Sodium metal crystallizes in a body centred cubic lattice wi

Question

Sodium metal crystallizes in a body centred cubic lattice with unit cell edge length of 4 mathop Alimits^o . The radius of sodium atom is __________ x 10^-1 mathop Alimits^o (Nearest integer)

DivineJEE · Accepted Answer

In a body-centered cubic (BCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by :

\sqrt{3}a = 4r

Given the unit cell edge length (a) of sodium metal as 4 Å :

a = 4 ~\mathop A\limits^o

We can now solve for the radius (r) of the sodium atom :

4r = \sqrt{3}a

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

r = \frac{\sqrt{3} \times 4 ~\mathop A\limits^o}{4}

r = \sqrt{3} ~\mathop A\limits^o

Now, we can approximate the numerical value :

r \approx 1.732 ~\mathop A\limits^o

To express the radius as a multiple of 10⁻¹

\mathop A\limits^o

:

r = 1.732 \times 10^{1} \times 10^{-1} ~\mathop A\limits^o = 17.32 \times 10^{-1} ~\mathop A\limits^o

So, the radius of the sodium atom is 17.32 × 10⁻¹

\mathop A\limits^o

.

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