FCC type gold contain atom
1 8 × 8 + 6 × 1 2 = 4 \frac{1}{8}\times8+6\times\frac{1}{2}=4 8 1 × 8 + 6 × 2 1 = 4
Mass of unit cell of FCC type
4 × 197 6.023 × 1 0 23 = 1.30 × 1 0 − 21 4\times\frac{197}{6.023\times10^{23}}=1.30\times10^{-21} 4 × 6.023 × 1 0 23 197 = 1.30 × 1 0 − 21 g
Density of gold =19.3g/cm3
Density=m a s s o f u n i t c e l l v o l u m e o f u n i t c e l l \frac{mass of unit cell }{volume of unit cell} v o l u m eo f u ni t ce ll ma sso f u ni t ce ll
Volume of unit cell =m a s s o f u n i t c e l l d e n s i t y o f g o l d \frac{mass of unit cell }{density of gold} d e n s i t yo f g o l d ma sso f u ni t ce ll
V = 1.30 × 1 0 − 21 19.3 = 6.73 × 1 0 − 23 c m 3 V=\frac{1.30\times10^{-21}}{19.3}=6.73\times10^{-23}cm^3 V = 19.3 1.30 × 1 0 − 21 = 6.73 × 1 0 − 23 c m 3 V = a 3 = 6.73 × 1 0 − 23 c m 3 V=a^3=6.73\times10^{-23}cm^3 V = a 3 = 6.73 × 1 0 − 23 c m 3
a = 4.06 × 1 0 − 8 c m a=4.06\times10^{-8}cm a = 4.06 × 1 0 − 8 c m
For FCC type unit cell
a = 8 r a=\sqrt{8}r a = 8 r
r = a 8 = 4.06 × 1 0 − 8 8 = 1.43 × 1 0 − 8 c m r=\frac{a}{\sqrt8}=\frac{4.06\times10^{-8}}{\sqrt8}=1.43\times10^{-8}cm r = 8 a = 8 4.06 × 1 0 − 8 = 1.43 × 1 0 − 8 c m
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