This question has been asked before, but there doesn't seem to be a decent answer.

Many sources state that " For cubic crystals with lattice constant a, the spacing d between adjacent (ℓmn) lattice planes is:

$$ {\displaystyle d_{\ell mn}={\frac {a}{\sqrt {\ell ^{2}+m^{2}+n^{2}}}}}$$ "


Could someone please explain what "adjacent" means in this case (Is it planes that share the same side, is it parallel planes, are these panes in the same unit cell or neighbouring cells etc)? Better yet, does anyone know of a sketch explaining this ? I am really at a loss here and this has been driving me nuts the whole day

  • $\begingroup$ Two distinct parallel planes that don't have any other planes between them. $\endgroup$
    – lemon
    Jul 20, 2016 at 19:00
  • $\begingroup$ That are perpendicular to the (l,m,n) direction... $\endgroup$
    – Jon Custer
    Jul 20, 2016 at 23:04

2 Answers 2


enter image description here

I could only find this poor-quality picture. It should give you an idea, anyway. For example, consider the first picture in the first row: $(l,m,n)=(1,0,0)$ in that case, and it is easy to verify that the distance between the grey planes is


In the second case, $(l,m,n)=(1,1,0)$, and you can see that

$$d=\frac a {\sqrt 2}$$


  • $\begingroup$ Thank you. So, if I understand this correctly, the above formula gives the distance between two neighbouring planes within the same set of planes? $\endgroup$
    – user57927
    Jul 21, 2016 at 10:02
  • $\begingroup$ @user57927 Exactly. $\endgroup$
    – valerio
    Jul 21, 2016 at 10:15
  • 1
    $\begingroup$ Okay, that clears one of the issues up but it now appears that I have a poor understanding of Miller indexes all together. I do not see how all planes in the first picture of the second row are part of the same set. Following the standard procedure for identifying Miller indexes I would have indexed the right most plane as (100) and the one just left of it (200). Any good resources to clear this up? $\endgroup$
    – user57927
    Jul 21, 2016 at 10:25
  • $\begingroup$ @user57927 I think it is pretty clear from the picture. You choose a set of atoms arranged on a plane of your choice, then move orthogonally with respect to it until you find another set arranged on a plane, etc. In the picture you are referring to, those planes happen to be separated by a distance of $a/2$, so they are $(200)$ planes (I am assuming that you know the definition of Miller index). $\endgroup$
    – valerio
    Jul 21, 2016 at 10:48
  • $\begingroup$ I once made this about diffraction, drawing in two dimensions to keep things simple: homepage.lnu.se/staff/pkumsi/1FY805/Laue.html $\endgroup$
    – user137289
    May 23, 2017 at 21:48

In answer to the question: adjacent planes are planes that are closest to one another when distance is measured along the normal to the plane. It is important to understand that every lattice point has exactly one of the infinite set of planes described by the Miller indices $(h k \ell)$ passing through it. (I will use $(h k \ell)$ instead of $(\ell m n)$ like the OP.) I agree that many explanations out there seem to lack some important information, so here is a somewhat rigorous treatment.

Disregarding some special cases, Miller indices are defined as follows. First, find the intersections of the plane in question along the three crystal axes $\pmb{a}, \pmb{b}, \pmb{c}$ in terms of multiples of the lattice constants, i.e., $m a, n b, o c$ for integers $m, n, o$. Then take the reciprocals of $m, n, o$ and find three integers $h, k, \ell$ having the same ratio, and whose greatest common divisor is 1. As an example, consider the plane that intersects the $\pmb{a}$ axis at the second lattice site, the $\pmb{b}$ axis at the third lattice site, and the $\pmb{c}$ axis at the first lattice site. The reciprocals of $2, 3, 1$ are $\frac{1}{2}, \frac{1}{3}, 1$, which have the same ratio as $3, 2, 6$. The plane is thus called $(h k \ell) = (326)$.

To find the distance between adjacent planes, it helps to use the ``reciprocal lattice vectors'', which may be defined as $$ \pmb{a^*} = V^{-1}\pmb{b} \times \pmb{c} \, , \quad \pmb{b^*} = V^{-1}\pmb{c} \times \pmb{a} \, , \quad \pmb{c^*} = V^{-1}\pmb{a} \times \pmb{b} \, $$ where $V = \pmb{a} \cdot ( \pmb{b} \times \pmb{c})$ is the volume of the unit cell. By construction, these have the convenient property that, for example, $\pmb{a} \cdot \pmb{a^*} = 1$, while $\pmb{a} \cdot \pmb{b^*} = 0$, and so on. It turns out that the vector $ \pmb{H} = h \pmb{a^*} + k \pmb{b^*} + \ell \pmb{c^*} $ is normal to the $(h k \ell)$ plane. This can be demonstrated by showing that the dot products of $\pmb{H}$ with two non-colinear vectors in the $(h k \ell)$-plane, for example, $n \pmb{b} - m \pmb{a}$ and $o \pmb{c} - n \pmb{b}$, are zero.

Consider now the plane $P_0$ that passes through the lattice point at the origin and is defined by $ \pmb{H} \cdot \pmb{r} = 0 \, , $ where $ \pmb{r} = x \pmb{a} + y \pmb{b} + z \pmb{c} $ for coordinates $x, y, z$. Because of the convenient properties of the reciprocal lattice vectors described above, we can rewrite $\pmb{H} \cdot \pmb{r} = 0$ as $h x + k y + \ell z = 0$. The lattice points are those $\pmb{r}$ for which $x, y, z$ are integers, call them $p, q, s$, i.e., we have $h p + k q + \ell s = 0$. The origin is the trivial case, where $p = q = s = 0$.

We now wish to find the closest plane, call it $P_1$, by moving from the origin along the positive $\pmb{H}$ direction. The equation of $P_1$ is $\pmb{H} \cdot \pmb{r} = \delta$, or $h p + k q + \ell s = \delta$ for some delta. The geometrical interpretation of the dot product means that $P_1$ should possess the smallest value of $\delta$ possible. Furthermore, because $h, k, \ell$ and $p, q, s$ are all integers, so too must be $\delta$. The smallest possible integer value of $\delta$ is 1. We are guaranteed to be able to find $p, q, s$ satisfying $h p + k q + \ell s = 1$ because of Bezout's identity, which says that for two integers $a$ and $b$ (not the same $a$ and $b$ as above, but we are running out of variable names) with greatest common factor $f$ (written $\mathrm{gcd}(a,b)=f$), there exist integer $x$ and $y$ (again, not the $x$ and $y$ above) such that $ax + by = f$. This generalizes to more than one pair of integers. Thus, we can always find $p, q, s$ such that $h p + k q + \ell s = 1$ because $\mathrm{gcd}(h, k, \ell) = 1$.

Now that we know $\delta$, we wish to find the distance between $P_0$ and $P_1$ measured along $\pmb{H}$. This can be accomplished first by traveling along $\pmb{a}$ from the origin until we encounter $P_1$, i.e., finding $x$ so that $H \cdot (x \pmb{a}) = 1$. This has solution $x = \frac{1}{h}$, so that the vector $\pmb{v} = \frac{1}{h} \pmb{a}$ reaches from $P_0$ at the origin to $P_1$ along the $\pmb{a}$ direction. Finally then, the planar spacing $d$ is the projection of $\pmb{v}$ along the $\pmb{H}$ direction. That is $$ d = \pmb{v} \cdot \frac{\pmb{H}}{|\pmb{H}|} = \frac{1}{|\pmb{H}|} \, . $$

For the special case of the primitive cubic lattice, the lattice vectors are all orthogonal with lattice constant $a$, i.e. $\pmb{a} = a \hat{\pmb{x}}$ and so on, and the reciprocal lattice vectors are $\pmb{a^*} = \frac{1}{a} \hat{\pmb{x}}$ and so on. Therefore $|\pmb{H}| = \frac{1}{a} \sqrt{h^2 + k^2 + \ell^2}$, giving $$ d = \frac{a}{\sqrt{h^2 + k^2 + \ell^2}} \, . $$


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