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Diamond has face-centred cubic lattice. There are two atoms at $\left ( 0,0,0 \right )$ and $\left ( \frac{1}{4},\frac{1}{4},\frac{1}{4} \right )$ coordinates. The ratio of other carbon-carbon bond distance to the edge of the unit cell is $\sqrt{\frac{x}{16}}$ , then what is x?

(Response should be in integer only lke 78).
Option: 1
3

Option: 2
1

Option: 3
4

Option: 4
7

Answers (1)

best_answer

A distance formula needs to use for 3d-

$d_{c-c}=\sqrt{\frac{1}{16}+\frac{1}{16}+\frac{1}{16}}=\sqrt{\frac{3}{16}}$

x = 3

Posted by

Rakesh

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