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A Metalcrystallisesinto twocubicfacesnamelyface centered(fcc)and bodycentered(bcc). Whoseunitcelledge lengthare 3.5 Å respectively.Findthe ratioof thedensitiesof fcc andbcc.

Answer»


Solution :Edge lengthof unitcell of fcc METAL`=3.5 Å = 3.5 Å = 3.5 xx 10^(-8)` cm
Edgelengthof unitcell of thebcc metal`=3 Å = 3 xx 10^(-8) cm`
Density `d= (z xx M)/(a^(3) xx N_(A))`
Where z= NUMBER of the Fe atomsin theunitcell
M= ATOMICMASS ofmetal
a= Edgelengthof unitcell
`N_(A) =` Avogadro number
`:.` For fcc unitcell = z =4
Forbcc unitcell= z= 2
`:. ("Densityof fccunit cell ")/("Densityof bccunit cell") = (d_("fcc"))/(d_("bcc"))`
`(z_("fcc")xx M)/(a_("fcc")^(3)) xx (N_(A))/(z_("bcc")xxM)(a_("bcc")^(3))xx N_(A)`
`=(z_("fcc"))/(z_("bcc")).(a_("bcc"))/(a_("fcc")^(3))`
`=(4)/(2) (3xx10^(-8))/(3.5xx 10^(-8))`
`= 1.26`


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