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If the length of the unit cell is 5 A, the smallest distance in A between the two neighboring metal atoms in a face center cubic lattice is :

\[\begin {array} {1 1} (a)\;2.50 & \quad (b)\;5.00 \\ (c)\;7.07 & \quad (d)\;3.535 \end {array}\]
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(d) 3.535
answered Nov 7, 2013 by pady_1
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