Question

If the edge length of the unit cell of a metal which form BCC
unit cell is 300 pm then the radius of the metal atom is
O 145.1 pm
O 129.9 pm
O 75 pm
O 106 pm​

Answer 1

Answer:

129.9 pm second option

Explanation:

for bcc

use 4r=$\sqrt3$   a    

in a use 300 in pm only

then by using  this  4r=$\sqrt3$   a    you will get r

Answer 2

Given:

The edge length of the unit cell of a metal which form BCC unit cell is 300 pm.

To find:

Radius of metal atom ?

Calculation:

In BCC lattice, there is atoms at each corner and another at body centre.

So, using PYTHAGORAS' THEOREM:

• Let body diagonal be d, edge length be a and face diagonal be r:

$\rm {d}^{2} = {a}^{2} + {r}^{2}$

• Now, face diagonal expressed as a² + a².

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

$\rm \implies {d}^{2} = 3 {a}^{2}$

• Now, radius of the atoms be R:

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

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

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

$\rm \implies R \approx \: 129.9 \: pm$

So, radius of atom is approximately 129.9 pm.