Solid State: Diagram-Based Questions (7)

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Tags Solid State

Question

A metal crystallises in a face-centred cubic (FCC) structure with edge length a=4.0 A˚a = 4.0 \text{ Å}. Find: (a) the radius of the metal atom, (b) the number of atoms per unit cell, (c) the packing efficiency.

Solution — Step by Step

In FCC: 88 corner atoms (each shared by 88 cells = 11) plus 66 face-centred atoms (each shared by 22 = 33). Total =1+3=4= 1 + 3 = 4.

In FCC, atoms touch along the face diagonal. The face diagonal has length a2a\sqrt{2} and contains 44 radii (r+2r+rr + 2r + r):

4r=a2  ⟹  r=a24=a224r = a\sqrt{2} \implies r = \tfrac{a\sqrt{2}}{4} = \tfrac{a}{2\sqrt{2}} r=4.022=4.02.828≈1.414 A˚r = \tfrac{4.0}{2\sqrt{2}} = \tfrac{4.0}{2.828} \approx 1.414 \text{ Å}

Volume of 44 atoms: 4×43πr34 \times \tfrac{4}{3}\pi r^3.

Volume of cell: a3a^3.

PE=4×(4/3)πr3a3×100%\text{PE} = \tfrac{4 \times (4/3)\pi r^3}{a^3} \times 100\%

Substitute r=a2/4r = a\sqrt{2}/4:

r3=a3⋅2264=a3232⋅2=a3232⋅2r^3 = \tfrac{a^3 \cdot 2\sqrt{2}}{64} = \tfrac{a^3 \sqrt{2}}{32 \cdot 2} = \tfrac{a^3 \sqrt{2}}{32 \cdot 2}

Cleaner: r3=(a24)3=a3⋅2264=a3232r^3 = \left(\tfrac{a\sqrt{2}}{4}\right)^3 = \tfrac{a^3 \cdot 2\sqrt{2}}{64} = \tfrac{a^3\sqrt{2}}{32}.

PE=4⋅(4/3)π⋅a32/32a3=16π296=π26\text{PE} = \tfrac{4 \cdot (4/3)\pi \cdot a^3\sqrt{2}/32}{a^3} = \tfrac{16\pi\sqrt{2}}{96} = \tfrac{\pi\sqrt{2}}{6} PE=3.1416×1.41426≈0.7405=74.05%\text{PE} = \tfrac{3.1416 \times 1.4142}{6} \approx 0.7405 = 74.05\%

Final answers: r≈1.414 A˚r \approx \mathbf{1.414 \text{ Å}}, atoms per cell =4= \mathbf{4}, packing efficiency =74%= \mathbf{74\%}.

Why This Works

FCC is the closest packing of spheres in 3D. The 74%74\% packing efficiency is the maximum possible for identical spheres — proved by Kepler in 1611, finally rigorously confirmed only in 1998!

The face-diagonal touch condition is what makes FCC distinct from BCC (body-diagonal touch) and SC (edge touch). Memorise these three:

  • SC: 2r=a2r = a, PE =52.4%= 52.4\%
  • BCC: 4r=a34r = a\sqrt{3}, PE =68%= 68\%
  • FCC: 4r=a24r = a\sqrt{2}, PE =74%= 74\%

Alternative Method

Compute the density: ρ=ZMNAa3\rho = \tfrac{ZM}{N_A a^3}, where ZZ = atoms per cell, MM = molar mass. Useful for cross-checking experimental data — JEE often gives density and asks you to identify the cell type.

Common Mistake

Using the body diagonal (a3a\sqrt{3}) for FCC. That’s the BCC condition. In FCC, atoms touch along the face diagonal, not the body diagonal. Drawing a quick figure of the face plane prevents this error.

JEE Main 2024 Shift 1 had: “Iron crystallises as FCC with a=3.6 A˚a = 3.6 \text{ Å}. Find atomic radius.” Direct application of r=a2/4r = a\sqrt{2}/4.

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