Question

Copper crystallises in fcc lattice has density 8.930 calculate radius of copper

Answer 1

Solution : Given,

Number of atom in unit cell of FCC (Z) = 4

Density = 8.93g/cm^38.93g/cm

3

Atomic mass of copper(M) = 63.5 g/mole

Avogadro's number (N_{A})=6.022\times 10^{23} mol^{-1}(N

A

)=6.022×10

23

mol

−1

Formula used :

\rho=\frac{Z\times M}{N_{A}\times a^{3}}ρ=

N

A

×a

3

Z×M

.............(1)

where,

\rhoρ = density

Z = number of atom in unit cell

M = atomic mass

(N_{A})(N

A

) = Avogadro's number

a = edge length of unit cell

Now put all the values in above formula (1), we get

8.93g/cm^3=\frac{4\times (63.5g/mol)}{(6.022\times 10^{23}mol^{-1}) \times a^3}8.93g/cm

3

=

(6.022×10

23

mol

−1

)×a

3

4×(63.5g/mol)

a=3.6147\times 10^{-8}cm=361.47pma=3.6147×10

−8

cm=361.47pm

(1pm=10^{-10}cm)(1pm=10

−10

cm)

Therefore, the length of the unit cell is, 361.47 pm