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Find the distance between the body-centred atom and one corner atom in sodium, a = 0.424 nm. |
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Answer» Solution :The atom at the centre of the cube is SUPPOSED to touch the two nearest corner atoms. The longest diagonal at which all these three atoms LIE will be of LENGTH 4r.`therefore` distance between the centre atom and the corner atom=2r `=2 xx (sqrt(3)a)/4 = (2 xx sqrt(3) xx 0.424)/4 NM = 0.367 nm` |
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