Sodium metal crystallises in a body centred cubic lattice with unit cell edge length of $$4$$ A. The radius of sodium atom is _____ $$\times 10^{-1}$$ A. (Nearest integer)

Sodium crystallizes in a BCC lattice with unit cell edge length $$a = 4$$ Angstrom. We need to find the radius of the sodium atom.

First, recall the relationship between atomic radius and edge length in BCC.

In a body-centred cubic (BCC) structure, atoms touch along the body diagonal. The body diagonal of a cube with edge length $$a$$ has length $$\sqrt{3}a$$. Along this diagonal, there are 4 atomic radii (the corner atom contributes $$r$$, the body-centre atom contributes $$2r$$, and the opposite corner contributes $$r$$):

$$ 4r = \sqrt{3}a $$

Next, solve for $$r$$.

$$ r = \frac{\sqrt{3}a}{4} = \frac{\sqrt{3} \times 4}{4} = \sqrt{3} \approx 1.732\;\text{Angstrom} $$

Now, express in the required form.

$$ r = 1.732\;\text{A} = 17.32 \times 10^{-1}\;\text{A} $$

Rounding to the nearest integer: $$17 \times 10^{-1}$$ A.

The answer is 17.