Question

Show that in a cubic crystal the distance between adjacent planes with Miller indices $(h k l)$ is given by $d_{h k l}=a /\left(h^{2}+k^{2}+l^{2}\right)^{1}$, where $a$ is the lattice constant.

          Show that in a cubic crystal the distance between adjacent planes with Miller indices $(h k l)$ is given by $d_{h k l}=a /\left(h^{2}+k^{2}+l^{2}\right)^{1}$, where $a$ is the lattice constant.

Added by Alfonso R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Urvashi Arora

01/23/2022

Step 1

Miller indices $(h k l)$ are a notation to describe the orientation of crystallographic planes in a crystal lattice. They are inversely proportional to the intercepts that the plane makes with the crystallographic axes, scaled by the lattice constants. Show more…

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Show that in a cubic crystal the distance between adjacent planes with Miller indices $(h k l)$ is given by $d_{h k l}=a /\left(h^{2}+k^{2}+l^{2}\right)^{1}$, where $a$ is the lattice constant.