# If I break a crystal, would the “broken” pieces also have the same crystal shape? [closed]

If I've a crystal of some particular geometrical shape and i hammer it and it breaks into small pieces. Will the small pieces have the same shape as of the parent crystal?

• No, it won't. $\ce{}$ – Nilay Ghosh Jun 30 '20 at 4:49
• Thank you, and this cost me my whole chemistry stack fortune. – Kashmiri Jun 30 '20 at 5:08
• @M.Farooq I would disagree with you on this one. The question has many signs of a one-liner "lazy" request that shows no research whatsoever and lacks clarity (and punctuation). What is "the same shape", exactly? Preservation of crystal symmetry, morphology, linear/angular parameters of the bulk sample, their ratio, or something else? OP might want to have a look at the old video by Alan Holden demonstrating what cleavage is (if that's what the question is about, and sorry, no hammering there). – andselisk Jun 30 '20 at 7:45
• @andselisk I do apologize if my post is vague, I'm not a chemistry student. However I do want to clarify that there may be people whose native language is not English in any way. To them writing in English is itself a challenge, not to mention punctuation. It's akin, asking an Englishman to write correct Chinese. – Kashmiri Jun 30 '20 at 9:05
• @dear andselisk, it's okay and I agree with what you said. Cheers :) – Kashmiri Jun 30 '20 at 9:19

In most cases, a random hammering on a crystal will smash the later into randomly shaped, even irregular pieces of matter. As shown in the video suggested by @andselisk however, if you aim parallel to the cleavage planes you may obtain fragments which are similar to each other. This similarity however is not necessarily about the shape of the objects (as in «a cube yielded smaller cubes»), nor in the ratio of the side lengths of the smaller fragments. The similarity you may find is that the constant angle two faces enclose for one crystallographic phase of a given compound as illustrated below: