K-beta line for the X-ray spectra of Molybdenum with Moseley

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Discussion Overview

The discussion revolves around calculating the Kβ line emitted energy of Molybdenum (Mo 42) using Moseley's law. Participants explore the implications of different screening constants and their effects on the calculated energy values, addressing theoretical and practical aspects of the calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the Kβ line energy using Moseley's law and notes a discrepancy between their result (20.321 keV) and the tabulated value (19.607 keV), attributing the difference to the approximation of the screening constant.
  • Another participant suggests trying a different screening constant (σ = 1.72) for the Kβ transition, providing additional values for Kα and Kβ energies and their corresponding screening constants.
  • A participant expresses uncertainty about how to calculate the screening constant when the emitted energy is unknown, asking for clarification on the method.
  • Another participant proposes that theoretical estimates of screening can be made if the exact electronic charge distribution is known, but notes that experimental estimates may be more reliable due to the complexities of electronic shell charge distributions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct screening constant or the method for calculating it, indicating that multiple competing views and uncertainties remain in the discussion.

Contextual Notes

Participants reference different screening constants and their calculations, highlighting the dependence on assumptions and the potential for variations in results based on the chosen values.

super_adun
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This should be easy but I'm totally stuck here.

So I'm calculating the Kβ line (emitted energy) of Molybdenum (Mo 42). I'm using Moseley's law and am moving from the M-shell to the K-shell. Recall Moseley's law is E = 13.6 eV (Z-σ)2 *1/n2. The effective formula for the energy emitted is thus E = 13.6 eV (Z-σ)2 *(1- (⅓2) since I am moving from n = 3 to n = 1. σ is the screening constant of n =3 of Molybdenum. Using a screening constant of 1 (I know this is incorrect, but bare with me) I get that the difference in energy is 20.321 keV. The answer according to all tables I've seen is 19.607 keV.

The only thing I could think was that the error was due to the approximation of the screening constant. Solving the equation backwards for the "table value"of K-beta gives an Zeff of 40.27. I can't fit the screening constant into that equation, since it is 4.15, as calculated with Slater's rules.

Could anybody tell me what I am missing here? I'm super thankful for all the help I can get.
 
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super_adun said:
The only thing I could think was that the error was due to the approximation of the screening constant. Solving the equation backwards for the "table value"of K-beta gives an Zeff of 40.27. I can't fit the screening constant into that equation, since it is 4.15, as calculated with Slater's rules.

Could anybody tell me what I am missing here? I'm super thankful for all the help I can get.

you may try a screening constant sigma(31) =1,72 for the Kbeta transition and check if you get a correct value
Mo z=42 E (Kalpha)17.48 sigma(21) 0.60
E (K beta)19.61 sigma(31) 1.72

details you can see at :
http://www.ld-didactic.de/literatur/hb/e/p6/p6354_e.pdf
 
drvrm said:
you may try a screening constant sigma(31) =1,72 for the Kbeta transition and check if you get a correct value
Mo z=42 E (Kalpha)17.48 sigma(21) 0.60
E (K beta)19.61 sigma(31) 1.72

details you can see at :
http://www.ld-didactic.de/literatur/hb/e/p6/p6354_e.pdf
Thank you, this would make it exactly right. I'm still not entirely sure how the screening constant (σ) is calculated if the energy E is unknown. How would you do it?
I tried with Slater's rules but I couldn't get it right. Thanks a lot for the help!
 
super_adun said:
I'm still not entirely sure how the screening constant (σ) is calculated if the energy E is unknown. How would you do it?

i have a hunch that a theoretical estimate of screening of the nuclear charge can be made exactly if you know the 'exact' electronic charge distribution- and electronic shell charge distribution is not 'realistic' enough to give correct picture except perhaps s-shells which are spherically symmetric -therefore the experimental estimates of sigma can be used and an average effect of screening is used instead
 

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