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