Question

The Molybdenum with an atomic radius 0.1363 nm and atomic weight 95.95 has a BCC unit cell structure. Calculate its theoretical density (g/cm^3).

Answer 1

Given :

atomic radius, r = 0.1363nm = 0.1363×10⁻⁹m

atomic wieght, M = 95.96

Cell structure is BCC (Body Centred Cubic)

For BCC, we know that no. of atoms per unit cell, z = 2

and atomic radius, r =
`(a√(3) )/(4)`

so, a =
`(4r)/(√(3))`

m = mass of each atom in a unit cell

mass of an atom =
`(M)/(N_(A) )`,

where,
`N_(A)` is Avagadro Number = 6.02×10^{23}

volume of unit cell = a^{3}

density, ρ =
`(mass of unit cell)/(volume of unit cell)`

density, ρ =
`(z* M)/(a^(3)* N_(A))`

ρ =
`(2* 95.95)/(((4* 0.1363* 10^(-9))/(√(3)))^(3)* (6.23* 10^(23)))`

ρ = 10.215gm/
`cm^(3)`