The void space in a primitive unit cell is:

Question

The void space in a primitive unit cell is: (A) 48% void space (B) 24% void space (C) 96% void space (D) 50% void space

Tardigrade Admin · Accepted Answer

a = 2r Volume of the cube =a3=(2 r)3=8r3 Packing fraction =( textVolume of one atom/ textVolume of the cube) =(((4/3) π r3)/8 r3)=(π /6)=0.52 Void fraction = 1 - 0.52 = 0.48 Void space =48 %

Q. The void space in a primitive unit cell is:

48% void space

24% void space

96% void space

50% void space

$a = 2 r$
Volume of the cube $= a^{3} = (2 r)^{3} = 8 r^{3}$
Packing fraction $= \frac{Volume of one atom}{Volume of the cube}$
$= \frac{( \frac{4}{3} π r ^{3} )}{8 r ^{3}} = \frac{π}{6} = 0.52$
Void fraction $= 1 - 0.52 = 0.48$
Void space $= 48 %$

Q. The void space in a primitive unit cell is:

48% void space

24% void space

96% void space

50% void space

a=2r Volume of the cube =a3=(2r)3=8r3 Packing fraction =Volume of the cubeVolume of one atom​ =8r3(34​πr3)​=6π​=0.52 Void fraction =1−0.52=0.48 Void space =48%

$a = 2 r$
Volume of the cube $= a^{3} = (2 r)^{3} = 8 r^{3}$
Packing fraction $= \frac{Volume of one atom}{Volume of the cube}$
$= \frac{( \frac{4}{3} π r ^{3} )}{8 r ^{3}} = \frac{π}{6} = 0.52$
Void fraction $= 1 - 0.52 = 0.48$
Void space $= 48 %$