Question

The radius of an atom of gold \((\mathrm{Au})\) is about \(1.35 \AA\). (a) \(\mathrm{Ex}\) press this distance in nanometers (nm) and in picometers (pm). (b) How many gold atoms would have to be lined up to span \(1.0 \mathrm{~mm} ?\) (c) If the atom is assumed to be a sphere, what is the volume in \(\mathrm{cm}^{3}\) of a single Au atom?

Short answer

(a) 1.35 Å = 0.135 nm = 135 pm. (b) Approximately 3.7 * 10^6 gold atoms would line up to span 1.0 mm. (c) The volume of a single gold atom is 1.29 * 10^-28 cm^3.

Step-by-step solution

Part (a) Î Angstrom to Nanometer and Picometer Conversion

To convert from Angstroms \(\mathrm{(Å)}\) to nanometers \(\mathrm{(nm)}\) and picometers \(\mathrm{(pm)}\), the student must know the following conversion factors: 1 Angstrom (\(\mathrm{Å}\)) = 0.1 nanometers (\(\mathrm{nm}\)) 1 Angstrom (\(\mathrm{Å}\)) = 100 picometers (\(\mathrm{pm}\)) Given that the radius of a gold atom is 1.35 Å, the student can use these conversion factors to convert between the units: 1.35 Å * 0.1 nm/Å = 0.135 nm 1.35 Å * 100 pm/Å = 135 pm

Part (b) - Number of Gold Atoms to Span 1.0 mm

To calculate how many gold atoms would line up to span 1.0 mm, the student needs to divide the given distance (1.0 mm) by the diameter of a single gold atom (which is double the radius). First, convert 1.0 mm to picometers: 1.0 mm * 10^6 nm/mm * 10^3 pm/nm = 10^9 pm Next, calculate the diameter of a gold atom in picometers: Diameter = 2 * Radius = 2 * 135 pm = 270 pm Finally, divide the given distance by the diameter of the gold atom: Number of Gold Atoms = 10^9 pm / 270 pm ≈ 3.7 * 10^6 gold atoms

Part (c) - Volume of a Single Gold Atom

To calculate the volume of a single gold atom, assuming it's a sphere, the student can use the formula for the volume of a sphere: Volume = \(\frac{4}{3}\pi r^{3}\) Given the radius of a gold atom in picometers, the student must first convert it to cm before calculating the volume: 1.35 Å * 10^-10 cm/Å = 1.35 * 10^-10 cm Now, the student can plug the radius in cm into the formula for the volume of a sphere: Volume = \(\frac{4}{3}\pi (1.35 * 10^{-10})^{3}\) = 1.29 * 10^-28 cm^3

Key concepts

Angstrom

The angstrom (\( \AA \)) is a unit of length widely used in fields like chemistry, physics, and biology to measure atomic and molecular scales. It is notably convenient when dealing with dimensions less than a nanometer, such as the size of atoms or small molecules.

• 1 angstrom equals 10^-10 meters or 0.1 nanometers.
• It is equivalent to 100 picometers (pm), providing a link to other units commonly used at atomic scales.

The angstrom offers a middle ground between nanometers and picometers, making it ideal for describing atomic radii, like the 1.35 \( \AA \) radius of a gold atom, found in many textbooks and academic papers. Its popularity stems from a simple conversion to nanometers and picometers, enabling multi-level precision in scientific calculations.For instance, if an atom's radius is 1.35 \( \AA \), this can be converted to 0.135 nm and 135 pm, illustrating the small differences between these influencers.

Nanometer

The nanometer (nm) is another unit of length commonly used in scientific disciplines to measure wavelengths, biological structures, and nanomaterials. A single nanometer can be visualized as one-billionth of a meter.

• 1 nm is equal to 10 angstroms, highlighting how close these units sit together in the realm of atomic-scale measurements.
• In our exercise, the 1.35 \( \AA \) gold atom radius is converted to 0.135 nm, demonstrating the translation from one precision-focused measurement to another.

Nanometers allow scientists to express sizes of atoms and small molecules in integers rather than fractions, simplifying calculations and the comparison of very small objects. This ease of use underlies their widespread adoption in nanotechnology and materials science, all working to precisely understand the structures, behaviors, and interactions on these crucial small scales.

Picometer

A picometer (pm) is an extremely small unit of length, used to measure atomic radii and internuclear distances of atoms and subatomic particles. It is defined as one-trillionth of a meter.

• 1 pm is equal to 0.01 angstroms, making it an even finer unit of measure for atomic-level detail.
• In the discussion of gold atoms, 1.35 \( \AA \) is equal to 135 pm, reflecting the detailed nature of atomic and subatomic lengths.

Picometers provide an exceptionally precise means of measurement, crucial when exploring atomic radii and their positioning within molecules. Their usage is particularly prevalent in quantum physics and chemistry, where such finesse is essential for accurate modeling of atomic structures, enabling better predictions and analyses of atomic interactions.

Gold Atom

Gold atoms, denoted by the symbol Au, play a vital role in various scientific fields due to their unique properties and atomic simplicity. Often studied due to their stable electron configuration and conductivity, gold atoms are integral to both physics and materials science.

• The atomic radius of a gold atom is approximately 1.35 \( \AA \), providing a substantial base for understanding gold's physical characteristics.
• In nanometer terms, this radius is 0.135 nm, and in picometers, it is 135 pm.

Gold's significance is heightened by its real-world applicability, from electronics to biochemical applications, showcasing how a single type of atom can impact multiple industries. Understanding the dimensions, such as the radius or volume of a gold atom, allows scientists to deeper explore its potential uses and adaptability in advanced technologies.

Volume of Sphere

The volume of a sphere is an essential geometric calculation, given by the formula \( \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere. This formula is crucial for calculating the space an atom occupies, as many elements, including gold, are frequently modeled as spheres at atomic scales.

• For a gold atom with a radius of 1.35 × 10^-10 cm, the calculated volume is 1.29 × 10^-28 cm³.
• This computation uses the centimeter equivalent of the radius for precision in volume measurement.

Understanding the volume of an atomic scale sphere allows for predictive insights about density, packing, and potential interactions with other matter, crucial for material science and chemistry. These calculations provide foundational knowledge in the study of atomic and nanoparticle properties, assisting in advancing applications in areas such as nanotechnology and materials engineering.