Final answer:
To calculate the lattice constant 'a' of a metal M with a face-centered cubic (fcc) unit cell and a given atomic radius of 179 pm, we use the relation between the face diagonal and the lattice constant in an fcc lattice, resulting in approximately 506 pm for the lattice constant.
Step-by-step explanation:
The student has asked to calculate the lattice constant 'a' for a metal M that crystallizes in a face-centered cubic (fcc) unit cell given the atomic radius 'r' is 179 pm.
In a face-centered cubic (fcc) lattice, atoms at the corners of the cell touch the atoms on the centers of the faces along the diagonal of the face. Hence, the diagonal 'd' of the face can be represented as d = 4r, where 'r' is the radius of the atom. The lattice constant 'a', which is the edge length of the unit cell, can be related to the face diagonal through the equation: d = a\(√2). Substituting 'd' with '4r', we have 4r = a\(√2), solving for 'a' gives us a = √(4r)\(√2).
By substituting the value of 'r' (179 pm = 179 × 10-12 m), we get the lattice constant 'a' as follows:
a = 4 × 179 pm / √2
a = 716 pm / √2
a = 506.196 pm
Since the measurement was given with three significant digits, we should round our answer to three significant digits as well:
a ≈ 506 pm
Therefore, the lattice constant of the crystal of metal M is approximately 506 pm.