Question

1. Show that the atomic packing factor for FCC is 0.74.
2. Show that the atomic packing factor for BCC is 0.68.
3. Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its theoretical density with the experimental value, 7.87 g/cm3.
4. Rhodium has an atomic radius of 0.1345 nm and density of 12.41 g/cm3. Determine whether it has an FCC or BCC crystal structure.
5. The unit cell for tin has tetragonal symmetry, with a and b lattice parameters of 0.583 and 0.318 nm, respectively. If the density, atomic weight, and atomic radius are 7.30 g/cm3, 118.69 g/mol, and 0.151 nm, respectively, compute the atomic packing factor.

Answer 1

1. $APF$ is the fraction of solid sphere volume in unit cell.

$APF=$ $Total volume of sphere\over Total unit cell volume$ $=$ $V_s\over V_c$

$V_s=(4).$ $4\over3$ $\Pi R^3=$ $16\over 3$ $\Pi R^3$

$V_c=$ $16R^3$ $\sqrt{2}$

$APF=$ $V_s\over V_c$ $=$ ${{16\over3}\Pi R^3\over 16R^3\sqrt{2}}$ $=0.74$

2. $BCC=0.68$

In terms of $a\infin$ $R$

Atoms per unit cell

$a=$ $4\over \sqrt{3R}$

Coordination $=no.8$

$APF=0.68$

3. In finding the density of $Fe$

$\rho=$ $_nA_{Fe}\over V_cN_A$

For $BCC, n=2$ atoms /unit cell, and

$V_c=$ $($ $4R\over V3$ $)^3$

$\rho=$ $_nA_{Fe}\over {({4R\over V3})}^3N_A$

${(2atoms/unit)(55.85g/mol)}\over[(4)(0.124\times 10^{-7}/unit cell(6.023\times10^{23}atoms/mol)]$ $=7.90g/cm^3$

The actual value of $Fe=7.87g/cm^3$

4. $Rd=0.1345nm$

$\rho=12.41g/cm^3$

$A_w=102.91g/mol$

$\rho=$ $nA_w\over V_cN_A$

$V_{c_{BCC}}=a^3_{BCC}=$ ${({4R\over V3})}^3$ $n_{BCC}=2$

$V_{c_{BCC}}=a^3_{FCC}=(2R\sqrt{2})^3$ $n_{FCC}=4$

For FCC, $\rho=$ $4(102.91g/mol)\over{{({(4)\times (0.1345\times 10^{-7}cm^3\over \sqrt{3}})}}$ $\times 6.023\times 10^23atoms/mol$

$=11.41g/cm^3$

It is $FCC$

5. $a=0.583nm$

$b=0.318nm$

$\rho=7.30g/cm^3$

$A_w=118.69g/mol$

$A_R=0.151nm$

$n=$ $\rho V_cN_A\over A_{sn}$

$=(7.3g/cm^3)(5.83^2)(3.18\times 10^{-24})(6.023\times 10^{23}atoms/mol)$

$=4atoms/ unit cell$

$\therefore$ $APF=$ $V_s\over V_c$ $=$ $(4)({4\over 3}\Pi R^3)\over(a)^2(b)$

$=$ $4({4\over 3}\Pi(1.52\times 10^{-8}cm^3))\over(a5.83\times 10^{-8}cm^2)(3.18\times 10^{-8})$

$=0.544$