Question

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate (a) the lattice parameter; and (b) the atomic radius of potassium.

Answer 1

For a: The edge length of the crystal is 533.5 pm

For b: The atomic radius of potassium is 231.01 pm

Step-by-step explanation:

• For a:

To calculate the lattice parameter or edge length of the crystal, we use the equation:

`\rho=(Z* M)/(N_(A)* a^(3))`

where,

`\rho` = density =
`0.855g/cm^3`

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal = 39.09 g/mol

`N_(A)` = Avogadro's number =
`6.022* 10^(23)`

a = edge length of unit cell = ?

Putting values in above equation, we get:

`0.855=(2* 39.09)/(6.022* 10^(23)* (a)^3)\\\\\a=5.335* 10^(-8)cm=533.5pm`

Conversion factor:
`1cm=10^(10)pm`

Hence, the edge length of the crystal is 533.5 pm

• For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

`R=(√(3)a)/(4)`

where,

R = radius of the lattice = ?

a = edge length = 533.5 pm

Putting values in above equation, we get:

`R=(√(3)* 533.5)/(4)=231.01pm`

Hence, the atomic radius of potassium is 231.01 pm