Question

For which set of crystallographic planes will a first-order diffraction peak occur at a diffraction angle of 44.53° for FCC nickel (Ni) when monochromatic radiation having a wavelength of 0.1542 nm is used? The atomic radius for Ni is 0.1246 nm.

Answer 1

Answer:

The set of planes responsible for this diffraction peak is the [111] set.

Explanation:

We need to calculate the interplanar spacing, dₕₖₗ for nickel.

dₕₖₗ = nλ/2 sin θ

where θ = half of the diffraction angle = 44.53°/2 = 22.265°, n is the order of reflection = 1 and λ is the wavelength of the monochromatic radiation = 0.1542 nm.

dₕₖₗ = 1×0.1542/2sin 22.265°

dₕₖₗ = 0.2035 nm

But, interplanar spacing, dₕₖₗ is related to the plane (hkl), by the relation

√(h² + k² + l²) = a/dₕₖₗ

a is the lattice parameter.

Since Nickel has an FCC structure, a = 2R√2, R is given as 0.1242 nm

√(h² + k² + l²) = 2R√2/dₕₖₗ = (2×0.1246√2)/0.2035 = 1.732

(h² + k² + l²) = 1.732² = 3

The only 3 integers the values of h, k and l which fit properly into the equation is [111]

Therefore, the answer is [111].

Hope this Helps!!