<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:
![\rho=\frac{Z\times M}{N_{A}\times a^{3}}](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%7BZ%5Ctimes%20M%7D%7BN_%7BA%7D%5Ctimes%20a%5E%7B3%7D%7D)
where,
= density = ![0.855g/cm^3](https://tex.z-dn.net/?f=0.855g%2Fcm%5E3)
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal = 39.09 g/mol
= Avogadro's number = ![6.022\times 10^{23}](https://tex.z-dn.net/?f=6.022%5Ctimes%2010%5E%7B23%7D)
a = edge length of unit cell = ?
Putting values in above equation, we get:
![0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm](https://tex.z-dn.net/?f=0.855%3D%5Cfrac%7B2%5Ctimes%2039.09%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%5Ctimes%20%28a%29%5E3%7D%5C%5C%5C%5C%5Ca%3D5.335%5Ctimes%2010%5E%7B-8%7Dcm%3D533.5pm)
<u>Conversion factor:</u> ![1cm=10^{10}pm](https://tex.z-dn.net/?f=1cm%3D10%5E%7B10%7Dpm)
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:
![R=\frac{\sqrt{3}a}{4}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B%5Csqrt%7B3%7Da%7D%7B4%7D)
where,
R = radius of the lattice = ?
a = edge length = 533.5 pm
Putting values in above equation, we get:
![R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B%5Csqrt%7B3%7D%5Ctimes%20533.5%7D%7B4%7D%3D231.01pm)
Hence, the atomic radius of potassium is 231.01 pm