Auger-effect, photoelectron energy

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In summary, the problem is that the book gives the KE and velocity of the electron, while the question asks for the kinetic energy.
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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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  • #2
Your answer looks to be correct, except that you've calculated the KE, when the question asks for the velocity.
 
  • #3
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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  • #4
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.
 
  • #5
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 ;)
 
  • #6
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.
 

1. What is the Auger-effect?

The Auger-effect is a phenomenon in which an atom or molecule in an excited state releases energy by emitting an electron rather than a photon.

2. How does the Auger-effect differ from photoemission?

The Auger-effect differs from photoemission in that the emitted electron is not caused by the absorption of a photon, but rather by the energy released from an electron transitioning between energy levels.

3. What is the photoelectron energy?

The photoelectron energy is the kinetic energy of an electron that is emitted from a material due to the absorption of a photon. It is determined by the difference in energy between the initial and final states of the photoelectron.

4. How does the photoelectron energy relate to the Auger-effect?

In the Auger-effect, the energy released from an electron transition is used to eject another electron, resulting in a cascading effect. The kinetic energy of the emitted electron in this process is known as the Auger-electron energy, and it is related to the initial and final energy levels of the electrons involved in the transition.

5. What are the applications of the Auger-effect and photoelectron energy?

The Auger-effect and photoelectron energy are used in various scientific fields, including surface science, materials science, and nanotechnology. They are also commonly used in analytical techniques such as Auger electron spectroscopy and X-ray photoelectron spectroscopy to study the composition and properties of materials.

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