The angle between (100) and (110) planes of a | vedclass

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(C) The angle $\phi$ between two planes $(h_1 k_1 l_1)$ and $(h_2 k_2 l_2)$ in a cubic system is given by the formula:$\cos \phi = \frac{h_1 h_2 + k_1

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) The angle $\phi$ between two planes $(h_1 k_1 l_1)$ and $(h_2 k_2 l_2)$ in a cubic system is given by the formula:$\cos \phi = \frac{h_1 h_2 + k_1 k_2 + l_1 l_2}{\sqrt{h_1^2 + k_1^2 + l_1^2} \sqrt{h_2^2 + k_2^2 + l_2^2}}$For the planes $(100)$ and $(110)$:$h_1=1, k_1=0, l_1=0$ and $h_2=1, k_2=1, l_2=0$Substituting these values:$\cos \phi = \frac{(1 	imes 1) + (0 	imes 1) + (0 	imes 0)}{\sqrt{1^2 + 0^2 + 0^2} \sqrt{1^2 + 1^2 + 0^2}}$$\cos \phi = \frac{1}{\sqrt{1} 	imes \sqrt{2}} = \frac{1}{\sqrt{2}}$$\phi = \cos^{-1}(\frac{1}{\sqrt{2}}) = 45^{\circ}$

The angle between $(100)$ and $(110)$ planes of an $FCC$ lattice is: (in $^{\circ}$)

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