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

Question

An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radiusis: (A) (√3/4)× 288 pm (B) (√2/4)× 288 pm (C) (4/√3)× 288 pm (D) (4/√2)× 288 pm

Tardigrade Admin · Accepted Answer

In bcc unit cell, atoms lying along a body diagonal touch each other Length of body diagonal =√3a √3a=4r <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/e69068d0646ac6b79aceca8bb3140d64-.png" /> r=(√3/4)a As a=288 pm ∴ r=(√3/4)×288Pm

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

$\frac{3}{4} \times 288 p m$

$\frac{2}{4} \times 288 p m$

$\frac{4}{3} \times 288 p m$

$\frac{4}{2} \times 288 p m$

In bcc unit cell, atoms lying along a body diagonal touch each other
Length of body diagonal $= 3 a$
$3 a = 4 r$

$r = \frac{3}{4} a$
As $a = 288 p m$
$∴ r = \frac{3}{4} \times 288 P m$

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

43​​×288pm

42​​×288pm

3​4​×288pm

2​4​×288pm

In bcc unit cell, atoms lying along a body diagonal touch each other Length of body diagonal =3​a 3​a=4r r=43​​a As a=288pm ∴r=43​​×288Pm

$\frac{3}{4} \times 288 p m$

$\frac{2}{4} \times 288 p m$

$\frac{4}{3} \times 288 p m$

$\frac{4}{2} \times 288 p m$

In bcc unit cell, atoms lying along a body diagonal touch each other
Length of body diagonal $= 3 a$
$3 a = 4 r$

$r = \frac{3}{4} a$
As $a = 288 p m$
$∴ r = \frac{3}{4} \times 288 P m$