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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Chemistry
Chapter
Solid State

BITSAT 2011 · Chemistry · Solid State

The fraction of total volume occupied by the atoms present in a...

The fraction of total volume occupied by the atoms present in a simple cube is

Options

  1. Aπ32\frac{\pi}{3 \sqrt{2}}
  2. Bπ42\frac{\pi}{4 \sqrt{2}}
  3. Cπ4\frac{\pi}{4}
  4. Dπ6\frac{\pi}{6}

Correct answer

D. π6\frac{\pi}{6}

Step-by-step solution

Number of atoms per unit cell =1=1 Atoms touch each other along edges. Hence r=a2(r=r=\frac{a}{2}(\mathrm{r}= radius of atom and a=\mathrm{a}= edge length) Therefore %=43πr3(2r)3=π6\%=\frac{\frac{4}{3} \pi r^{3}}{(2 r)^{3}}=\frac{\pi}{6}
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