Problem 19 - Chemistry: The Central Science in SI Units, Global Edition

Question

The radius of an atom of tungsten \((\mathrm{W})\) is about \(2.10 \AA\) (a) Express this distance in nanometers (nm) and in picometers (pm). (b) How many tungsten atoms would have to be lined up to create a wire of \(2.0 \mathrm{~mm}\) (c) If the atom is assumed to be a sphere, what is the volume in \(\mathrm{m}^{3}\) of a single \(\mathrm{W}\) atom?

Step-by-Step Solution

Answer

The radius of a tungsten atom is \(0.210 \mathrm{nm}\) or \(210 \mathrm{pm}\). To create a 2mm long wire, approximately \(4.76 \times 10^6\) W atoms are needed. Assuming the atom is a sphere, the volume of a single W atom is approximately \(4.086 \times 10^{-29} \mathrm{m}^{3}\).

Step 1: 1. Convert radius to nanometers (nm) and picometers (pm)

To convert the given radius in angstroms to nanometers and picometers, we will use the conversion factors: 1 angstrom (Å) = 0.1 nanometers (nm) 1 angstrom (Å) = 100 picometers (pm) Given radius: 2.10 Å Convert to nanometers: \(2.10 \mathrm{Å} \times \frac{0.1 \mathrm{nm}}{1 \mathrm{Å}} = 0.210 \mathrm{nm}\) Convert to picometers: \(2.10 \mathrm{Å} \times \frac{100 \mathrm{pm}}{1 \mathrm{Å}} = 210 \mathrm{pm}\) So, in nanometers the radius is \(0.210 \mathrm{nm}\) and in picometers the radius is \(210 \mathrm{pm}\).

Step 2: 2. Calculate the number of W atoms required to create a 2 mm long wire

To find the number of tungsten atoms required to create a 2 mm long wire, we'll first convert the length of the wire to the same units as the radius (in picometers), and then divide the length of the wire by the diameter of a single W atom (which is twice the radius). Convert wire length to picometers: \(2.0 \mathrm{mm} \times \frac{10^{9} \mathrm{pm}}{1 \mathrm{mm}} = 2 \times 10^{9} \mathrm{pm}\) Diameter of a single W atom = 2 × radius = \(2 \times 210 \mathrm{pm} = 420 \mathrm{pm}\) Number of W atoms in a 2 mm wire: \(\mathrm{Number\ of\ W\ atoms}=\frac{2\times 10^{9} \mathrm{pm}}{420 \mathrm{pm}}=4.76 \times 10^6\) So, approximately \(4.76 \times 10^6\) W atoms are needed to create a 2 mm long wire.

Step 3: 3. Calculate the volume of a single W atom

To calculate the volume of a single tungsten atom, we assume it to be a sphere and use the formula: \(V = \frac{4}{3} \pi r^{3}\) Here, V represents the volume, r represents the radius, and \(\pi\) is a constant with an approximate value of 3.14159. The radius must be in meters for the volume to be in \(\mathrm{m}^{3}\). Convert radius to meters: \(r = 2.10 \mathrm{Å} \times \frac{0.1 \mathrm{nm}}{1 \mathrm{Å}} \times \frac{1 \times 10^{-9} \mathrm{m}}{1 \mathrm{nm}} = 2.10 \times 10^{-10} \mathrm{m}\) Volume of a single W atom: \(V = \frac{4}{3} \pi (2.10 \times 10^{-10} \mathrm{m})^{3}\) \(V = 4.086 \times 10^{-29} \mathrm{m}^{3}\) Thus, the volume of a single W atom is approximately \(4.086 \times 10^{-29} \mathrm{m}^{3}\).

Key Concepts

Nanometer Conversion: Understanding Atomic Radius

Converting units is fundamental in understanding and expressing the size of atomic structures. When it comes to measuring the atomic radius, we often need to convert between different units to make the data more manageable. One commonly used unit is the nanometer (nm). A nanometer is equal to one billionth of a meter. Specifically, in the context of atomic radius, we use the conversion factor: 1 angstrom (Å) is equal to 0.1 nanometers (nm). This conversion is crucial because it allows scientists to handle atomic dimensions, which are incredibly small and hard to conceptualize, using a more standardized metric system. For example, if the radius of a tungsten atom is given as 2.10 Å, we can convert it to nanometers by multiplying 2.10 by 0.1, resulting in 0.210 nm. This straightforward conversion simplifies the process of comparing and understanding atomic dimensions in various scientific and engineering contexts.
This skill is especially useful in fields such as nanotechnology and molecular chemistry, where precise measurements at the nanoscale are necessary to develop and understand new materials.

Picometer Conversion: Precision at the Atomic Scale

In the realm of atomic measurements, the picometer (pm) serves as an even smaller unit to express these tiny dimensions. 1 picometer is equal to one trillionth of a meter, a conversion especially valuable when dealing with atomic and subatomic scales. Understanding how to convert from angstroms, a common unit for atomic radii, to picometers helps in visualizing these minute distances in a familiar metric scale. The conversion factor here is that 1 angstrom (Å) equals 100 picometers (pm).
For instance, using the previous example of a tungsten atom with a radius of 2.10 Å, this converts to 210 pm when multiplied by 100. Recognizing these quantities in picometers provides an extra degree of precision and is crucial for applications in fields such as physics and materials science. Here, the accuracy of atomic dimensions can impact the understanding and manipulation of materials at the deepest level. Building accurate models and simulations of atomic interactions often relies on such precise measurements.

Volume of Atoms: Calculating an Atomic Dimension

The concept of atomic volume offers a valuable perspective on the space an atom occupies. When atoms are assumed to be spherical, we use the sphere's volume formula to calculate this, remembering that atomic dimensions need converting into meters for the volume to be expressed in cubic meters (\(\mathrm{m}^{3}\)).
The formula for the volume of a sphere is given by \(V = \frac{4}{3} \pi r^3\), where \(r\) is the radius in meters and \(\pi\) is approximately 3.14159. Using our example of a tungsten atom with a radius of 2.10 Å, we first convert this radius into meters: \(2.10 \mathrm{Å} = 2.10 \times 10^{-10} \mathrm{m}\).
Substituting this value into the formula yields \[V = \frac{4}{3} \pi (2.10 \times 10^{-10} \, \mathrm{m})^3 = 4.086 \times 10^{-29} \, \mathrm{m}^3\].
This calculation provides a sense of scale, illustrating just how small individual atoms are and how much space they occupy in a given material. Understanding atomic volume is essential in fields like chemistry and solid-state physics, where reactions and properties depend heavily on the spatial arrangement of atoms.