calculate the density of silver which crystallises in the fcc structure the distance between the nearest silver atoms in this structure is 287pm

Asked by Akshaya Parthasarathy | 2nd Jul, 2011, 12:00: AM

Expert Answer:

For fcc   a= 1.414 d  where d= distance between the nearest silver atoms
            a= (1.414 x 287) pm
            a= 405.8 pm =405.8 x 10-10 cm
Also  for fcc  z=4 
      d  = (Z X M) /(a3 x NA)
           = 4 X107.87 /[(405.8 X10-10)3 X6.022X 1023]
           = 10.73 g cm-3

Answered by  | 4th Jul, 2011, 10:19: AM

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