Calculating Atomic Spacing in Copper Using Density and Atomic Mass
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Reference
Serway & Jewett's "Physics for Scientists and Engineers with Modern Physics 9th Edition" is a comprehensive textbook that covers fundamental physics principles, including mechanics, thermodynamics, and modern physics concepts. This renowned text emphasizes problem-solving strategies and real-world applications, making it an essential resource for science and engineering students.
Question
8. The mass of a copper atom is 1.06 × 10⁻²⁵ kg, and the density of copper is 8,920 kg/m³. (a) Determine the number of atoms in 1 cm³ of copper. (b) Visualize the one cubic centimeter as formed by stacking up identical cubes, with one copper atom at the center of each. Determine the volume of each cube. (c) Find the edge dimension of each cube, which represents an estimate for the spacing between atoms.
Explain Question
This problem explores the atomic structure of copper by relating macroscopic properties (density) to microscopic properties (atomic mass). We need to find three key elements: the number of atoms in a cubic centimeter, the volume per atom assuming a cubic arrangement, and the atomic spacing. The question requires unit conversion skills and understanding of three-dimensional geometry.
Explain Solve
Part (a) requires converting the density to atoms per cubic centimeter. We'll divide the density by the mass per atom, considering unit conversions from m³ to cm³. For part (b), we'll take the reciprocal of the atoms per volume to find the volume per atom. Finally, in part (c), we'll calculate the cube root of this volume to find the atomic spacing.
Summary
This problem demonstrates how macroscopic properties like density can be used to understand microscopic structure. The solution involves unit conversions, volume calculations, and cube root operations to determine atomic spacing in copper crystal structure.
Solve
The solution involves three main steps:
(a) Number of atoms per cm³ = (8,920 kg/m³) ÷ (1.06 × 10⁻²⁵ kg) × (1 m³/10⁶ cm³) = 8.42 × 10²² atoms/cm³
(b) Volume per atom = 1 ÷ (8.42 × 10²² atoms/cm³) = 1.19 × 10⁻²³ cm³/atom
(c) Atomic spacing = ∛(1.19 × 10⁻²³ cm³) = 2.28 × 10⁻⁸ cm
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