Respuesta :
Answer:
For a: The edge length of the unit cell is 314 pm
For b: The radius of the molybdenum atom is 135.9 pm
Explanation:
- For a:
To calculate the edge length for given density of metal, we use the equation:
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density = [tex]10.28g/cm^3[/tex]
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal (molybdenum) = 95.94 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]10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm[/tex]
Conversion factor used: [tex]1cm=10^{10}pm[/tex]
Hence, the edge length of the unit cell is 314 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 = 314 pm
Putting values in above equation, we get:
[tex]R=\frac{\sqrt{3}\times 314}{4}=135.9pm[/tex]
Hence, the radius of the molybdenum atom is 135.9 pm
