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 Sodium metal crystallises in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately

Answer:

Explanation:

From this figure

      $(AC)^{2}=(AB)^{2}+(BC)^{2}$

     $(AC)^{2}=a^{2}+a^{2}=2a^{2}$

  Also,         $(AD)^{2}=(AC)^{2}+(DC)^{2}$

               $(4r)^{2}=(2a)^{2}+a^{2}$

      $16r^{2}=3a^{2}$

   $r=\frac{\sqrt{3}}{4}a$

Now, when Na metal crystallise in bcc unit cell with unit cell edge, a= 4.29Å

   We have the formula for radius

i.e,            $r=\frac{\sqrt{3}}{4}\times4.29$ Å  =1.86Å