# Auger-effect, photoelectron energy

1. Oct 7, 2008

### Antti

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.

2. Oct 7, 2008

### Gokul43201

Staff Emeritus
Your answer looks to be correct, except that you've calculated the KE, when the question asks for the velocity.

3. Oct 7, 2008

### Antti

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.

Last edited: Oct 7, 2008
4. Oct 7, 2008

### Gokul43201

Staff Emeritus
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. Oct 8, 2008

### Antti

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. Oct 9, 2008

### Gokul43201

Staff Emeritus
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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