Question

how to find packing efficiency of bcc​ .....?

Answer 1

Explanation:

Body-Centred Cubic Structures

In body-centred cubic structures, the three atoms are diagonally arranged. To find the packing efficiency we consider a cube with edge length a, face diagonal length b and cube diagonal as c.

BCC structure

In △ EFD according to Pythagoras theorem

b2 = a2 + a2

b2 = 2a2

b = √2 a

Now in △ AFD according to Pythagoras theorem

c2 = a2 + b2 = a2 + 2a2

c2 = 3a2

c = √3 a

If the radius of each sphere is ‘r’ then we can write

c = 4r

√3 a = 4r

r = √3/ 4 a

As there are two atoms in the bcc structure the volume of constituent spheres will be

2 × (4/3) π r3

Therefore packing efficiency of the body-centred unit cell is 68%.

Metals like iron and chromium fall under bcc category.

♥ Hope it will help you ♥

Answer 2

${\boxed{\sf \red{Packing\: Efficiency=\dfrac{Z×Volume\:of\:Sphere}{Total\: Volume\:of\: Crystal}}}}$

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Here ,

Z for bcc is = 2

Z Refered to as Number of atoms in a unit cell.

$\Large\underline{\underline{\sf \pink{Packing\: Efficiency\:Of\:BCC:}}}$

$\implies{\sf Packing\: Efficiency=\dfrac{2×\dfrac{4}{3}×πr^3}{a^3}×100 }$

Here ,

a = Edge Lenght

r = Radius of atom

For bcc r = $\sf{\dfrac{\sqrt{3}}{4}a}$

$\implies{\sf \dfrac{2×\dfrac{4}{3}×π×\dfrac{3\sqrt{3}}{64}×a^3×100}{a^3}}$

$\implies{\sf 2×\dfrac{3.14}{3}×\dfrac{3×1.732}{16}×100 }$

$\implies{\sf \dfrac{3263}{48} }$

$\implies{\sf ≈68\% }$

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⛬ Packing Efficiency for BCC is 68%