Single Choice

If a is the edge length, In BCC, the distance between two nearest atoms will be:

A$$\sqrt 3$$a
B$$\displaystyle \frac{\sqrt 3}{2}$$a
Correct Answer
C$$\displaystyle \frac{\sqrt 3}{4}$$a
D$$\displaystyle \frac{a}{2}$$

Solution

In BCC, there will be atoms at the body centre and at corners. So the distance between two nearest atoms is nothing but distance between point A and O as shown in the above image.

It is given by : $$AF= \sqrt {\left ( AD \right )^2 + \left ( FD\right )^2}= \sqrt {\left ( a \right )^2 + \left ( a\sqrt2 \right )^2} = a\sqrt3$$

$$AO=AF/2=\sqrt3a/2$$

So, the correct option is B.

SIMILAR QUESTIONS

Solid State

Number of atoms per unit cell in B.C.C. is:

Solid State

CsBr crystallizes in a body-centered cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being $$6.02\times 10^{23}$$ mol$$^{-1}$$, the density of CsBr is:

Solid State

The radius of the largest sphere which fits properly at the centre of the edge of body centred cubic unit cell is:(Edge length is represented by '$$a$$')

Solid State

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

Solid State

Body centred cubic lattice has a co-ordination number of:

Solid State

The intermetallic compound $$LiAg$$ crystallizes in a cubic lattice in which both $$Li$$ and $$Ag$$ atoms have coordination numbers of $$8$$. To what crystal class does the unit cell belong?

Solid State

The low density of alkali metals is due to

Solid State

What is the theoretical density of crystalline $$K$$?

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