Problem 32 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Potassium has BCC structure with nearest neighbour distance \((2.5 \times \sqrt{3}) \dot{A}\). Its density will be \(\left(\mathrm{K}=39, N_{\mathrm{A}}=6 \times 10^{23}\right)\) (a) \(1.040 \mathrm{~kg} / \mathrm{m}^{3}\) (b) \(104 \mathrm{~kg} / \mathrm{m}^{3}\) (c) \(520 \mathrm{~kg} / \mathrm{m}^{3}\) (d) \(1040 \mathrm{~kg} / \mathrm{m}^{3}\)

Step-by-Step Solution

Answer

1040 kg/m^3

Step 1 - Calculate the edge length of the unit cell

In a BCC structure, the nearest neighbour distance is the diagonal of the cube that passes through the body center of the cube. Since the nearest neighbour distance is given by \(2.5 \times \sqrt{3}\textrm{ Å}\), we can use this to find the edge length \(a\) of the unit cell. The body diagonal is \(\sqrt{3}a\) and is equal to four times the radius of an atom because there are two half radii at each end plus the full radius of the atom in the body center. However, since nearest neighbour distance is between the centers of two adjacent atoms, only three radii are considered. Therefore, \(\sqrt{3}a = 4r\) where \(r = \frac{2.5}{2} = 1.25Å\). Solving for \(a\) gives us \(a = \frac{2.5 \times \sqrt{3}}{\sqrt{3}} = 2.5Å\).

Step 2 - Calculate the volume of the unit cell

The volume \(V\) of the unit cell can be calculated using the edge length \(a\) found in Step 1. The formula for the volume of a cube is \(V = a^3\). Substituting the value of \(a\) gives us \(V = (2.5Å)^3 = 15.625Å^3\). However, we need the volume in cubic meters for density calculation, so we convert \(Å^3\) to \(m^3\) by using the conversion \(1Å = 10^{-10}m\). Therefore, \(V = 15.625 \times (10^{-10})^3 m^3 = 15.625 \times 10^{-30} m^3\).

Step 3 - Calculate the mass of the unit cell

Since potassium has a BCC structure, there are 2 atoms per unit cell (1 atom at each corner shared by 8 cells, and 1 atom at the body center). The molar mass of potassium (K) is 39 grams per mole, and Avogadro's number \((N_{A})\) is \(6 \times 10^{23}\) atoms per mole. The mass of a single atom of potassium is \(\frac{39 g/mol}{6 \times 10^{23} atoms/mol}\). The mass of the unit cell is then \(2 \times \frac{39 g/mol}{6 \times 10^{23} atoms/mol}\). Converting grams to kilograms gives us the mass in kg: \(\frac{2 \times 39 \times 10^{-3} kg/mol}{6 \times 10^{23} atoms/mol}\).

Step 4 - Calculate the density of potassium

Density \(\rho\) is mass per unit volume. So, using the mass of the unit cell from Step 3 and the volume of the unit cell from Step 2, we find the density as \(\rho = \frac{\text{mass}}{\text{volume}} = \frac{2 \times 39 \times 10^{-3} kg/mol}{15.625 \times 10^{-30} m^3 \times 6 \times 10^{23}}\). Simplifying this expression gives us the density in \(\mathrm{kg/m^3}\).

Step 5 - Simplify the density calculation

Plugging the values into the density formula from Step 4, we get \(\rho = \frac{2 \times 39 \times 10^{-3}}{15.625 \times 10^{-30} \times 6 \times 10^{23}} kg/m^3\). The \(10^{-3}\) and \(10^{-30}\) terms can combine to form \(10^{-3 + 30}\) or \(10^{27}\), and the \(6\) in the denominator can be taken to the numerator by multiplying the other terms by \(\frac{1}{6}\). This simplifies to \(\rho = \frac{2 \times 39 \times 10^{27}}{15.625 \times 6 \times 10^{23}} = \frac{78 \times 10^{27}}{93.75 \times 10^{23}} = \frac{78}{93.75} \times 10^{27-23} = \frac{78}{93.75} \times 10^4 = 0.832 \times 10^4 = 8320 kg/m^3\).

Key Concepts

Body-Centered Cubic (BCC) Structure

When we talk about the body-centered cubic (BCC) structure in chemistry, we're referring to a specific arrangement of atoms within a crystal. In this structure, each corner of a cubic unit cell has an atom, and there's also an atom in the very center of the cube. This setup results in a total of 2 atoms per unit cell. The beauty of the BCC structure is its simplicity and efficiency in packing atoms in a crystal lattice.

Apart from being a fundamental concept in solid-state chemistry and materials science, understanding the BCC structure is crucial when it comes to calculating physical properties of materials, such as their density. The BCC structure is also characterized by geometrical relations that allow us to calculate distances between atoms and, with additional data, determine the unit cell's spatial dimensions.

Nearest Neighbour Distance

In the context of crystal structures, 'nearest neighbour distance' refers to the closest distance between the centers of two adjacent atoms. This distance is particularly important in BCC crystals because it helps determine the lattice parameter, which is the physical dimension of the unit cell edge. For a BCC lattice, the nearest neighbour distance is equal to the cube's body diagonal divided by \( \sqrt{3} \) because the body diagonal contains the radius of one atom located in the corner, the entirety of the central atom's diameter, and another radius from the opposite corner's atom.

To illustrate, imagine stretching a straight line from the corner of a cube through its center, where an atom sits, to the opposite corner. The length of this line in a BCC structure is crucial for calculating other properties of the material. In the exercise provided, this understanding is applied to derive the edge length of the unit cell which in turn is used in density calculations.

Avogadro's Number

Avogadro's number, denoted as \( N_A \), is a fundamental constant in chemistry that represents the number of constituent particles, typically atoms or molecules, in one mole of a substance. Its value is approximately \( 6.022 \times 10^{23} \) particles per mole. This number is vital when converting between the number of atoms and the amount of substance in moles.

Understanding Avogadro's number is essential when performing calculations involving the mass of atoms in a unit cell, as in our exercise. In a BCC structure, even though there are atoms at each of the eight corners and one atom in the center, each corner atom is shared among eight unit cells. This sharing means that only one-eighth of each corner atom is actually within a given unit cell, resulting in a total of two atoms per BCC unit cell. Avogadro's number enables us to translate the molar mass of an element, given typically in grams per mole, into the mass of a single atom, which is crucial for calculating the mass of atoms within the unit cell and, subsequently, the density of the material in question.