Question

an x-ray diffractometer scan on a pure Ca sample results in a third peak at 2(theta) = 20.743421135575524 degrees.

Determine the x-ray target material and the x-ray K(alpha) wavelength, (lambda). Assume K(beta) radiation is fully attenuated.

Please show complete steps.

Answer 1

Angle is (theta) = 20.743421135575524

The target material is tungsten.

third peak is the third order spectrum as n=3

so bragg law is

3 (lambda) = 2 a sin theta    ( a is interplanar distance for calcium which is 5.6404x10^-10m)

Lambda = 2(5.6404x10^-10m) sin 20.7434 / 3

             = 1.3222x10^-10m

             = 0.1322nm