1
$\begingroup$

My textbook says the following in a section on Miller indices:

Adjacent planes $(hkl)$ in a simple cubic crystal are spaced a distance $d_{hkl}$ from each other, with $d_{hkl}$ given by:

$$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \tag{2.12}$$

where $a$ is the lattice constant. Equation $2.12$ provides the magnitude of $d_{hkl}$ and follows from simple analytical geometry. To generalize this expression, notice that for a plane $(hkl)$ with $hx + ky + lz = a$, the distance from any point $(x_1, y_1, z_1)$ to this plane is:

$$d_{hkl} = \left| \frac{hx_1 + ky_1 + lz_1 - a}{(h^2 + k^2 + l^2)^{1/2}} \right| \tag{2.13}$$

Hence when that point is at origin $(0, 0, 0)$, we find Equation $2.12$ back.

Example 2.1: With $a = \pu{5 Å}$, we find $d = a = \pu{5 Å}$ for $(100)$ planes and $d = \frac{a}{\sqrt{2}} = \pu{3.535 Å}$ for $(110)$ planes.

I'm confused as to how the author calculated sample 2.1. For instance, if we take $a = \pu{5 Å}$ for $(100)$ planes, then we get

$$d_{hkl} = \left| \frac{x_1 - \pu{5 Å}}{(1 )^{1/2}} \right| = \left| x_1 - \pu{5 Å} \right|$$

So how did the author get $d = a = \pu{5 Å}$?

I would greatly appreciate it if people could please take the time to clarify this.

$\endgroup$
  • 1
    $\begingroup$ He/she must have calculated w.r.t the origin so that $$(x_1,y_1,z_1) \equiv (0,0,0)$$. Then $$d_{100} = \bigg|\frac{0-a}{\sqrt1}\bigg| = a$$ and $$d_{110} = \bigg|\frac{0-a}{\sqrt{1+1}}\bigg| = \frac{a}{\sqrt2} = \frac{5}{\sqrt2} \approx 3.535$$ $\endgroup$ – Ak19 Jul 21 at 12:22
  • $\begingroup$ @Ak19 I thought so. Thanks for the clarification. $\endgroup$ – The Pointer Jul 21 at 12:27
  • 1
    $\begingroup$ You're welcome! $\endgroup$ – Ak19 Jul 21 at 12:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.