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Solution for an element having atomic mass 60 amu.Has fcc unit cell.The edge length of the unit cell 4×10+2 pm .Find the density of the unit cell

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An element having atomic mass 60 has face-centered cubic unit cells.The edge length of the unit cell is 400pm.Find the density of the element.

6.2gcm−35.2gcm−37.2gcm−38.2gcm−3

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

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

Mass of the unit cell =Number of atoms in the unit cell × Mass of each atom

Number of atoms in the fcc unit cell =8×18+6×12

⇒4

Mass of one atom =Atomic massAvogadro number

⇒606.023×1023

∴ Mass of the unit cell =4×606.023×1023

∴ Density of unit cell =Mass of unit cellVolume of unit cell

⇒4×606.023×1023×164×10−24

⇒6.2gcm−3

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