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An element having atomic mass 63.1 g/mol...

An element having atomic mass 63.1 g/mol has face centered cubic unit cell with edge length `3.608 xx 10^(-8)` cm. Calculate the density of unit cell [Given `N_(A) = 6.022 xx 10^(23)` atoms/mol].

Text Solution

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IT is a percentage of total space filled by the particles in a crystal.
Edge length or side of a cube =a, radius of a particle = r
Particles touch each other along the edges .
`therefore a=2r`,volume of cell =`a^(3)=8r^(3)`.
A simple cubic unit cell contains only 1 atom.
Volume occupied =`4/3pir^(3)`
Packing efficiency `=`volume of one atom //volume of the unit cell `xx100 %` =`(4//3pi r^(3))/(8r^(3))"xx100=52.4%`

(b) `d=(Z.M)/(a^3.Na)`=`(4 atoms xx63.1 g//mol) //(3.608xx10^(-8)cm)^(3) xx(6.022 xx 10^(23)) `atom =`8.92 g cm^(-3)`
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