Lithium (6.941 g/mol) packs into a body-centered cubic (BCC) unit cell. If the unit cell has a known volume of 0.04350 nm^3, what is the atomic radius of Li in units of nanometers? Show all your work.
The Correct Answer and Explanation is:
To find the atomic radius of lithium (Li) in a body-centered cubic (BCC) unit cell, we need to use the information provided: the molar mass of lithium (6.941 g/mol), the volume of the unit cell (0.04350 nm³), and the fact that lithium crystallizes in a BCC structure.
Step-by-step solution:
- Understand the relationship between the BCC structure and the unit cell:
In a BCC unit cell, there are two atoms per unit cell (one at the center and one-eighth of an atom at each of the eight corners). The relationship between the atomic radius RRR and the unit cell edge length aaa in a BCC structure is: Diagonal of the unit cell=4R\text{Diagonal of the unit cell} = 4RDiagonal of the unit cell=4R The diagonal in terms of the edge length aaa is given by the Pythagorean theorem: Diagonal=3a\text{Diagonal} = \sqrt{3}aDiagonal=3a So, equating the two expressions for the diagonal: 3a=4R\sqrt{3}a = 4R3a=4R Rearranging to solve for aaa: a=4R3a = \frac{4R}{\sqrt{3}}a=34R - Use the volume of the unit cell to find the edge length aaa:
The volume VVV of the unit cell is related to the edge length aaa by the equation: V=a3V = a^3V=a3 Given that the volume VVV is 0.04350 nm³, we can solve for aaa: a=V3=0.04350 nm33≈0.353 nma = \sqrt[3]{V} = \sqrt[3]{0.04350 \, \text{nm}^3} \approx 0.353 \, \text{nm}a=3V=30.04350nm3≈0.353nm - Solve for the atomic radius RRR:
Now that we have aaa, we can substitute it into the equation for aaa in terms of RRR from the BCC geometry: a=4R3a = \frac{4R}{\sqrt{3}}a=34R Substituting a=0.353 nma = 0.353 \, \text{nm}a=0.353nm: 0.353=4R30.353 = \frac{4R}{\sqrt{3}}0.353=34R Solving for RRR: R=0.353×34≈0.153 nmR = \frac{0.353 \times \sqrt{3}}{4} \approx 0.153 \, \text{nm}R=40.353×3≈0.153nm
Final Answer:
The atomic radius of lithium (Li) in a BCC unit cell is approximately 0.153 nm.
Explanation:
We used the relationship between the unit cell geometry in a BCC structure and the volume of the unit cell to find the edge length aaa. Using this, we derived the atomic radius RRR by applying the known relationship between aaa and RRR in the BCC structure. This method leverages both the geometric and volumetric properties of the unit cell.
