The density of copper metal is `8.95" g "cm^(-3)`. If the redius of copper atom be `127. 8` pm, is the copper unit cell simple cubic, body - centred or face- centred cubic ?
(Given : atomic mass of Cu = 63. 5 g/mol)

Explanation

In the case, edge length is not given. Therefore, the problem can be solved indirectly by assuming the nature of unit cell and then finding out the density for verification. Let us assume that the unit cell for copper is face centred (f.c.c.) with value of Z = 4. For the unit cell,
Radius (r) = 127.8 pm
Edge length (a) = `2sqrt2r=2xx1.414xx127.8"pm"=361.42"pm"=361.42`
Atomic mass of copper (M) =`63.54"g mol"^(-1)`
Avogadro's no. `(N_(0))=6.022xx10^(23)"mol"^(-1)`
`"Density"(rho)=(ZxxM)/(a^(3)xxN_(0)xx(10^(-30)cm^(3)))=(4xx(63.54gmol^(-1)))/((361.42)^(3)xx(6.022xx10^(23)mol^(-1))xx(10^(-30)cm^(3)))`
`=(4xx(63.54g mol^(-1)))/((4.72xx10^(7))xx(6.022xx10^(23)mol^(-1))xx(10^(-30)cm^(3)))=8.942"g cm"^(-3)`
Since the density of the unit cell is almost the same as given, this means that it is f.c.c type unit cell.
Note: In case the calculated density fails to tally with given value, the problem can be repeated by considering the other alternatives i.e, b.c.c. (having z = 2) or simple cubic (having z = 1).