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Question

The atomic radius of aluminium is 1.431 Å. The interplanar spacing of (111) planes in aluminium, in Å, is ________

Aluminum Interplanar Spacing Calculation

This solution explains how to calculate the interplanar spacing for Aluminium's (111) planes using its atomic radius.

Relevant Formulas

The interplanar spacing ($d_{hkl}$) for cubic crystals is:

$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$

Where '$a$' is the lattice parameter and '$h, k, l$' are Miller indices.

For Face-Centered Cubic (FCC) structures like Aluminium, the relationship between atomic radius ($r$) and lattice parameter ($a$) is:

$a = 2\sqrt{2} r$

Calculation Steps

Given Data:

  • Atomic Radius ($r$) = 1.431 Å
  • Miller Indices ($hkl$) = (111)

Step 1: Determine Lattice Parameter ($a$)

Use the FCC formula:

$a = 2\sqrt{2} \times 1.431 \text{ Å} \approx 4.048 \text{ Å}$

Step 2: Calculate Interplanar Spacing ($d_{111}$)

Apply the interplanar spacing formula:

$d_{111} = \frac{a}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{a}{\sqrt{3}}$

$d_{111} \approx \frac{4.048 \text{ Å}}{\sqrt{3}} \approx 2.337 \text{ Å}$

Result

The calculated interplanar spacing for Aluminium's (111) planes is approximately 2.337 Å.

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Important Questions from Crystal Structure Density Atomic Packing Factor

  1. Match the crystal systems in Column I with the corresponding axial lengths (a, b, c) and interaxial angles ($\alpha$, $\beta$, $\gamma$) provided in Column II
    Column IColumn II
    (P) Tetragonal(1) $a \neq b \neq c$, $\alpha = \beta = \gamma = 90^\circ$
    (Q) Rhombohedral(2) $a = b \neq c$, $\alpha = \beta = \gamma = 90^\circ$
    (R) Orthorhombic(3) $a \neq b \neq c$, $\alpha = \gamma = 90^\circ \neq \beta$
    (S) Monoclinic(4) $a = b = c$, $\alpha = \beta = \gamma \neq 90^\circ$
  2. The coordination number for an octahedral site in pure copper is __________.
  3. The lattice parameter of face-centered cubic iron ($\gamma$-Fe) is 0.3571 nm. The radius (in nm) of the octahedral void in $\gamma$-Fe is _______________

  4. For a bcc metal the ratio of the surface energy per unit area of the (100) plane to that of the (110) plane is ________
  5. Pure iron transforms from body centered cubic (BCC) to face centered cubic (FCC) crystal structure at $912 \text{ °C}$. If the lattice parameter of the BCC phase is $0.293 \text{ nm}$ and that of the FCC phase is $0.363 \text{ nm}$, the associated volume change is ________ (in % to one decimal place)
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