Question

if the edge length of
the unit Cell
of a
metal which form Bcc unit cell
is 300pm
then

the radius of the metal atom is​

Answer 1

Given:

Edge length of the unit Cell of a metal which form Bcc unit cell is 300pm.

To find:

Radius of metal atom?

Calculation:

In case of BCC, we can use PYTHAGORAS' THEOREM as follows:

• Let body diagonal be b ,face diagonal be d and edge length be a.

$\rm \: {b}^{2} = {d}^{2} + {a}^{2}$

• Now, face diagonal is represented as (a² + a²).

$\rm \implies\: {b}^{2} = ( {a}^{2} + {a}^{2}) + {a}^{2}$

$\rm \implies\: {b}^{2} = 3{a}^{2}$

• Now, body diagonal is represented as 4r, where 'r' is radius of atom.

$\rm \implies\: {(4r)}^{2} = 3{a}^{2}$

$\rm \implies\: r = \dfrac{ \sqrt{3} a}{4}$

Now, putting the values:

$\rm \implies\: r = \dfrac{ \sqrt{3} \times 300}{4}$

$\rm \implies\: r = 129.9 \: pm$

So, radius of atom is 129.9 pm.