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.
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.