Introduction:
The fcc (face-centered cubic) lattice is a common crystal structure in which atoms are arranged in a repeating pattern. In this lattice, each atom is surrounded by 12 nearest neighbors and 6 second-nearest neighbors. The distance between nearest neighbors is known as the nearest-neighbour distance, and the distance between second-nearest neighbors is known as the second-neighbour distance. The ratio of these two distances can be calculated to determine the answer to the given question.

Calculation:
Let's assume the nearest-neighbour distance as 'a'. Since the fcc lattice has a cubic structure, the second-neighbour distance will be along the body diagonal of the cube, which is equal to the edge length of the cube. Let's assume the second-neighbour distance as 'b'.

Deriving the Value:
To find the ratio of the second-neighbour distance to the nearest-neighbour distance, we need to calculate 'b/a'.

1. Finding 'a':
In an fcc lattice, the atoms are arranged in such a way that each face of the cube has an atom at each corner and one at the center. The face diagonal of the cube is equal to 'a'. By using the Pythagorean theorem, we can calculate 'a' as:
a² = (edge length of the cube)² + (edge length of the cube)²
a² = 2(edge length of the cube)²
a = √2(edge length of the cube)

2. Finding 'b':
The body diagonal of the cube is equal to 'b'. By using the Pythagorean theorem, we can calculate 'b' as:
b² = (edge length of the cube)² + (edge length of the cube)² + (edge length of the cube)²
b² = 3(edge length of the cube)²
b = √3(edge length of the cube)

Calculating the Ratio:
Now we can substitute the values of 'a' and 'b' in the ratio formula:
b/a = (√3(edge length of the cube)) / (√2(edge length of the cube))
b/a = √3/√2
b/a = √(3/2)

Answer:
Therefore, the ratio of the second-neighbour distance to the nearest-neighbour distance in an fcc lattice is √(3/2), which is approximately equal to 1.225. Thus, the correct option is d) √2.