Question

How much, if any, would the (111) peak of aluminum shift in an x-ray diffraction experiment assuming the octahedral interstitials are filled with lithium ions (radii of Al = 125 pm, Li+ = 90 pm)?

(value of wavelength not given)

Answer 1

The density of aluminum = a/(h² + k² + l²)^1/2

Where 'a' is the edge length of Al

And h, k, l are the miller indices.

The fact is that aluminum crystallizes in a face-centered cubic lattice (FCC).

i.e. a = 2(2)^1/2r, where 'r' is the radius of aluminum.

i.e. a = 2*1.1414*125 pm ~ 354 pm

Therefore, the density of aluminum (111) = 354/(1² + 1² + l²)^1/2 ~ 204 pm

Now, lithium crystallized in BCC lattice.

i.e. for lithium, a = 4r/(3)^1/2 = 4*90/1.732 ~ 208 pm

i.e. density of lithium (111) = 208/(1² + 1² + l²)^1/2 ~ 120 pm

The density of Lithium (120 pm) is less than the density of Aluminum (204 pm)

Hence, the (111) peak of aluminum shift when the octahedral interstitials are filled with lithium ions = 204 - 120 = 84 pm