Question

You are given a material to analyze and are told that the interplanar spacing is
approximately 1 Å. You are trying to decide which x-ray target to use to characterize this
material: copper or tungsten. Calculate the K emission wavelength for each material. Which
target should you use and why?

Answer 1

Answer:

Braggs Law: 2dsin(theta) = n*lambda

d=interplanar spacing; given 1 angstrom

Copper: Face Centered Cubic

Tungsten(W): Body-Centered Cubic

Assuming, n=1

Therefore, Braggs Law can be written as;

2dsin(theta) = lambda-------------------- (1)

Where lambda is the wavelength of the material (Here: Copper or Tungsten), theta is the Braggs Angle.

From equation (1);

1. lambda      =2sin(theta) (angstroms)(Where theta is the Braggs angle for Copper)
2. lambda      =2sin(theta) (angstroms)(Where theta is the Braggs angle for Copper)

We also know;

d=a/square root (h²+k²+l²)

By substituting in equation 1, we get;

Sin² (theta) = lambda²*(h²+k²+l²)/a², where a is the lattice parameter

a= 3.57A^o (Cu standard value)

a= 3.16A^o (Tungsten standard value)

lambda Mainly depends upon a/square root (h²+k²+l²); neglecting theta as it gives slight variation

For FCC: h²+k²+l² =3, 4,8,11 and so on. Let it be 3, then

      lambda      = 2.04 (Cu)

For BCC: h²+k²+l² = 2, 4 , 6 , 8 and so on. Let it be 2, then

lambda      = 2.23 (Tungsten)

Tungsten has higher wavelength so that the Energy would be less compared to Copper. Higher the Energy, more number of X-rays are generated. Therefore, Copper must be used as material for X-Ray