Question

Given Bragg's Law, at what angle θ will you expect to see the Ni 111 Bragg peak in our diffractometer? Assume we use the Cu-Ka X-ray wavelength of 1.54 Angstroms. (We actually record data versus 20, but give your answer as e in degrees here)

Answer 1

Bragg's Law 2d sin theta = n lambda 2a sin theta/squareroot h^2 + k^2 + l^2 = n lambda For Ni, a = 0.352 nm So, For peak at (hkl) = (111) 2 times 0.352 times 10^-9 times sin theta/squareroot 3 = 1.54 times 10^-10 m sin theta = 1.54 times squareroot 3 times 10^-1/2 times 0.352 = 0.38 theta = 22.26 degree