Q.Potassium has a bcc structure with nearest neighbor distance $4.52\ \text{\AA}$. Its atomic weight is 39. Its density will be

Step 1. For a bcc structure, the nearest-neighbour distance equals $2r$, and $2r$ relates to the edge length by $\sqrt{3}a = 4r$, so $a = \dfrac{4r}{\sqrt{3}} = \dfrac{2\times(\text{nearest-neighbour distance})}{\sqrt{3}}$.

Step 2. With nearest-neighbour distance $= 4.52\times10^{-8}$cm: $a = \dfrac{2\times4.52\times10^{-8}}{1.732} = 5.22\times10^{-8}$cm.

Step 3. $a^3 = (5.22\times10^{-8})^3 \approx 1.422\times10^{-22}\,\text{cm}^3$. …

Convert the nearest-neighbour distance to the bcc edge length via a = 2d/root3, then apply the s …

• Treating the given 'nearest neighbor distance' as if it were the edge length a itself, skipping the 2d/root3 conversion -- this gives a badly wrong …