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In Diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of unit cell is 356 pm then diameter of carbon atom is ________?

I got the answer but is it with correct method? Anything wrong here (in the attachment first one is purely wrong as no answer but why)

Is it wrong attempt in Second attempt? So in order to find answer I thought this as assuming in octahedral void which has distance 2(radius){assumption} and as the atoms are in alternate Tetrahedral voids so again +2(radius) and as at corner (both) (radius) + (radius) and this equals √3(a) {a = unit cell edge length }

Answer is 154.07

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  • $\begingroup$ Any help here please $\endgroup$ – TreaV Jun 10 '17 at 16:21
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It is true that the answer is √3/8 * a. But you already knew that. I am not smart enough to understand how you arrived at that. I agree that not only is 1+1 = 2 but that 2r + 4r + 2r = 8r. But I hope you already knew that as well. Restating a√3/8 = r as a√3 = 8r is also correct. After that, you have lost me completely. I never re-visit questions, so if you are sure that you can correctly justify your bald assertions that 2r + (2r)*2 + 2r = √3*a, then your solution is one of the correct ways to arrive at the answer. As for me, I had to recall enough trigonometry to determine that if the edge was "a" then the distance to any face's center from a vertex is a½√2 and that the tetrahedral angle was Θ = arc-cosine(-1/3). From there we use the three points: a vertex, a face center, and a tetrahedral occupancy to form a triangle with sides D, D and a½√2 as well as the angle between sides D and D of Θ. Simple decomposition of the triangles into right triangles gives D = a√2/(4*sin(Θ/2)) which is indeed a√3/4.

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  • $\begingroup$ Digonal of a cube is a* sqrt{3} so i applied that and imagined it as containg atoms in digonal $\endgroup$ – TreaV Jun 10 '17 at 16:56

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