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JEE Main 20139 AprilOnline
Exam
JEE Main
Year
2013
Date
9 April
Mode
Online
Subject
Chemistry
Chapter
Solid State

JEE Main 2013 · 9 April · Online · Chemistry · Solid State

An element having an atomic radius of 0.14 ~nm crystallizes in an...

An element having an atomic radius of 0.14 nm0.14 \mathrm{~nm} crystallizes in an fccf_{c c} unit cell. What is the length of a side of the cell ?

Options

  1. A0.56 nm0.56 \mathrm{~nm}
  2. B0.24 nm0.24 \mathrm{~nm}
  3. C0.96 nm0.96 \mathrm{~nm}
  4. D0.4 nm0.4 \mathrm{~nm}

Correct answer

D. 0.4 nm0.4 \mathrm{~nm}

Step-by-step solution

For a fcc unit cell r=2a4a=4r2=22×0.14=0.39≈0.4 nm.\begin{aligned} & r=\frac{\sqrt{2} a}{4} \\ & a=\frac{4 r}{\sqrt{2}}=2 \sqrt{2} \times 0.14=0.39 \approx 0.4 \mathrm{~nm}. \end{aligned}
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