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Q8) An element  having  atomic mass 60 has FCC unit cells  .the edge length of the unit cell is 400pm .find the density of the element 


Solution

Unit-cell edge length  =   400 pm  =  400×10-10 cm

Volume of unit cell  =  (400×10-10cm)3  =  64×10-24 cm3

Mass of one atom  = Atomic mass / Avogadro number  ⇒  60 / (6.023×1023)

The unit cell,  contains 4 atoms, So, Number of atoms in the fcc unit cell = 4   

∴ Mass of the unit cell  =  4×60 / (6.023×1023)

∴  Density of unit cell =  Mass of unit cell / Volume of unit cell

             4 x 60                   1
⇒    -----------------   x   ---------- 
         6.023× 1023          64× 1024

⇒ 6.2 gcm−3  Ans.

 

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