Q22 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Chemistry

The angle between (100) and (110) planes of FCC lattice is

  • A
    $90^{\circ}$
  • B
    $0^{\circ}$
  • Correct answer
    C
    $45^{\circ}$
  • D
    $120^{\circ}$

Solution

Angle between (100) and (110) planes of fcc lattice is $\cos \phi=\frac{h_1 h_2+k_1 k_2+l_1 l_2}{\sqrt{h_1^2+k_1^2+l_1^2 \sqrt{h_2^2+k_2^2+l_2^2}}}$ $\cos \phi=\frac{(1 \times 1)+(0 \times 1)+(0 \times 0)}{\sqrt{(1)^2+(0)^2+(0)^2} \sqrt{(1)^2+(1)^2+(0)^2}}$ $\cos \phi=\frac{1}{\sqrt{2}} \Rightarrow \phi=45^{\circ}$

Hence, the correct answer is option: 'C'.

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