Problem 20 - Chemistry The Central Science

Question

An atom of rhodium (Rh) has a diameter of about \(2.7 \times 10^{-8} \mathrm{cm} .\) (a) What is the radius of a rhodium atom in angstroms (A) and in meters \((\mathrm{m}) ?\) (b) How many Rh atoms would have to be placed side by side to span a distance of 6.0\(\mu \mathrm{m}\) ? (c) If you assume that the Rh atom is a sphere, what is the volume in \(\mathrm{m}^{3}\) of a single atom?

Step-by-Step Solution

Answer

The radius of a rhodium atom is \(1.35 \times 10^{-8}\) cm, which is equivalent to 1.35 Å and \(1.35 \times 10^{-10}\) m. To span a distance of 6.0 µm, approximately 22,222 Rh atoms are needed side by side. The volume of a single Rh atom, assuming it is a sphere, is approximately \(1.03 \times 10^{-29}\) m³.

Step 1: Convert the diameter to radius in angstroms and meters

First, we convert the diameter of the rhodium atom to radius by dividing it by 2. Radius (in cm) = Diameter / 2 = \((2.7\times10^{-8}\mathrm{cm})\) / 2 Next, convert the radius from centimeters to angstroms and meters. 1 cm = \(10^{8}\) Å (angstroms) and 1 cm = 0.01 m (meters).

Step 2: Calculate the number of Rhodium atoms needed to span 6.0 µm

We are given the distance, d, to be spanned as 6.0 μm (micrometers). First, convert the distance from μm to cm. 1 μm = \(10^{-4}\) cm d (in cm) = 6.0 μm * \(10^{-4}\) Now, divide the distance by the diameter of a single rhodium atom to find the number of atoms needed to span the distance: Number of atoms = Distance / Diameter

Step 3: Calculate the volume of a single Rhodium atom

We are given the assumption that rhodium atom is a sphere. The volume formula for a sphere is: Volume = \(\frac{4}{3}\pi r^{3}\) Using the radius in meters, calculate the volume of a single rhodium atom in m³.

Key Concepts

Atomic Radius

The concept of the atomic radius is crucial in understanding the size of an atom. The atomic radius is half the distance of the full diameter of the atom. It essentially measures how large an atom's electron cloud extends in space. For the rhodium atom, given a diameter of \(2.7 \times 10^{-8} \text{ cm}\), we first determine its radius by dividing the diameter by 2. The calculated radius will be\(1.35 \times 10^{-8} \text{ cm}\).
Next, when computing the radius in different units, such as in angstroms or meters, a few conversions are necessary:

• 1 cm = \(10^{8}\) Å (angstroms)
• 1 cm = 0.01 m (meters)

By employing these conversions, the radius can be expressed as \(1.35 \times 10^{-8} \text{ cm} = 1.35 \, Å \) and \(1.35 \times 10^{-10} \text{ m}\). Understanding these unit conversions helps to grasp atomic dimensions more clearly.

Unit Conversions in Chemistry

In chemical calculations, converting units is often necessary to ensure consistency and accuracy in results. For example, when dealing with the size of atoms or molecules, units like centimeters, meters, micrometers, and angstroms are commonly used.
Here's a simple guide to some of these conversions:

• 1 micrometer (μm) is \(10^{-4}\) cm. Thus, to convert 6.0 μm into centimeters, multiply by \(10^{-4}\), resulting in \(6.0 \times 10^{-4}\) cm.
• 1 angstrom (Å) is \(10^{-8}\) cm, used for atomic-scale distances.

These conversions are crucial because they allow measurements to be seamlessly translated into various units, providing a clear comparison across different scales. For example, determining how many atoms fit into a certain distance requires converting that distance into a consistent unit, like comparing the 6.0 μm space with the atom's diameter in cm.

Volume of Spheres

Assuming atoms are spherical simplifies calculations of their volumes. Spheres have a simple, well-known formula for volume, making it easier to calculate the space an atom occupies. The formula for the volume \(V\) of a sphere is given by:\[V = \frac{4}{3}\pi r^{3}\]Where \(r\) is the radius of the sphere.
For a rhodium atom with a radius calculated in meters (\(1.35 \times 10^{-10} \text{ m}\)), its volume can be determined by plugging the value into the sphere's volume formula.
Application of this formula helps to estimate how much space each atom occupies, which is vital in diverse fields such as material science and chemistry. It reflects the true scale of atomic and molecular computations, helping us understand the actual size of atoms in contrast to macroscopic objects. Such calculations illuminate how seemingly tiny units combine to form substantial materials.