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Calculate atomic packing factor for FCC crystal structure ?
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written 3.5 years ago by |
1) For FCC, we know that, effective number of atoms/unit cell i.e. n=4.
2) Volume of 1 atom = $\dfrac {4}{3} \Pi r^{3} \\$ 3) Volume of 1 unit cell = $a^{3}=(2\sqrt{2}r)^{3}$ $\because a=2\sqrt{2}r$ 4) By definition – APF= $\dfrac {n\times volume\ of\ 1\ atom}{volume\ of\ unit\ cell} \$
$=\dfrac {4 \times \left (\dfrac {4}{3}\Pi r^{3} \right ) }{(2\sqrt{2}r)^{3}}=0.74 \\$ Note that FCC has the height value of APF 5) Void space – Void space = $(1-APF)\times 100$ $=(1-0.74) \times 100=26\%$ 6) Density $(\rho)$ – $\rho=\dfrac {n\left (\dfrac {M}{N_{A}} \right )}{a^{3}} \$
$=\dfrac {4\times \left (\dfrac {M}{N_{A}} \right )}{a^{3}}\$
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