Question

The density of vanadium, which has the BCC structure, is 6.1 . The atomic weight of vanadium is 50.941 . Use Avogadro's number: 6.02210. Calculate the lattice parameter (Enter your answer to three significant figures.) = the atomic radius of vanadium (Enter your answer to three significant figures.) =

Answer 1

Answer:

a ≅ 3.03 × 10⁻⁸cm

R  ≅ 1.31  × 10⁻⁸cm

Explanation:

The formula for Density per unit cell is given as;

ρ = $\frac{n*M}{N_{A}*a^{3}}$

where;

ρ= density per unit cell = 6.1g/cm³

$N_A$= avogadro number = 6.022 × 10²³ atom/mol

n = the number of  atom per unit cell in a BCC lattice which is usually = 2

M = atomic weight given as = 50.941g/mol

$a^3$ = lattice parameter = ???

From the above equation;

we can make subtitute for the $a^3$  to be;

$a^3$ = $\frac{n*M}{N_A*p}$

a = $(\frac{n*M}{N_A*p})^{1/3}$

a = $\frac{2 atom*50.941g/mol}{6.022*10^{-23}atom/mol*6.1g/cm^3}$

a = 3.026 × 10⁻⁸cm

a ≅ 3.03 × 10⁻⁸cm (to  3 significant figure)

To calculate the atomic radius of vanadium in BCC Lattice:

We use the formula R = $\frac{\sqrt{3} * a}{4}$

R = $\frac{\sqrt{3} *3.03*10^{-8}cm}{4}$

R = 1.312 × 10⁻⁸cm

R ≅ 1.31  × 10⁻⁸cm (to  3 significant figure)