19
$\begingroup$

I often find that when talking to experimentalists outside of the field of nanomaterials, it is nearly impossible to explain Miller indices concisely. As this often comes up during talks from other students or even faculty, how would you suggest it be explained assuming they have only a generic chemistry background?

$\endgroup$
9
  • 3
    $\begingroup$ I find the generalization in the title and question rather surprising. Is it intended to target some specific group of experimentalists? Because crystallographers and others using scattering techniques tend to be very familiar with Miller indices. (After all, Miller introduced his notation in the context of crystallography.) $\endgroup$
    – Anyon
    Nov 4, 2020 at 22:06
  • $\begingroup$ I agree that the title change is not warranted because the problem is even relevant when talking to other computational chemists unfamiliar with solid state chemistry (surprisingly common). $\endgroup$ Nov 4, 2020 at 22:55
  • 1
    $\begingroup$ @Anyon I have slightly reverted the title to reflect what I meant more accurately. $\endgroup$ Nov 4, 2020 at 23:23
  • $\begingroup$ Also, I work with nanomaterials and didn't need Miller indices so far :) $\endgroup$
    – Camps
    Nov 5, 2020 at 3:03
  • 1
    $\begingroup$ @TristanMaxson - I fully agree that any materials scientist or solid state physicist should at least know what Miller indices are, in general. Chemistry less so. $\endgroup$
    – Jon Custer
    Nov 5, 2020 at 16:07

4 Answers 4

19
$\begingroup$

You can see it with VESTA software. For example, we can see the different lattice planes of NaCl crystal.

  • [001] plane of NaCl:

enter image description here

  • [101] plane of NaCl:

enter image description here

  • [111] plane of NaCl:

enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ This can be used together with the explanation given by @jason-m-gray in order to visualize them. $\endgroup$
    – Camps
    Nov 4, 2020 at 12:29
  • $\begingroup$ So the Miller index is just a normal vector to the plane? $\endgroup$
    – user1499
    Nov 5, 2020 at 18:57
  • $\begingroup$ @Servaes More or less this tends to be correct but there are different ways they are used where this doesnt always seem true. I have accepted this answer since it really comes down to a visual is required. It might not be possible to explain quickly without one. $\endgroup$ Nov 7, 2020 at 14:43
13
$\begingroup$

Assuming a generic chemistry background I wouldn't assume that knowledge of crystal structure would be too in depth at an undergraduate level. It is definitely encountered, but depending on the type of chemistry you want to go into, you probably never deal with solid state chemistry. I would first explain briefly how crystals are described by periodic lattices, with basis vectors that describe translations of atoms to build the structure. Then I would say that Miller indices are related to planes that intersect the basis vectors. Since you want to talk about planes that you can visually picture inside of the unit cell (for convenience), you talk about intersections at fractional distances along the basis vectors. The indices are the reciprocals of the fractions, once again for convenience. Then go on to say that if the plane never intersects a basis vector, you say it has an infinite intersection distance, and the reciprocal is 0. Diffraction peak locations of light is unique for crystals of the same structure, because diffraction occurs at the miller planes.

$\endgroup$
5
$\begingroup$

I would propably explain that there are different planes within a crystal, show some of them in an animation or pyhsical prop and depending on the depth of the presentation just omit the numbering and details.

$\endgroup$
4
$\begingroup$

Take care in the figure posted by Jack that the [hkl] notation actually represents the vector plane, that is the direction perpendicular to the plane. The plane are indexed as (hkl).

for example, the first figure should be read as (001) plane of NaCl, whereas [001] represents the direction along the c-axis.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .