# Calculating distance between adjacent planes (hkl) in a simple cubic crystal

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.

• 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 $$d_{110} = \bigg|\frac{0-a}{\sqrt{1+1}}\bigg| = \frac{a}{\sqrt2} = \frac{5}{\sqrt2} \approx 3.535$$
– Ak.
Jul 21 '19 at 12:22
• @Ak19 I thought so. Thanks for the clarification. Jul 21 '19 at 12:27
• You're welcome!
– Ak.
Jul 21 '19 at 12:27