Suppose the mass of a single Ag atom is 'm'. Ag metal crystallises in fcc lattice with unit cell of length 'a'. The density of Ag metal in terms of 'a' and 'm' is-

Options

A. `(4m)/a^(3)`
B. `(2m)/a^(3)`
C. `(m)/a^(3)`
D. `(m)/(4a^(3))`

Answer

A

Explanation

Volume of unit cell = `a^(3)`, no. of atoms present in fcc unit cell = `4xxm`. Thus, the density of the Ag metal = `(4m)/a^(3)`.