Respuesta :
solution:
[tex]calculate the total number of unit cells in BCC structure\\
atoms=atoms at center+atoms at corners\\
atoms=1+(8\times\frac{1}{8})\\
atoms= 2 atoms/unitcell\\
equate the length of the cube diagonal in terms of unit cell length,s and radius of atom R.\\
\sqrt{3s}=4R\\
s=\frac{4}{\sqrt{3}}R\\
consider the volume of unit cell\\
v=s^3\\
=(\frac{4}{\sqrt{3}}R)\\
=\frac{64}{3\sqrt{3}R^3}\\
consider the density of vanadium atom\\
p=\frac{nAv}{Vn}\\[/tex]
Here, Avogadro’s number is N, number of atoms in unit cell is n, and atomic mass of the vanadium is , and volume of the unit cell is V.
[tex]substitute \frac{64}{3\sqrt{3}R^3} for V.\\
6.022\times10^23 atoms/mol for N,\\
2 for n 5.96g/km^3 for p.\\
and 50.9g/mol for the radius of the vanadium atom is 0.132nm.[/tex]
Answer: The radius of vanadium atom is 132.07 pm
Explanation:
To calculate the edge length of metal, we use the equation:
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density of metal = [tex]5.96g/cm^3[/tex]
Z = number of atom in unit cell = 2 (BCC
M = atomic mass of metal = 50.9 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]5.96g/cm^3=\frac{2\times 50.9}{6.022\times 10^{23}\times (a)^3}\\\\a^3=2.836\times 10^{-23}cm^3\\\\a=\sqrt[3]{2.836\times 10^{-23}}=3.05\times 10^{-8}cm^3=305pm[/tex]
To calculate the radius, 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 = 305 pm
Putting values in above equation, we get:
[tex]R=\frac{\sqrt{3}\times 305}{4}=132.07pm[/tex]
Hence, the radius of vanadium atom is 132.07 pm
