JEE Main 2023 — Solid State Question with Solution

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 :

$$a=4\stackrel{o}{A}$$

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

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

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

$$r=\frac{\sqrt{3}\times 4\stackrel{o}{A}}{4}$$

$$r=\sqrt{3}\stackrel{o}{A}$$

Now, we can approximate the numerical value :

$$r\approx 1.732\stackrel{o}{A}$$

To express the radius as a multiple of 10⁻¹ $\stackrel{o}{A}$ :

$$r=1.732\times 10^1\times 10^{-1}\stackrel{o}{A}=17.32\times 10^{-1}\stackrel{o}{A}$$

So, the radius of the sodium atom is 17.32 × 10⁻¹ $\stackrel{o}{A}$.