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

Explanation

`a^3=(MxxZ)/(rhoxxN_0xx10^(-30))=(93 "g mol"^(-1)xx2)/("8.55 g cm"^(-3)xx6.02xx10^23 "mol"^(-1)xx10^(-30))=3.61xx10^7=36.1xx10^6`
`therefore a=(36.1)^(1//3)xx10^2 "pm"=3.304xx10^2 "pm=330.4 pm"`
`[x=(36.1)^(1//3), log x=1/3 "log"36.1 =1/3 xx"1.5575=0.519 or x =antilog 0.519=3.304"]`
For body-centred cubic, `r=sqrt3/4 a` =0.433 a =0.433 x 330.4 pm = 143.1 pm