Why Is My Photoelectron Energy Calculation Off in Auger-Effect Problem?

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SUMMARY

The forum discussion centers on solving problem 18.9 from Haken & Wolf's "The Physics Of Atoms And Quanta," 7th edition, which involves calculating the velocity of a photoelectron released during an Auger process in tungsten. The user initially calculated the kinetic energy of the photoelectron using formulas 18.3 and 18.6, arriving at a value of 6.765 * 10^{-15} Joules, which is close but not correct compared to the book's answer of 5.57 * 10^{-15} Joules. The discussion emphasizes the importance of correctly interpreting the problem's requirements, specifically distinguishing between kinetic energy and velocity.

PREREQUISITES
  • Understanding of Auger processes in atomic physics
  • Familiarity with the Rydberg constant and its application in atomic calculations
  • Knowledge of kinetic energy formulas and their derivations
  • Ability to interpret and solve problems from physics textbooks
NEXT STEPS
  • Review the derivation of the Rydberg formula and its application in atomic transitions
  • Learn about Moseley's law and its significance in determining atomic properties
  • Practice additional Auger effect problems from various physics resources
  • Study the solved examples in Haken & Wolf, particularly part 4 on page 334
USEFUL FOR

Students and educators in atomic physics, particularly those studying Auger processes and energy calculations, as well as anyone seeking to deepen their understanding of the principles outlined in Haken & Wolf's textbook.

Antti
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This problem is straight from Haken & Wolf - The Physics Of Atoms And Quanta, 7th edition. It's problem 18.9 on page 336. If you happen to have the book that is.

Problem text:

The L_{I} absorption edge in tungsten is at 1.02 Å. Assume that a K_{\alpha} photon is "absorbed" by one of the 2s electrons by an Auger process. Determine the velocity of the photoelectron released.

Attempted solution:

I've used formulas 18.3 and 18.6 in the book. I can use

\bar{\nu}_K_\alpha = \frac{3}{4} R (Z-1)^{2} (18.3)

to get the wave number for the K_\alpha line. I inserted the Rydberg constant and tungsten atomic number, R = 10973731 and Z = 74, and got the wave number 4.386 * 10^{10} inverse meters. This corresponds to the energy = 8.7124 * 10^{-15} Joules. Next I use

E_{kin} = hv_K_\alpha - E_L (18.6)

I set hv_K_\alpha = (the energy I calculated) = 8.7124 * 10^{-15} and E_L = (energy of the ebsorption adge given in the problem) = 1.9475 * 10^{-15}. Subtracting the second energy from the first gives

E_{kin} = 8.7124 * 10^{-15} - 1.9475 * 10^{-15} = 6.765 * 10^{-15} Joules

The correct answer is 5.57 *10^{-15} Joules according to the book. So I'm not that far off but I can't figure out what I've missed.
 
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Your answer looks to be correct, except that you've calculated the KE, when the question asks for the velocity.
 
Okay, seems like my crappy first post is no longer editable... EDIT: fixed! :)

Gokul: The book gives both the kinetic energy and velocity of the electron in the answer. I have only compared the first value since the velocity can't turn out right if the kinetic energy is wrong. So my answer isn't correct. I should have been more clear about this, sorry.
 
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For the KE, I get almost exactly the same number that you got (6.8*10^{-15}J), and I think Moseley's law is more than sufficiently accurate for 2 sig figs.
 
I see. Well that obviously raises the question about how they got the answer in the book. I think you can understand that I'd rather trust the authors than you ;)
 
I understand.

To make sure you've understood the principle properly, and feel confident that you have, you should work through part 4 of the solved example on pg. 334 and make sure you get the correct answer. You could also try more problems from other books.
 

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