Niobium crystallizes in a body centred cubic structure. If density is 8.55 `"gcm"^(-3)`, calculate atomic radius of niobium, given that its atomic mass is `92.9mu`.

Explanation

`M=92.9"g mol"*d=8.55"g cm"^(-3)`
Z = 2 (for b.c.c. lattice) a = ?
`a^(3)=(MxxZ)/(d xxN_(A))`
`=((92.9"g mol"^(-1))xx2)/(8.55"cm"^(-3)xx6.02xx10^(23)"mol"^(-1))`
`=3.61xx10^(-23)=36.1xx10^(24)`
`a=(36.1)^(1//3)xx10^(8)"cm"`
`=3.304xx10^(-8)"cm"`
`=3.304xx10^(-8)xx10^(8)"pm"`
= 33.04 pm
`[x=(36.1)^(1//3),"log"x=1/(3)"log"36.1=1/(3)xx1.5575=0.519rArrx="antilog"0.519=3.304]`
For body-centred cubic, `r=sqrt(3)/(4)a=0.433a`
`=0.433xx330.4"pm"=143.1"pm"`