Q84 / 200 · NEET 2008 (Screening) · Chemistry

If the lattice parameter for a crystalline structure is $3.6 Å$, then the atomic radius in fcc crystal is

  • A
    $1.81 Å$
  • B
    $2.10 Å$
  • C
    $2.92 Å$
  • Correct answer
    D
    $1.27 Å$

Solution

Atomic radius is the distance between two nearest neighbours. For fcc crystal Atomic radius $=\frac{1}{2 \sqrt{2}} \times$ lattice parameter or $r=\frac{a}{2 \sqrt{2}}$ or $r=\frac{3.6}{2 \sqrt{2}} Å=1.27 Å$

Hence, the correct answer is option: 'D'.

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