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

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Homework Help Overview

The problem involves calculating the velocity of a photoelectron released during an Auger process in tungsten, specifically related to the Kα photon absorption. The context is based on a problem from Haken & Wolf's textbook, focusing on the LI absorption edge and the associated energies involved in the process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations of kinetic energy and the relationship to velocity, with some questioning the accuracy of the original poster's energy values and the method used to derive them. There is also mention of Moseley's law and its relevance to the problem.

Discussion Status

Participants are actively engaging with the calculations presented, with some providing feedback on the approach taken. There is a recognition of the discrepancy between the calculated kinetic energy and the expected value from the textbook, prompting further exploration of the underlying principles and calculations.

Contextual Notes

There is an acknowledgment that the original poster's calculations may not align with the textbook's answers, leading to questions about the methods and assumptions used in the problem-solving process. The discussion reflects a focus on ensuring understanding of the principles involved rather than simply arriving at the correct numerical answer.

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 [tex]L_{I}[/tex] absorption edge in tungsten is at 1.02 Å. Assume that a [tex]K_{\alpha}[/tex] 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

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

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

[tex]E_{kin} = hv_K_\alpha - E_L[/tex] (18.6)

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

[tex]E_{kin} = 8.7124 * 10^{-15} - 1.9475 * 10^{-15} = 6.765 * 10^{-15}[/tex] Joules

The correct answer is [tex]5.57 *10^{-15}[/tex] 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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