Question

An element has a body centered cubic (bcc)
structure with a cell edge of 288 pm. The atomic
radius is :​

Answer 1

SQUARE ROOT 3 BY 4 * 288 PM

Answer 2

It has given that, an element has a n

body centered cubic (bcc) structure with a cell edge of 288 pm.

To find : The atomic radius is ....

solution : in body centered cubic structure,

eight atoms occur at corners of a cube and one atom occurs at centre of cube.

so, diagonal of cube = r + 2r + r, where r is atomic radius.

⇒√3 × edge length of cube = 4r

⇒√3 × 288 pm = 4r

⇒r = √3/4 × 288 pm

⇒r = √3 × 288/4

⇒r = 72√3 pm = 124.7 pm

Therefore atomic radius of element is 124.7 pm