Problem 17 - Chemistry: The Central Science

Question

The radius of an atom of krypton \((\mathrm{Kr})\) is about \(1.9 \AA .\) (a) Express this distance in nanometers (nm) and in picometers (pm). (b) How many krypton atoms would have to be lined up to span \(1.0 \mathrm{~mm} ?(\mathrm{c})\) If the atom is assumed to be a sphere, what is the volume in \(\mathrm{cm}^{3}\) of a single \(\mathrm{Kr}\) atom?

Step-by-Step Solution

Answer

The radius of a krypton atom is approximately 0.19 nm and 190 pm. About 2,631,579 krypton atoms need to be lined up to span 1.0 mm. Assuming the krypton atom is a sphere, its volume is approximately \(7.194 \times 10^{-23}\, cm^{3}\).

Step 1: Converting the radius to nanometers and picometers

Here we have the radius given in Angstroms (Å), which is equivalent to \(1 \times 10^{-10}\) meters. We will convert it to nanometers (nm) and picometers (pm). To convert from Angstroms to nanometers, we know that 1 Å = 0.1 nm. So we'll multiply the given radius by this conversion factor: \(1.9\, Å \times \frac{0.1\, nm}{1\, Å} = 0.19\, nm\) Next, we'll convert from Angstroms to picometers. We know that 1 Å = 100 pm. Multiply the given radius by this conversion factor: \(1.9\, Å \times \frac{100\, pm}{1\, Å} = 190\, pm\) Now we have the radius expressed in nanometers (0.19 nm) and picometers (190 pm).

Step 2: Determining the number of krypton atoms to span 1.0 mm

To find out how many krypton atoms need to be lined up to span 1.0 mm, we need to first convert the length from mm to nanometers since our given radius is in nanometers. 1 mm = \(1 \times 10^{3}\) µm = \(1 \times 10^{6}\) nm Now we can divide the given length (in nm) by the diameter of a single krypton atom (twice the radius) to find the number of atoms needed: \(\frac{1.0 \times 10^{6}\, nm}{2 \times 0.19\, nm} = \frac{1.0 \times 10^{6}\, nm}{0.38\, nm} \approx 2,631,579\) So, approximately 2,631,579 krypton atoms would have to be lined up to span 1.0 mm.

Step 3: Calculating the volume of a single krypton atom

To find the volume of a single krypton atom, we'll assume it's spherical and use the formula for the volume of a sphere: \(V = \frac{4}{3}\pi r^{3}\) First, let's convert the radius to cm since we need the volume in \(\mathrm{cm}^{3}\). We know that: 1 nm = \(1 \times 10^{-7}\) cm So, 0.19 nm = \(0.19 \times 10^{-7}\) cm Now, we'll plug in the radius in cm into the formula: \(V = \frac{4}{3}\pi (0.19 \times 10^{-7}\, cm)^{3} = \frac{4}{3}\pi (5.727 \times 10^{-23}\, cm^{3}) \approx 7.194 \times 10^{-23}\, cm^{3}\) Thus, the volume of a single krypton atom is approximately \(7.194 \times 10^{-23}\, cm^{3}\).

Key Concepts

Unit Conversion

Unit conversion is the process of converting a quantity expressed in one set of units to another, without changing its overall value. It's essential in scientific calculations because measurements can be given in many different units depending on the context.

In our example, we have the radius of a krypton atom expressed in Angstroms (\( \text{Å} \)). An Angstrom is a convenient unit for measuring atomic scales and is equal to \(1 \times 10^{-10}\) meters.

To express this radius in nanometers (nm) and picometers (pm), we use conversion factors:

• 1 Angstrom (\( \text{Å} \)) = 0.1 nanometers (nm)
• 1 Angstrom (\( \text{Å} \)) = 100 picometers (pm)

These factors help translate the radius from Angstroms to more standard measurement units used in other fields of study.

Converting measurements using these factors ensures precision and ease when performing calculations that might require differing units.

Spherical Geometry

Spherical geometry is the branch of geometry concerned with the properties and relations of figures on the surface of a sphere. Unlike Euclidean geometry that deals with flat surfaces, spherical geometry addresses curved surfaces.

Atoms, such as krypton, are often idealized as spheres. This idealization simplifies calculations of properties like volume, surface area, and density. When performing operations that involve atoms, assuming a spherical shape allows us to use formulas from spherical geometry, such as those for computing the volume or surface area of a sphere.

Understanding spherical geometry is vital when studying the structure of atoms because it provides a practical model that approximates the behavior and characteristics of atoms in a tangible way.

Atomic Radius

The atomic radius is one of the most fundamental measurements that can be taken of an atom. It represents the size of an atom and is typically defined as the distance from the nucleus to the furthest orbital of electrons.

In the case of krypton, the atomic radius is given in Angstroms (\(1.9\, \text{Å}\)). This helps us understand how large the atom is in a way that can be easily visualized when compared with other atomic radii.

Understanding the atomic radius is crucial for:

• Predicting how atoms will interact in chemical reactions.
• Understanding the physical properties of elements.
• Determining the packing of atoms in solid forms.

Atomic radius measurements are foundational in studying chemistry and physics because they directly influence the behavior and properties of elements and compounds.

Volume Calculation

Volume calculations are essential in physics and chemistry because they allow us to comprehend the space that a particular object occupies. For atoms, especially when assumed to be ideal spheres, the volume is calculated using the formula for a sphere's volume: \[ V = \frac{4}{3} \pi r^3 \]Where:

• \( r \) is the radius of the sphere.
• \(\pi\) is a constant roughly equal to 3.14159.

For the krypton atom, converting the radius to centimeters (\( 0.19 \times 10^{-7} \text{ cm}\) ) precedes volume calculation because the desired volume is in cubic centimeters (\( \text{cm}^3 \)).

This conversion gives an approximate volume for a krypton atom as \(7.194 \times 10^{-23} \text{ cm}^3 \), helping to quantify its space within a sample or in reactions. Accurate atomic volume calculations assist in predicting behaviors like gas pressure and diffusion rates in wider analyses of material characteristics.