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.(adsbygoogle = window.adsbygoogle || []).push({});

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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# Homework Help: Auger-effect, photoelectron energy

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