Question

Lithium crystallizes as body centered cubic crystals. If the length of the side of unit cell is 350 \(\mathrm{pm}\), the atomic radius of lithium is: (a) \(303.1 \mathrm{pm}\) (b) \(606.2 \mathrm{pm}\) (c) \(151.5 \mathrm{pm}\) (d) \(123.7 \mathrm{pm}\).

Short answer

The atomic radius of lithium is approximately 151.5 pm.

Step-by-step solution

Understand the Body-Centered Cubic (BCC) Unit Cell

In a BCC crystal structure, atoms are located at the corners of a cube and one atom is present at the body center. The body diagonal of the cube passes through the center atom and connects two opposite corners, thus passing through two radii from two corner atoms and the entire radius of the body-centered atom. Therefore, the body diagonal is equivalent to four atomic radii.

Relate Diagonal Length to Side Length and Atomic Radius

The body diagonal (d) in terms of the side length (a) of the cube and the atomic radius (r) is given by the equation: \(d = 4r\). Since the diagonal forms a right-angled triangle with the sides of the cube, by Pythagorean theorem, we have: \(d^2 = a^2 + a^2 + a^2\), which simplifies to \(d = a\sqrt{3}\).

Express Atomic Radius in Terms of Side Length

Combining the two relationships gives us: \(4r = a\sqrt{3}\). We need to solve this equation for atomic radius (r) in terms of the side length of the cube (a).

Calculate the Atomic Radius

Substitute the given side length (a = 350 pm) into the equation and solve for r: \(4r = 350\mathrm{pm} \sqrt{3}\). Therefore, \(r = \frac{350\mathrm{pm} \sqrt{3}}{4}\).

Evaluate the Atomic Radius

Calculate the exact value of the atomic radius using the approximate value \(\sqrt{3} \approx 1.732\). Replace \(\sqrt{3}\) with 1.732 in the equation and solve for r: \(r = \frac{350 \times 1.732}{4} \approx 151.5 \mathrm{pm}\).

Key concepts

Understanding the Body Centered Cubic Unit Cell

In the world of crystal structures, the body centered cubic (BCC) unit cell is one of the simplest yet significant patterns observed in metals like lithium. Imagine a straightforward cube, like a dice from a board game, with each corner serving as an anchor point for an atom. Now, place another atom right at the very heart of the cube—this is the additional feature that distinguishes the BCC structure from a simple cubic lattice.

Every corner atom is touching the central one along the cube's body diagonal, which is a straight line running from one corner of the cube, through the center, to the opposite corner. Notably, this diagonal is a key player when it comes to calculating atomic radii. An important fact to digest here is that the body diagonal hosts four atomic radii—two halves from the corner atoms and one full radius from the center atom. This understanding is central to grasping further calculations and piecing together the connection between the physical dimensions of the cube and the size of its constituent atoms.

Atomic Radius Calculation in Crystallography

When faced with a question about atomic radius, particularly within the context of crystal structures, we turn to the geometric relationships that define the unit cell. For a BCC structure, as we've mentioned, the body diagonal encompasses four atomic radii. The side length of the unit cell is also a piece of the puzzle when we consider the relationship between the diagonal and the atomic radius.

Through the formula we derived previously, \(4r = a\sqrt{3}\), the atomic radius \(r\) can be calculated if the side length \(a\) is known. This formula is the golden ticket to finding the elusive atomic radius. Our task is simply to substitute the provided side length into this equation and do the simple arithmetic. In this way, we peel back the layers of the BCC structure to reveal the size of the atoms that are snugly fit within its confines.

The Role of Pythagorean Theorem in Crystallography

The Pythagorean theorem is not just a staple of high school geometry—it's a vital tool in the crystallographer's arsenal. To understand how this theorem fits into the puzzle, remember the body diagonal can be considered the hypotenuse of a right-angled triangle within the cube. This hypotenuse is related to the sides of the cube through the equation \(d^2 = a^2 + a^2 + a^2\), which nicely simplifies down to \(d = a\sqrt{3}\) using the Pythagorean theorem.

Why is this important? Well, the step from conceptual understanding to practical calculation hinges on the ability to bridge lengths within the cube to the atomic radii we seek. The Pythagorean theorem unearths this relationship, offering a clear path from the edge of the unit cell, through its inner geometry, to the quantifiable details of its atomic components. And so, what begins as a journey through a maze of edges and angles ends with the precise value of an atomic radius—a value that is foundational to materials science and chemistry alike.