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

We are given that
Atomic mass of Nb = 93 u
Density (d) = 8.55 g `cm^(-3)`
We know that
Avogadro.s number `(N_A) = 6.02xx10^(23) mol^(-1)`
z for bcc unit cell =2
We are to find out the atomic radius (r) of Nb.
Using the relation,
`d=(zxxm)/(N_Axxa^3xx10^(-30))`
`a^3=(zxxM)/(N_Axxd xx 10^(-30))=(2xx93u)/(6.02xx10^(23)mol^(-1)xx8.55gcm^(-3)xx10^(-30))`
`a^3=3.61xx10^(7)`
`a=root(3)(3.61xx10^7)=root(3)(3.61xx10^6)=330pm`
For bee, we know that
Radius (r) = `(sqrt3)/4a=(sqrt3)/4xx330=142.89pm`