Niobium crystallises in a body-centred cubic structure. If its density is 8.55 g `cm^(-3)`, calculate the atomic radius of niobium using its atomic mass 93 u.

Explanation

`a^(3)=(M xx Z)/(d xx N_(A)) " " d=(Z xx M)/(a^(3)xx N_(A))`
`rArr 8.55 =(2xx93)/(a^(3)xx 6.023 xx 10^(23))`
`a=((2xx93)/(8.55 xx 6.023 xx 10^(23)))^(1//3)`
`a=3.306 xx 10^(-8)cm`
Now, `r=(sqrt(3))/(4) xx 3.306 xx 10^(-8)`
`=(1.732 xx 3.306xx10^(-8))/(4)`
`=1.4315xx10^(-8)cm`
`=14.31`pm