Chapter 5: Problem 43
Calculate the atomic radius of each of the following elements (b) tantalum, bcc structure, density \(16.654 \mathrm{~g} \cdot \mathrm{cm}^{-3}\).
Short Answer
Expert verified
The atomic radius of tantalum in a BCC structure is determined using its edge length calculated from its density and molar mass, then applying the BCC geometry to relate edge length and atomic radius.
Step by step solution
01
Recall the relationship between unit cell edge length and atomic radius in BCC
For a body-centered cubic (BCC) structure, the atoms at the corners of the cube touch the atom in the center along the cube's body diagonal. The body diagonal is equal to \( \sqrt{3}a \) if \( a \) is the cube edge length. Since the body diagonal consists of 4 atomic radii in a BCC cell, the atomic radius \( r \) can be found by \( r = \frac{\sqrt{3}a}{4} \).
02
Express the volume in terms of edge length
The volume \( V \) of the unit cell can be expressed as \( V = a^3 \) where \( a \) is the edge length of the cube.
03
Calculate the mass of a unit cell
In a BCC structure, there are 2 atoms per unit cell (8 corner atoms each contributing 1/8th of its volume and 1 center atom contributing its entire volume). The mass of a unit cell can be calculated using the formula \( m = 2 \times M/N_A \) where \( M \) is the molar mass of the element, and \( N_A \) is Avogadro's number. For tantalum \( (Ta) \) with molar mass approximately 180.95 g/mol, the mass of the unit cell is therefore calculated based on this data.
04
Find the cube edge length (a)
The density \( \rho \) can be related to the mass of the unit cell and its volume by \( \rho = m/V \). Using this relationship, we can solve for the edge length \( a \) since \( m \) is known from the previous step and the density is given as \( 16.654 \text{g/cm}^3 \). This yields \( a = \sqrt[3]{m/\rho} \).
05
Calculate the atomic radius (r)
Finally, using the relationship from Step 1, the atomic radius \( r \) can be found by inserting the value of \( a \) into the equation \( r = \frac{\sqrt{3}a}{4} \) to obtain the atomic radius of tantalum in the BCC structure.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Body-Centered Cubic Structure
The body-centered cubic (BCC) structure is one of the fundamental crystalline arrangements found in metals and other solid materials. In a BCC structure, each unit cell—which is the smallest repeating unit of the crystal lattice—consists of one atom at each of the eight corners of a cube and one atom right at the center, hence the term ‘body-centered’. These atoms are not separate; they represent the positions that a repeating pattern of the same atom occupies within the crystal.
Understanding the BCC arrangement is crucial when studying material properties, as it directly affects the density and mechanical characteristics of the material. In the case of the BCC structure, atoms touch along the body diagonal of the cube, which helps in determining the relationship between the atoms’ radii and the cube's edge length. For students calculating atomic radius, comprehending this geometric relationship aids in visualizing how the atoms fit within the unit cell and interact with each other.
Understanding the BCC arrangement is crucial when studying material properties, as it directly affects the density and mechanical characteristics of the material. In the case of the BCC structure, atoms touch along the body diagonal of the cube, which helps in determining the relationship between the atoms’ radii and the cube's edge length. For students calculating atomic radius, comprehending this geometric relationship aids in visualizing how the atoms fit within the unit cell and interact with each other.
Unit Cell Volume
Volume plays a pivotal role in understanding the unit cell, as it denotes the space an individual cell occupies. It can reveal critical information about the structure when combined with density and mass. In a BCC crystal structure, the unit cell has a simple cubic geometry, making the calculation of its volume straightforward: it's the cube of the edge length, symbolically represented as V = a^3, where a is the edge length of the cube.
This volume, when multiplied by the density, yields the mass of the unit cell, a vital step in calculating the atomic radius from a known density. For educational purposes, it is crucial for students to internalize the relationship between edge length and volume—not only can it help figure out the number of atoms within a cell but also allows them to calculate other properties like atomic mass when combined with density and Avogadro's number.
This volume, when multiplied by the density, yields the mass of the unit cell, a vital step in calculating the atomic radius from a known density. For educational purposes, it is crucial for students to internalize the relationship between edge length and volume—not only can it help figure out the number of atoms within a cell but also allows them to calculate other properties like atomic mass when combined with density and Avogadro's number.
Avogadro's Number
Avogadro's number, N_A, is a fundamental constant in chemistry and physics, representing the number of atoms, ions, or molecules in one mole of a substance. The value is approximately 6.022 x 10^23 per mole, establishing a bridge between the microscopic world of atoms and the macroscopic world of grams and centimeters that we measure in the lab.
When solving problems in crystallography, Avogadro's number enables the conversion from molar mass (the mass of one mole of an element) to the mass of a single atom. This conversion is particularly useful when calculating the mass of a unit cell in a crystal lattice, as it requires knowing the mass of individual atoms to determine properties like atomic radius. The ability to use Avogadro’s number to switch between the macroscopic and microscopic is a cornerstone of chemistry that underpins many calculations and understanding it can help students with a vast range of topics, from gas laws to stoichiometry and beyond.
When solving problems in crystallography, Avogadro's number enables the conversion from molar mass (the mass of one mole of an element) to the mass of a single atom. This conversion is particularly useful when calculating the mass of a unit cell in a crystal lattice, as it requires knowing the mass of individual atoms to determine properties like atomic radius. The ability to use Avogadro’s number to switch between the macroscopic and microscopic is a cornerstone of chemistry that underpins many calculations and understanding it can help students with a vast range of topics, from gas laws to stoichiometry and beyond.