The unit cell for tin (Sn) has tetragonal symmetry, with a and b lattice parameters of 0.583 and 0.318 nm, respectively. If its density, atomic weight, and atomic radius are 7.30 g/cm3, 118.69 g/mol, and 0.151 nm, respectively. Determine its atomic packing factor.

Respuesta :

Answer:

0.1334

Explanation:

The number of atoms per unit (n) is given by:

[tex]n=\frac{\rho a^2bN_a}{A} \\\\where\ a=5.83*10^{-8}cm,b=a=3.18*10^{-8}cm,\rho=7.3\ g/cm^3,\\N_a=Avogadro\ number=6/022*10^{23} mol^{-1},A=atomic\ weight\\=118.68 \ g/mol\\\\n=\frac{7.3* (5.83*10^{-8})^2*(3.18*10^{-8})*6.02*10^{23}}{118.69}\\\\n=4\ atoms/unit[/tex]

The atomic packing factor (APF) is:

[tex]APF=\frac{n(\frac{4\pi R^3}{3} )}{a^2b} \\\\But\ R=atomic\ radius=1.51*10^{-8}\ cm\\\\APF=\frac{4(\frac{4\pi (1.51*10^{-8})^3}{3} )}{(5.83*10^{-8})^2*3.18*10^{-8}}\\\\APF=0.1334[/tex]