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

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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.
 

Answers and Replies

  • #2
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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
 
  • #3
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!
 
  • #4
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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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