Question

An atom of tin (Sn) has a diameter of about \(2.8 \times 10^{-8} \mathrm{~cm}\). (a) What is the radius of a tin atom in angstroms \((\AA)\) and in meters \((\mathrm{m}) ?\) (b) How many Sn 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 tin atom is a sphere, what is the volume in \(\mathrm{m}^{3}\) of a singleatom?

Short answer

(a) The radius of a tin atom is \(1.4 \times 10^{0} \mathrm{~\AA}\) (angstroms) or \(1.4 \times 10^{-10} \mathrm{~m}\) (meters). (b) Approximately 21,429 tin atoms would have to be placed side by side to span a distance of \(6.0 \mu \mathrm{m}\). (c) Assuming the tin atom is a sphere, the volume of a single tin atom is \(1.44 \times 10^{-29} \mathrm{~m}^{3}\).

Step-by-step solution

Convert radius from cm to angstroms and meters)

To find the radius of a tin atom in angstroms and meters, we first have to find the radius in cm. Since we're given the diameter, divide the diameter by 2 to get the radius. Now, convert the radius from cm to angstroms (1 cm = 10^8 angstroms) and to meters (1 cm = 0.01 m). Radius in cm: \(\frac{2.8 \times 10^{-8} \mathrm{~cm}}{2}\)

Calculate the radius in angstroms and meters)

Next, multiply the radius in cm by the appropriate conversion factor to find the radius in angstroms and meters. Radius in angstroms: \(\frac{2.8 \times 10^{-8} \mathrm{~cm}}{2} \times \frac{10^{8} \mathrm{~\AA}}{1 \mathrm{~cm}}\) Radius in meters: \(\frac{2.8 \times 10^{-8} \mathrm{~cm}}{2} \times \frac{0.01 \mathrm{~m}}{1 \mathrm{~cm}}\)

Calculate the number of Sn atoms to span 6.0 μm)

To calculate the number of tin atoms required to span a distance of \(6.0 \mu \mathrm{m}\), divide the distance by the diameter of a single tin atom, remembering to convert the distance to cm. Number of Sn atoms: \(\frac{6.0 \times 10^{-4} \mathrm{~cm}}{2.8 \times 10^{-8} \mathrm{~cm}}\)

Calculate the volume of a single Sn atom)

Assuming that the tin atom is a sphere, calculate the volume using the formula \(V = \frac{4}{3}\pi r^{3}\) with the radius in meters. Volume in \(m^3\): \(\frac{4}{3}\pi \times (Radius~in~meters)^{3}\) Now calculating the results: a) Radius in angstroms: \(\frac{2.8 \times 10^{-8} \mathrm{~cm}}{2} \times \frac{10^{8} \mathrm{~\AA}}{1 \mathrm{~cm}} = 1.4 \times 10^{0} \mathrm{~\AA}\) Radius in meters: \(\frac{2.8 \times 10^{-8} \mathrm{~cm}}{2} \times \frac{0.01 \mathrm{~m}}{1 \mathrm{~cm}} = 1.4 \times 10^{-10} \mathrm{~m}\) b) Number of Sn atoms: \(\frac{6.0 \times 10^{-4} \mathrm{~cm}}{2.8 \times 10^{-8} \mathrm{~cm}} \approx 21429\) c) Volume in \(m^3\): \(\frac{4}{3}\pi \times (1.4 \times 10^{-10})^3 = 1.44 \times 10^{-29} \mathrm{~m}^{3}\)

Key concepts

Angstrom Conversion

Understanding the conversion of units to angstroms is crucial when studying atomic and molecular scales. An angstrom (\r)\r(A)) is a unit of length equal to 10^{-10} meters (\r)\r(10^{-10} \r) m)), typically used to express atomic and molecular dimensions. To convert from centimeters to angstroms, you multiply by 10^{8}, since 1 centimeter (\r)\r(cm)) is equivalent to 10^{8} angstroms (\r)\r(10^{8} \r) \r) A)). For instance, a tin atom with a diameter of \(2.8 \times 10^{-8} \r) \) cm has a radius of \(1.4 \times 10^{0} \r) \) angstroms when calculated by halving the diameter and applying the conversion factor. This measure enables scientists and students to work with more tangible numbers when comparing atomic sizes or calculating other properties.

Volume of an Atom

To conceptualize the space an individual atom occupies, we often approximate the atom as a sphere and calculate its volume. The volume is obtained by the formula for the volume of a sphere, \(V = \frac{4}{3}\pi r^3\), where \(r\) is the atom's radius. This equation emphasizes the significance of the radius, as the volume is directly proportional to the cube of the radius—a small change in the radius results in a much larger change in volume. In our tin atom example, after converting the radius to meters, which is a standard SI unit, the calculated volume of a single tin atom is \(1.44 \times 10^{-29} \) m^3). Although this number seems incredibly small, it's a fundamental concept when scaling up to a material's macroscopic properties, such as density and molar volume.

Atoms Arrangement

Arrangement of atoms tells us a lot about the structure and characteristics of materials. In calculations involving the number of atoms that fit into a given space, like how many tin atoms span 6.0 micrometers (µm), we confront the question of how these atoms are packed together. For a simplistic linear arrangement, which assumes atoms are perfectly lined up next to each other without any gaps, we divide the total length by the diameter of one atom. This provides an estimation of how many atoms would fit in a line across 6.0 µm of space. Calculating with our example, approximately 21429 tin atoms are needed to span 6.0 µm. This simple model does not take into account the more complex, efficient packing arrangements atoms might have in solid states, but rather provides a starting point for understanding the scale and density of atomic structures.