<u>Answer:</u>
<u>For a:</u> The edge length of the crystal is 533.5 pm
<u>For b:</u> The atomic radius of potassium is 231.01 pm
<u>Explanation:</u>
To calculate the lattice parameter or edge length of the crystal, we use the equation:

where,
= density = 
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal = 39.09 g/mol
= Avogadro's number = 
a = edge length of unit cell = ?
Putting values in above equation, we get:

<u>Conversion factor:</u> 
Hence, the edge length of the crystal is 533.5 pm
To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

where,
R = radius of the lattice = ?
a = edge length = 533.5 pm
Putting values in above equation, we get:

Hence, the atomic radius of potassium is 231.01 pm