Question

An element (atomic mass = 100 ) having BCC structure has unit cell edge length \(400 \mathrm{pm}\). The density of this element will be \(\left(N_{\Lambda}=6 \times 10^{23}\right)\) (a) \(5.2 \mathrm{~g} / \mathrm{m}\) (b) \(10.4 \mathrm{~g} / \mathrm{ml}\) (c) \(0.42 \mathrm{~g} / \mathrm{ml}\) (d) \(2.6 \mathrm{~g} / \mathrm{ml}\)

Short answer

The density of the element is 10.4 g/ml (Option b).

Step-by-step solution

- Convert Units

Convert edge length from picometers to centimeters. 1 pm = 1e-10 cm. Therefore, the edge length in cm is 400 pm * 1e-10 cm/pm = 4e-8 cm.

- Calculate the Volume of the Unit Cell

Since the unit cell is a cube, volume = edge length^3. Thus, the volume of the unit cell is \( (4 \times 10^{-8} \text{ cm})^3 \) cubic centimeters.

- Calculate the Number of Atoms per Unit Cell

For a BCC structure, there are 2 atoms per unit cell. (1 atom at each corner shared by 8 adjacent unit cells and 1 atom at the body center.)

- Determine Mass per Unit Cell

Mass of a single atom = atomic mass / Avogadro's number. Mass per unit cell = 2 atoms/unit cell * (100 g/mol) / (6e23 atoms/mol).

- Calculate Density of the Element

Density = mass / volume. Using the mass per unit cell from Step 4 and the volume from Step 2, calculate the density in g/cm^3, then convert to g/ml (1 cm^3 = 1 ml).

Key concepts

BCC structure

The body-centered cubic (BCC) structure is one of several ways in which atoms can be arranged in a crystal lattice. In the BCC structure, atoms are located at all eight corners of a cube and one atom is positioned at the center of the cube. This central atom is equidistant from all of the corner atoms and is fully contained within one unit cell.

One important aspect of the BCC structure is how we count the atoms in the unit cell. Although there's an atom at each corner, each of these corner atoms is shared among eight adjacent unit cells. Therefore, each contributes only 1/8th of an atom to the unit cell. Combining that with the one fully contained central atom gives us a total of two atoms per unit cell for BCC structures. Understanding this arrangement is crucial when we calculate properties like density, as we will see in further steps.

Unit Cell Volume

To envision the volume of a unit cell, imagine a tiny cube within a crystal lattice; this is the essential building block of the crystal structure. In the case of a BCC structure, the volume of the unit cell is calculated simply by cubing the edge length. However, it's key that all measurements are in consistent units when performing calculations.

For example, in our exercise, we first convert the edge length from picometers to centimeters because other quantities such as atomic mass and Avogadro's number are typically expressed in terms of grams and moles which correspond to centimeter-gram-second (CGS) units. The conversion ensures that when we calculate density, which is mass per volume, no discrepancies are introduced due to incompatible units.

Avogadro's Number

Avogadro's number, denoted by N_A, is a fundamental constant in chemistry that represents the number of constituent particles, usually atoms or molecules, in one mole of a substance. Its approximate value is 6.022 x 10²³ particles per mole.

When determining the mass of a single atom or molecule, Avogadro's number serves as a bridge between the macroscopic and microscopic worlds. It allows us to equate a known quantity in grams (the molar mass or atomic mass) to an exact number of atoms or molecules. When we mention the mass of an atom in the context of a BCC structure, we're typically gaining that information through the use of Avogadro's number in our calculations.

Atomic Mass

The atomic mass is a characteristic property of an element's atoms, essentially the 'weight' of the atom, and it is typically expressed in atomic mass units (amu) or grams per mole (g/mol). In chemistry and physics, atomic mass provides an estimation of the total mass of an atom and is roughly equivalent to the number of protons and neutrons in the atom (the mass of electrons is negligible).

For our calculation, we use the atomic mass to determine the mass of a single atom. By dividing the atomic mass (expressed in grams per mole) by Avogadro's number, we can calculate the mass of one atom. This step is pivotal for determining the mass of all atoms within our unit cell and subsequently the overall density of the material in question.