Question

The fraction of volume occupied by atoms in a body centered cubic unit cell is: (a) \(0.32\) (b) \(0.48\) (c) \(0.68\) (d) \(0.74\)

Short answer

0.68

Step-by-step solution

Understand the Geometry of a Body-Centered Cubic Unit Cell

A body-centered cubic (BCC) unit cell consists of one atom at each corner of the cube and one atom at the center of the cube. Each corner atom is shared among eight unit cells, and thus each corner contributes 1/8 of an atom to the unit cell. In total, there are 8 corner atoms, each contributing 1/8, and 1 center atom contributing entirely to the unit cell.

Calculate the Total Number of Atoms per Unit Cell

Calculate the effective number of atoms in the BCC unit cell by adding the contributions from corner atoms and the center atom. The corners provide a total of 8 * (1/8) = 1 atom, and the center provides 1 full atom, which adds up to 1 + 1 = 2 atoms per unit cell.

Determine the Volume of One Atom

The volume of one atom in a BCC structure can be approximated as a sphere. The atomic radius, r, is related to the side length of the cube, a, by the relation a = 4r/√3. The volume of a sphere is given by V_atom = (4/3)πr³.

Calculate the Volume of the Cube

The volume of the cubic unit cell, V_cell, is the cube of the side length, a. Therefore, V_cell = a³.

Calculate the Volume Fraction Occupied by Atoms

The total volume occupied by atoms in the cell is the volume of two atoms, since there are 2 atoms per unit cell. The volume fraction is then found by dividing the total volume occupied by atoms by the volume of the cell, (2 * V_atom) / V_cell.

Insert Values and Compute the Volume Fraction

Substituting the volume of one atom and the volume of the cubic unit cell, we get the fraction as ((2 * (4/3)πr³) / a³). Since a = (4r/√3), a³ = 64r³/3√3. Hence, the fraction is ((2 * (4/3)πr³) / (64r³/3√3)) which simplifies to (π/√3)/8 or approximately 0.68.

Key concepts

Volume Fraction in BCC

Understanding the volume fraction in a body-centered cubic (BCC) unit cell is crucial for students preparing for competitive exams like the Joint Entrance Examination (JEE) as it underpins many fundamental concepts in solid-state chemistry. The volume fraction represents the proportion of space within the unit cell that is actually occupied by atoms. To visualize this, imagine the BCC unit cell as a room, where furniture represents the atoms. The volume fraction tells us how much of the room is filled with furniture.

In the BCC structure, the atoms are positioned in a way that only a fraction of the cubic unit cell's volume is occupied by the spherical atoms. This is because the atoms do not completely fill the space, leading to voids or empty spaces within the unit cell. Knowing the volume fraction helps us to understand the density and properties of the material. In the BCC arrangement, the calculation of the volume fraction involves determining the effective number of atoms in the unit cell and the space each atom occupies, which is based on its atomic radius.

Atomic Packing in Crystals

Atomic packing in crystals refers to how the atoms are arranged within the crystal structure and is a significant aspect of materials science. Crystals are made up of atoms arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions. There are different types of packing in crystalline materials, such as face-centered cubic (FCC), hexagonal close-packed (HCP), and body-centered cubic (BCC), each with varying efficiencies of space usage.

In the context of atomic packing, the term 'packing efficiency' is often used. It is defined as the percentage of volume in a crystal structure that is actually occupied by the constituent particles. Different arrangements will have different packing efficiencies. For instance, in a BCC arrangement, the volume fraction occupied by atoms is comparatively lower than in an FCC structure. This difference in packing leads to different material properties, which is why understanding atomic packing is so essential for students, especially when studying the properties and behaviors of different materials.

Solid State Chemistry for JEE

Solid state chemistry is an integral part of the chemistry syllabus for the Joint Entrance Examination (JEE), an engineering entrance assessment in India. It deals with the study of the structure, properties, and behavior of solids at the atomic or molecular level. Solid state chemistry encompasses several concepts, including crystal lattices, unit cells, the type of points, calculations involving unit cells, and the properties arising from a material's structure.

Students are required to grasp these fundamental concepts in order to solve problems related to them, such as calculating the volume fraction in a BCC unit cell. Questions on this topic test the student's ability to link the unit cell geometry with numerical calculations, which are an essential part of the solid-state chemistry coursework. It's important for students to not only memorize formulas but also understand the underlying principles to successfully apply them to varied problems. High proficiency in this area can give students an edge when tackling the competitive JEE.