Question

Relation between Edge length and radius of particle in Body centred Unit cell

2

Answer 1

Answer:

For Body centered cubic cell with the relation between Edge length (a) and radius of particle (r).

AD=4r (since particles at body diagonal touch each other)

By Pythagoras' theorem

$AC^{2} = AB^{2} +BC^{2}$

$AC^{2} +a^{2} +a^{2} =2a^{2}$

$AC=\sqrt{2a}$     Equation 1

Again by Pythagoras' theorem in second half

$AD^{2} =AC^{2} +CD^{2}$

$AD^{2} =(\sqrt{Aa}) ^{2} +(a)^{2}$

$AD^{2} +2a^{2} +a^{2} =3a^{2}$

$AD=\sqrt{3a}$      Equation 2

From relation 1 and 2 we get,

$\sqrt{3a} =4r$

$a=\frac{4}{\sqrt{3} } r$

so the relation is $\sqrt{3a}=4r$