UPSEE Solved Model Paper 7

Given : Edge length of a unit cell = 4.53 $\overset{0}{A}$ = 4.53 × $10^{-8}$ cm Molecular weight = 24 Density, d = 1.74 $\text{g/cm}^3$ Avogadro's number = 6 × $10^{23}$ We have to find the radius of the metal. Now, Density is given as: ρ = $\frac{z \times \text{Molecular weight}}{a^3 \times \text{Avogadro's number}}$ ... (i) where, a = Edge length z = No. of atoms in a unit cell From equation (i): z = $\frac{\rho \times a^3 \times \text{Avogadro's number}}{\text{Molecular weight}}$ ... (ii) On substituting all the values in equation (ii), we get z = z = z = $\frac{97.05}{24}$ z = 4.043 z ~ 4 As, the value of 'z' is 4. thus, metal crystallises in face centred cubic (fcc) lattice Now. For fcc structure, radius is related to edge length as r = $\frac{a}{2\sqrt{2}}$ ... (iii) Substituting all the values in equation (iii), we get r = $\frac{4.53 \times 10^{-8}}{2 \times 1.414}$ cm r = $\frac{4.53 \times 10^{-8}}{2.828}$ cm r = 1.60 × $10^{-8}$ cm or r = 160 × $10^{-10}$ cm or r = 160 × $10^{-12}$ m r = 160 pm Hence, option 'B' is correct