Question

Copper has the fcc crystal structure. Assuming an atomic radius of 130 pm for copper atom (Cu = 63.54):

(1) What is the length of the unit cell of Cu?

(2) What is the volume of the unit cell?

(3) How many atoms belong to the unit cell?

(4) Find the density of Cu?

• A. (a) ⟶367pm, (b) ⟶ 4.94 x 10^-23 cm³, (c) ⟶ 4 , (d) ⟶ 8.54 gm/cm³
• B. (a) ⟶ 376pm, (b) ⟶ 4.94 x 10^-23 cm³, (c) ⟶ 4 , (d)⟶ 9.54 gm/cm³
• C. (a) ⟶ 367pm, (b) ⟶ 8.94 x 10^-23 cm³, (c) ⟶4 , (d) ⟶ 10.54 gm/cm³
• D. (a) ⟶ 376pm, (b) ⟶ 8.94 x 10^-23 cm³, (c) ⟶4 , (d) ⟶ 12.54 gm/cm³

Answer 1

The correct option is A

(a) ⟶367pm, (b) ⟶ 4.94 x 10^-23 cm³, (c) ⟶ 4 , (d) ⟶ 8.54 gm/cm³

As we know $\rho =\frac{n\times M_m}{N_A\times a^3},$

(a) For fcc structure

$4r=\sqrt{2}aa=2\sqrt{2}r=2\sqrt{2}\times 130pm=367.64pm$

(b) Volume of unit cell

$a^3=\frac{n\times M_m}{N_a\times \rho }=\frac{4\times 63.54}{6.023\times {10}^{23}\times 8.54}g/m^3=4.94\times {10}^{-23}c{m}^3$

(c) n = 4

(d) $\rho =\frac{4\times 63.54}{6.023\times {10}^{23}\times (3.67\times {10}^{-8c{m}^3})}=8.54gm/c{m}^3$