X-Ray Diffraction HWK Problems

[UzumakiNaruto]

Homework Statement

1. The intensities of Kalpha and Kbeta of a Cu X-ray tube have a ratio of about 10/1. It is desirable to increase the ratio to 1000/1 by using a Ni foil as a critical absorber. What should be the thickness of the Ni foil? How much attenuation in intensity has the Cu Kalpha suffered?

2. Suppose FeAl (i.e. Fe50Al50) exists in two forms: ordered phase and disordered phase. In the ordered phase, Fe and Al form two interpenetrating sc sublattices. Each al atom is surrounded by 8 Fe atoms at the eight corners of a cube, and vice versa. In the disordered phase, Fe and Al randomly occupy the 8 corners and the body center of a cube. What would be x-ray diffraction patterns for the two phases?

Homework Equations

1. I = Ioexp[u*x] = Ioexp[(u/rho)*rho*x]
where I = attenuation of a photon intensity through a material.
u = absorption coefficient (in cm^-1) and (1/u) = one absorption length.
rho = density of the material (usually it is not u but u/rho (in units of cm^2/g that is tabulated.
1 barn = 10^-24 cm^2

2. Have no clue!

The Attempt at a Solution

1. I have no clue how to even do this. I can't find any equations relating the filter thickness to the information given. Any help is greatly appreciated! I only know we are supposed to use the attenuation equation for the second part of question 1. Thanks in Advance!

2. I also have no clue how do this! Please help! Thanks in Advance!

 

[Jhero]

For the first part of your question, we can use the Beer-Lambert Law to find the thickness of the Ni foil needed to increase the ratio of Kalpha to Kbeta to 1000/1. The Beer-Lambert Law states that the intensity of a beam of light (or in this case, X-rays) decreases exponentially as it passes through a material. The equation for this is:

I = I0 * e^(-ux)

Where:
I = intensity after passing through the material
I0 = initial intensity
u = absorption coefficient
x = thickness of the material

We can rearrange this equation to solve for x:

x = ln(I/I0) / -u

We know that the initial ratio of Kalpha to Kbeta is 10/1, so I0 = 10. We want to increase this to 1000/1, so I = 1000. We can also find the absorption coefficient for Ni at the energy of the Cu X-ray tube (8 keV) from a table, which is approximately 5.5 cm^-1.

Plugging in these values, we get:

x = ln(1000/10) / -5.5 = 4.9 cm

So the thickness of the Ni foil needed to increase the ratio to 1000/1 is approximately 4.9 cm.

As for the second part of the question, we can use the Bragg Equation to find the x-ray diffraction patterns for the two phases of FeAl. The Bragg Equation is:

nλ = 2dsinθ

Where:
n = order of diffraction
λ = wavelength of the x-rays
d = spacing between crystal planes
θ = angle of diffraction

For the ordered phase, the spacing between crystal planes can be found using the formula:

d = a / √(h^2 + k^2 + l^2)

Where:
a = lattice parameter (distance between unit cells)
h, k, l = Miller indices

Since the Fe and Al atoms form two interpenetrating sc sublattices, we can use the lattice parameter for simple cubic structures, which is equal to the length of one side of the unit cell. For simplicity, we can assume that the lattice parameter is equal to 1 Å (10^-10 m).

For the ordered phase, we have h = k = l =