Given:
Length of edge of unit cell, a = 407 pm
Atomic mass of gold, M = 197 amu

Density of Gold:
Density of gold can be calculated using the formula:
Density = (mass of unit cell)/(volume of unit cell)

Mass of unit cell:
Mass of unit cell can be calculated as the product of the number of atoms in the unit cell and the mass of each atom.
The given crystal structure of gold is face-centered cubic (FCC). In FCC, there are 4 atoms per unit cell.
Therefore, mass of unit cell = 4 × atomic mass of gold

Mass of unit cell = 4 × 197 amu
Mass of unit cell = 788 amu

Volume of unit cell:
Volume of unit cell can be calculated using the formula:
Volume of unit cell = a^3
where a is the length of the edge of the unit cell.

Substituting the given values, we get:
Volume of unit cell = (407 pm)^3
Volume of unit cell = 68.06 × 10^-24 cm^3

Converting pm to cm, we get:
Volume of unit cell = 68.06 × 10^-24 cm^3

Substituting the values of mass and volume in the formula for density, we get:
Density = (mass of unit cell)/(volume of unit cell)

Density = 788 amu / 68.06 × 10^-24 cm^3

Converting amu to g and cm^3 to m^3, we get:
Density = (788 × 1.66 × 10^-27 kg) / (68.06 × 10^-24 m^3)
Density = 19.30 g/cm^3

Therefore, the density of gold is 19.30 g/cm^3.

Atomic Radius of Gold:
The atomic radius of gold can be calculated using the formula:
Volume of an atom = (4/3)πr^3

where r is the atomic radius.

Volume of an atom:
The volume of an atom in FCC can be calculated as:
Volume of an atom = (1/4) × (volume of unit cell/number of atoms in unit cell)

Substituting the given values, we get:
Volume of an atom = (1/4) × [(407 pm)^3 / 4]
Volume of an atom = 6.76 × 10^-23 cm^3

Converting pm to cm, we get:
Volume of an atom = 6.76 × 10^-23 cm^3

Substituting the value of volume in the formula for atomic radius, we get:
Volume of an atom = (4/3)πr^3

6.76 × 10^-23 cm^3 = (4/3) × π × r^3

Solving for r, we get:
r = [(3 × 6.76 × 10^-23 cm^3)/(4π)]^(1/3)

r = 1.439 Å
r = 143.9 pm

Therefore, the atomic radius of gold is 143.9 pm.