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!!
