Question

Sodium metal crystallizes in the bcc structure. The metal has atomic weight 23 and
density 971 kgm^−3. Calculate the cubic lattice parameter a and the shortest distance
between atoms in this structure.

Answer 1

Answer on Question #46466-Physics-Atomic Physics

Sodium metal crystallizes in the bcc structure. The metal has atomic weight 23 and density 971 kgm$^{\wedge}$–3. Calculate the cubic lattice parameter $a$ and the shortest distance between atoms in this structure.

Solution

For the BCC structure, the density is given by:

$$\rho = \frac {2 \frac {M _ {a t} \left(1 0 ^ {- 3} \frac {k g}{g}\right)}{N _ {A}}}{a ^ {3}}.$$

Thus the lattice parameter $a$ is:

a = \left(\frac {1}{5 0 0} \frac {\frac {g}{k g}} \frac {M _ {a t}}{N _ {A} \rho}\right) ^ {\frac {1}{3}} = \left(\frac {1}{5 0 0} \frac {\frac {g}{k g} (6 . 0 2 2 \cdot 1 0 ^ {2 3} m o l ^ {- 1}) \cdot 9 7 1 \frac {k g}{m ^ {3}}}\right) ^ {\frac {1}{3}} = 4. 2 8 \cdot 1 0 ^ {- 1 0} m.

The radius of the atom, $R$, and the lattice parameter, $a$, are related.

$$R = \frac {\sqrt {3}}{4} a.$$

The shortest distance between atoms in this structure is one diameter

$$d = 2 R = 2 \cdot \frac {\sqrt {3}}{4} a = \frac {\sqrt {3}}{2} a = \frac {\sqrt {3}}{2} (4. 2 8 \cdot 1 0 ^ {- 1 0} m) = 3. 7 1 \cdot 1 0 ^ {- 1 0} m.$$

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