1.

The unit cell of a metallic element of atomic mass 10^(8) and density 10.5 g/cm^(3) is a cube with edge length of 409 PM. The structure of the crystal lattice is - (A) fcc (B) bcc (C) hcp (D) None of these

Answer»

Solution :(A)
`RHO=(ZxxM)/(Nxxa^(3))`
Here,M=108, `N_(A)`=6.023xx10^(23)`
Put on these values and solving we GET- a=409 PM=`4.09xx10^(-8)` cm.
`rho=10.5 g//cm^(3)`
n=4= number of ATOMS per unit CELL So, The structure of the crystal lattice is fcc.


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