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

• I
In summary, the conversation discusses the calculation of the Kβ line of Molybdenum using Moseley's law and the effective formula for emitted energy. The disagreement in the energy values is attributed to the approximation of the screening constant (σ), and a suggested value of 1.72 is proposed for the correct value. There is also a discussion on the calculation of the screening constant and the possible use of experimental estimates for a more accurate result.

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.

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!

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

## What is the significance of the K-beta line in the X-ray spectra of Molybdenum?

The K-beta line in the X-ray spectra of Molybdenum is an emission line that corresponds to a specific energy level transition within the atom. This transition occurs when an electron from the K-shell is excited to the L-shell and then drops back down, releasing energy in the form of an X-ray photon. The energy of this line is specific to Molybdenum and can be used to identify the presence of this element in a sample.

## How is the K-beta line determined in the X-ray spectra of Molybdenum?

The K-beta line is determined using Moseley's law, which states that the square root of the frequency of an X-ray emission line is directly proportional to the atomic number of the element. By measuring the frequency of the K-beta line and applying this law, the atomic number of Molybdenum can be calculated.

## What is the relationship between the K-beta line and the K-alpha line in the X-ray spectra of Molybdenum?

The K-beta line is the second strongest emission line in the X-ray spectra of Molybdenum, after the K-alpha line. Both of these lines result from transitions in the K-shell of the atom, with the K-beta line corresponding to a higher energy transition than the K-alpha line. Together, these two lines provide a unique fingerprint for Molybdenum and can be used to identify its presence in a sample.

## What are the applications of studying the K-beta line in the X-ray spectra of Molybdenum?

Studying the K-beta line in the X-ray spectra of Molybdenum has many practical applications, such as in materials analysis, non-destructive testing, and medical imaging. Because the energy of this line is specific to Molybdenum, it can be used to identify and quantify the amount of this element in a sample, providing valuable information for various industries and scientific fields.

## How has the understanding of the K-beta line in the X-ray spectra of Molybdenum advanced scientific research?

The K-beta line in the X-ray spectra of Molybdenum has played a critical role in advancing scientific research in many areas. For example, it has been used in X-ray crystallography to determine the atomic structure of proteins and other molecules, leading to breakthroughs in fields such as biochemistry and drug development. Additionally, the study of the K-beta line has contributed to a better understanding of atomic and quantum physics, furthering our knowledge of the fundamental building blocks of matter.