Question

vanadium crystallizes in a body-centered cubic lattice and the density is 5.96g/cm3 what is the unit cell edge length in pm?

Answer 1

The unit cell edge length of vanadium in a body-centered cubic lattice is approximately 303.4 pm.

Step-by-step explanation:

Vanadium crystallizes in a body-centered cubic (BCC) lattice. In a BCC structure, there are atoms at the eight corners of the unit cell and one atom in the center. The edge length (a) of the unit cell can be calculated using the formula:

a = (4 * radius) / √3

Given that the density of vanadium is 5.96 g/cm3, we can use its molar mass (50.9415 g/mol) and Avogadro's number (6.022 × 10^23 atoms/mol) to calculate the radius of a vanadium atom using the formula:

density = (molar mass * number of atoms) / (volume of unit cell)

By rearranging the formulas and substituting the given values, we can find the unit cell edge length:

a = (√[3 * molar mass * Avogadro's number / (4 * density)]) / 10

Substituting the values gives:

a = (√[3 * 50.9415 * 6.022 *10^23 / (4 * 5.96)]) / 10

Simplifying the expression gives:

a ≈ 3.034 Å

To convert Ångströms (Å) to picometers (pm), multiply by 100:

a ≈ 3.034 * 100 = 303.4 pm

Answer 2

The correct answer is 305 pm

Step-by-step explanation:

The density of a unit cell (d) is given by the following equation:

`d= (zM)/(a^(3)N_(a) )`

Where M is the molar mass, Nₐ is the Avogadro number (6.023 x 10²³ atom/mol), a is the edge of unit cell and z is the number of atoms per unit cell. We have to determine the edge of the unit cell (a) and we know the molar mass from the Periodic Table (M= 50.9415) and the density (d= 5.96 g/cm³). For BCC, z= 2 atom. So, we introduce the data in the equation and we determine a:

⇒
`a= \sqrt[3]{(zM)/(dN_(a) ) }`

`a= \sqrt[3]{((2atom) (50.9415g/mol))/((5.96g/cm3)(6.023x10^(23)) ) }`

a= 3.05 x 10⁻⁸ cm= 3.05 x 10⁻¹⁰ m

We need the value in pm, so we convert from m to pm (1 pm= 10⁻¹²m):

3.05 x 10⁻¹⁰m x 1 pm/10⁻¹²m= 305.03 pm= 305 pm