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X-Ray emission and Bragg Scattering (K-Alpha and K-beta absorption)

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Homework Statement


Complete this sentence. When doing Bragg scattering of x-rays off of NaCl, the first order K-Alpha peak has an energy that is ___________ the energy of the second order K-Alpha peak.

Known Information
It is known that the Intensity of the K-Alpha is greater than the K-Beta intensity in the first order. This means that the energy required to produce K-Beta is greater.

At higher energies (greater than 30Kev) the Intensity will be smaller than the first order Intensities.

Homework Equations



[tex]\lambda[/tex] = 2 d sin [tex]\theta[/tex]


Emax = eV = hfmax=[tex]\frac{hc}{ \lambdamin }[/tex]

Please excuse the crude format error in the equation above (im a noob to the forum)

Where Emax is the maximum photon energy, [tex]\lambda[/tex] min is the minimum wavelength that is determined by the accelerating voltage V of the tube

The Attempt at a Solution



I attempted to combine the equations above to come up with an explanation to the problem above however all it yields is lattice spacing and the angle of incidence. What am I missing?

Google searches have been helpful in the understanding of how x-rays are made but little information is give to K-alpha and K-beta other than their definition and how they occur but nothing as to the first and second order energy difference between them.

What is the percentage or fraction that describes this difference between the first order and second order K-alpha peaks?

Which equations are necessary to find this besides the ones listed above?
 

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