Respuesta :
Answer:
For a: The edge length of the crystal is 533.5 pm
For b: The atomic radius of potassium is 231.01 pm
Explanation:
- For a:
To calculate the lattice parameter or edge length of the crystal, we use the equation:
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density = [tex]0.855g/cm^3[/tex]
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal = 39.09 g/mol
[tex]N_{A}[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]
a = edge length of unit cell = ?
Putting values in above equation, we get:
[tex]0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm[/tex]
Conversion factor: [tex]1cm=10^{10}pm[/tex]
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:
[tex]R=\frac{\sqrt{3}a}{4}[/tex]
where,
R = radius of the lattice = ?
a = edge length = 533.5 pm
Putting values in above equation, we get:
[tex]R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm[/tex]
Hence, the atomic radius of potassium is 231.01 pm
