Atomic Size and Dimensional Analysis: Silver and Sample Exercise 2.1

Atomic Scale Dimensional Analysis and Sample Exercise 2.1

Sample Exercise 2.1, titled Atomic Size, is derived from the textbook Chemistry: The Central Science, 14th Edition (2018), authored by Brown, LeMay, Bursten, Murphy, Woodward, and Stoltzfus. This exercise focuses on the practical application of dimensional analysis to bridge the gap between macroscopic measurements and the sub-microscopic world of atoms. The primary objective is to calculate an unknown quantity, specifically the total number of silver ($Ag$) atoms required to span a distance equivalent to the diameter of a standard currency coin, which is a U.S. dime. The methodology involves utilizing the physical dimensions of the atom as a conversion factor to relate distance to a discrete number of particles.

Comparative Size: Silver Atoms and the United States Dime

The calculation for Sample Exercise 2.1 begins with the identification of known physical constants and measurements. The diameter of a silver ($Ag$) atom is cited as $2.88\,\text{Å}$. This provides the essential conversion factor where $1\,Ag\text{ atom} = 2.88\,\text{Å}$. The macroscopic object in question is a U.S. dime, which possesses a diameter of $17.9\,mm$. To solve the problem, the distance is first converted from millimeters ($mm$) into the atomic unit of angstroms ($\text{Å}$). By converting the diameter of the dime ($17.9\,mm$) into angstroms and then applying the diameter of the individual silver atom as a divisor, the total number of atoms is determined. The solution demonstrates that approximately $62.2\,\text{million silver atoms}$ could be arranged side by side across the span of a single dime, illustrating the incredibly small scale of individual atoms relative to everyday objects.

Determinants of Atomic Volume and Practice Exercise 1

Practice Exercise 1 addresses the conceptual understanding of what fundamentally defines the size or volume of an atom. The exercise presents several variables and asks which one dictates atomic size: (a) the volume of the nucleus; (b) the volume of space occupied by the electrons of the atom; (c) the volume of a single electron, multiplied by the number of electrons in the atom; (d) the total nuclear charge; or (e) the total mass of the electrons surrounding the nucleus. While the nucleus accounts for nearly all of the atom's mass, it occupies an extremely small fraction of the atom's total volume. According to chemical principles, the physical size (volume) of an atom is determined by the volume of space occupied by the electrons of the atom (option b). This space is defined by the electron cloud and the probability distributions of the electrons within their orbitals, rather than the mass of the particles or the size of the nucleus itself.

Unit Conversions and Linear Alignment in Practice Exercise 2

Practice Exercise 2 shifts the focus to the carbon atom and requires the performance of specific unit conversions and alignment calculations. The diameter of a carbon atom is given as $1.54\,\text{Å}$. In part (a) of the exercise, this measurement must be expressed in picometers ($pm$). Given that $1\,\text{Å} = 100\,pm$, the diameter is calculated as $154\,pm$. In part (b), the problem poses a scenario involving a pencil line with a width of $0.20\,mm$. Students must determine how many carbon atoms could be aligned side by side across this width. This requires converting the line width ($0.20\,mm$) into a compatible unit such as angstroms ($2.0 \times 10^{6}\,\text{Å}$) or picometers ($2.0 \times 10^{8}\,pm$) and dividing by the diameter of the carbon atom ($1.54\,\text{Å}$ or $154\,pm$). This exercise reinforces the utility of angstroms and picometers in describing sub-atomic dimensions and highlights the precision required in chemical calculations.