Question

In an orthorhombic crystal, the lattice constants are 3.0 \AA, 3.2 \AA, and 4.0 \AA. The distance \(d_{101}\) between the successive (101) planes is \rule{1cm{0.15mm} \AA. (up to one decimal place)}

Hint:

Memorizing the interplanar spacing formulas for different crystal systems is very useful. - Cubic: \(\frac{1}{d^2} = \frac{h^2+k^2+l^2}{a^2}\) - Tetragonal: \(\frac{1}{d^2} = \frac{h^2+k^2}{a^2} + \frac{l^2}{c^2}\) - Orthorhombic: \(\frac{1}{d^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2}\) Notice how the cubic and tetragonal formulas are special cases of the orthorhombic one.

Answer 1

Correct Answer: 2.4

Step 1: Understanding the Concept:
The distance between adjacent parallel planes in a crystal lattice, known as the interplanar spacing, can be calculated from the Miller indices of the planes (\(hkl\)) and the lattice parameters of the unit cell. The formula depends on the crystal system.
Step 2: Key Formula or Approach:
For an orthorhombic crystal system (where the unit cell axes are mutually perpendicular, \(a \neq b \neq c\)), the interplanar spacing \(d_{hkl}\) is given by the formula: \[ \frac{1}{d_{hkl}^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2} \] Step 3: Detailed Explanation:
We are given the lattice constants: \(a = 3.0\) \AA \(b = 3.2\) \AA \(c = 4.0\) \AA And the Miller indices for the plane are (101), so \(h=1, k=0, l=1\). Substitute these values into the formula: \[ \frac{1}{d_{101}^2} = \frac{1^2}{(3.0)^2} + \frac{0^2}{(3.2)^2} + \frac{1^2}{(4.0)^2} \] \[ \frac{1}{d_{101}^2} = \frac{1}{9} + 0 + \frac{1}{16} \] Find a common denominator to add the fractions: \[ \frac{1}{d_{101}^2} = \frac{16}{144} + \frac{9}{144} = \frac{16+9}{144} = \frac{25}{144} \] Now, solve for \(d_{101}\) by inverting the expression and taking the square root: \[ d_{101}^2 = \frac{144}{25} \] \[ d_{101} = \sqrt{\frac{144}{25}} = \frac{12}{5} = 2.4 \] Step 4: Final Answer:
The distance \(d_{101}\) between the successive (101) planes is 2.4 \AA.