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1.1. A hypothetical metal has the BCC crystal structure, a density of 7.24 g/cm3...

Apr 7, 2024

1.1. A hypothetical metal has the BCC crystal structure, a density of 7.24 g/cm3, and an atomic weight of 48.9 g/mol. Determine atomic radius of this metal. [6]

Solution by Steps

step 1

Given the density formula for a crystal structure:

\text{density} = \frac{2 \times \text{atomic mass}}{\text{volume of unit cell}}

step 2

For a BCC crystal structure, there are 2 atoms per unit cell. The volume of a unit cell is given by:

\text{volume of unit cell} = \left(\frac{4}{3}\right)\pi r^3

step 3

Substitute the given density and atomic mass into the density formula:

7.24 \text{ g/cm}^3 = \frac{2 \times 48.9 \text{ g/mol}}{\text{volume of unit cell}}

step 4

Use Avogadro's number to convert atomic mass to mass per atom:

\text{mass per atom} = \frac{48.9 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}}

step 5

Substitute the mass per atom into the density formula and solve for the volume of the unit cell:

\text{volume of unit cell} = \frac{2 \times \text{mass per atom}}{7.24 \text{ g/cm}^3}

step 6

Calculate the volume of the unit cell using the values from step 4 and step 5

step 7

Substitute the volume of the unit cell back into the volume formula and solve for the atomic radius

r

\left(\frac{4}{3}\right)\pi r^3 = \text{volume of unit cell}

step 8

Solve for

r

to find the atomic radius of the metal

Answer

[Insert final answer here]

Key Concept

Density and Volume Relationship in Crystal Structures

Explanation

The atomic radius in a crystal structure can be determined by using the density of the material, the atomic mass, and Avogadro's number to find the volume of the unit cell, and then solving for the radius.