombs, and units derived from them) in the calculation, you should get energy in joules (J). Then convert to eV using the conversion factor 1 eV = 1.60e-19 J.
.
thanks, everything works but the conversion to eV

9ff35fae0af4bff7855be82159314730.png


this is the equation I used and the units are not Joules. The wiki article says to multiply by hc but that is something like 10^-28 so you can't get to 13.6 eV that way
 
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Well hc is approx 3*10^8 x 6.63*10^(-34) which is approx 2*10^(-25) which multiplied by 1.1*10^7 is approx 2.2*10^(-18) which devided by 1.6*10^(-19) is approx 13.6 in units of eV.

You may, of course, use a calculator for more precision, but I don't think it's necessary.
 
Last edited:
dextercioby said:
Well hc is approx 3*10^8 x 6.63*10^(-34) which is approx 2*10^(-25) which multiplied by 1.1*10^7 is approx 2.2*10^(-18) which devided by 1.6*10^(-19) is approx 13.6 eV.

You may, of course, use a calculator for more precision, but I think it's not necessary.

thanks, the last step was what I was missing
 

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