1.

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].

Answer»


Solution :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 gcm^(-3)`


Discussion

No Comment Found

Related InterviewSolutions