Question

The density (in g cm$^{-3}$) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is _____. Use: Atomic mass of metal = 105.6 amu and Avogadro’s constant = $6 \times 10^{23}$ mol$^{-1}$

Hint:

For ccp (fcc) lattices, use \(a = 2\sqrt{2} r\) and number of atoms per unit cell = 4 to find density using \(\rho = \frac{Z M}{N_A a^3}\), where \(Z=4\).

Answer 1

Step 1: Understand the lattice type and parameters.
A cubic close packed (ccp) lattice is the same as a face centered cubic (fcc) lattice. In an fcc lattice, atoms are located at each corner and the centers of all the faces of the cube. The relationship between the edge length \(a\) and the atomic radius \(r\) is: \[ a = 2 \sqrt{2} r \] Step 2: Calculate atomic radius \(r\). Given \(a = 400 \text{ pm} = 400 \times 10^{-10} \text{ cm} = 4.0 \times 10^{-8} \text{ cm}\), \[ r = \frac{a}{2 \sqrt{2}} = \frac{4.0 \times 10^{-8}}{2 \times 1.414} = \frac{4.0 \times 10^{-8}}{2.828} \approx 1.414 \times 10^{-8} \text{ cm} \] Step 3: Number of atoms per unit cell in fcc (ccp) lattice is 4. Step 4: Calculate the volume of the unit cell. \[ V = a^3 = (4.0 \times 10^{-8})^3 = 64 \times 10^{-24} = 6.4 \times 10^{-23} \text{ cm}^3 \] Step 5: Calculate mass of atoms in one unit cell. Atomic mass \(M = 105.6 \text{ amu}\) Mass of one atom in grams: \[ m = \frac{M}{N_A} = \frac{105.6}{6 \times 10^{23}} = 1.76 \times 10^{-22} \text{ g} \] Mass of 4 atoms in unit cell: \[ m_{\text{cell}} = 4 \times 1.76 \times 10^{-22} = 7.04 \times 10^{-22} \text{ g} \] Step 6: Calculate density \(\rho\). \[ \rho = \frac{\text{mass of unit cell}}{\text{volume of unit cell}} = \frac{7.04 \times 10^{-22}}{6.4 \times 10^{-23}} = 11 \text{ g cm}^{-3} \] % Final answer Density \(= 11 \text{ g cm}^{-3}\)