Sodium metal crystallises in a body centred cubic lattice with edge length of x Å. If the radius of sodium atom is 1.86 Å, the value of x is
Show Hint
Memorize the relationship between edge length (a) and atomic radius (r) for the main cubic lattices:
- Simple Cubic (SC): \(a = 2r\) (atoms touch along the edge)
- Body-Centered Cubic (BCC): \(a\sqrt{3} = 4r\) (atoms touch along the body diagonal)
- Face-Centered Cubic (FCC): \(a\sqrt{2} = 4r\) (atoms touch along the face diagonal)
In a body-centered cubic (BCC) lattice, the atoms touch along the body diagonal of the cube.
Let 'a' be the edge length of the cubic unit cell (given as 'x' in the problem).
Let 'r' be the radius of the atom.
The length of the body diagonal of a cube with edge length 'a' is \( \sqrt{a^2+a^2+a^2} = \sqrt{3a^2} = a\sqrt{3} \).
Along this body diagonal, there is one full atom at the center and one radius from each of the two corner atoms.
So, the total length of the body diagonal in terms of the atomic radius is \( r + 2r + r = 4r \).
By equating these two expressions for the length of the body diagonal, we get the relationship between the edge length and the atomic radius for a BCC lattice:
\( a\sqrt{3} = 4r \).
We need to find the value of the edge length, \( x = a \).
\( a = \frac{4r}{\sqrt{3}} \).
We are given the radius of the sodium atom, \( r = 1.86 \) Å.
Substitute this value into the formula.
\( a = \frac{4 \times 1.86}{\sqrt{3}} = \frac{7.44}{\sqrt{3}} \).
Using the value \( \sqrt{3} \approx 1.732 \):
\( a \approx \frac{7.44}{1.732} \approx 4.295 \).
Rounding to two decimal places, the value of x is 4.29 Å.
