Potassium bromide (KBr) has a density of 2.75×1032.75\$$times10^3 kg/m3 $$and the same crystal structure as NaCl. The mass of a potassium atom is 6.49×10−266.49\$$times10^{-26} kg$$, and the mass of a bromine atom is 1.33×10−251.33\$$times10^{-25} kg. $$Calculate the average spacing between adjacent atoms in a KBr crystal.

Ch 42: Molecules and Condensed Matter

Young & Freedman Calc - University Physics 15th Edition

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Young & Freedman Calc 15th Edition

Ch 42: Molecules and Condensed Matter

Problem 12a

Chapter 41, Problem 12a

Potassium bromide (KBr) has a density of $2.75 times 10 cubed$ kg/m³ and the same crystal structure as NaCl. The mass of a potassium atom is $6.49 \times 10^{- 26}$ kg, and the mass of a bromine atom is $1.33 \times 10^{- 25}$ kg. Calculate the average spacing between adjacent atoms in a KBr crystal.

Verified step by step guidance

Step 1: Understand the problem. The goal is to calculate the average spacing between adjacent atoms in a KBr crystal. This involves using the density of the crystal and the masses of the potassium and bromine atoms to determine the volume occupied by each atom pair in the crystal structure.

Step 2: Calculate the mass of one formula unit of KBr. Since KBr consists of one potassium atom and one bromine atom, the mass of one formula unit is the sum of the masses of a potassium atom and a bromine atom: m_{KBr} = m_{K} + m_{Br}. Use MathML:

m_{KBr} = m_{K} + m_{Br}

Step 3: Use the density of KBr to calculate the volume occupied by one formula unit. The density formula is ρ = m/V, where ρ is the density, m is the mass, and V is the volume. Rearrange to find the volume per formula unit: V_{unit} = m_{KBr} / ρ. Use MathML:

V_{unit} = m_{KBr} / ρ

Step 4: Relate the volume of one formula unit to the average spacing between adjacent atoms. In a cubic crystal structure like NaCl (which KBr shares), the volume of one formula unit corresponds to the cube of the average spacing between adjacent atoms: V_{unit} = a^3, where a is the average spacing. Solve for a: a = (V_{unit})^(1/3). Use MathML:

a = {(V_{unit})}^{1 / 3}

Step 5: Substitute the values for m_{K}, m_{Br}, and ρ into the equations to calculate V_{unit} and then find a. Ensure units are consistent throughout the calculation (e.g., kg, m^3). This will yield the average spacing between adjacent atoms in the KBr crystal.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Crystal Structure

Crystal structure refers to the orderly and repeating arrangement of atoms within a crystalline material. In the case of potassium bromide (KBr), it shares the same face-centered cubic structure as sodium chloride (NaCl), where each unit cell contains a specific number of atoms arranged in a three-dimensional lattice. Understanding this structure is crucial for calculating properties like atomic spacing.

Density and Mass Relationship

Density is defined as mass per unit volume and is a critical property in determining how closely packed atoms are in a material. For KBr, the given density allows us to calculate the volume occupied by a certain mass of the substance, which can then be used to find the average spacing between atoms. The relationship between mass, volume, and density is fundamental in solid-state physics.

Atomic Spacing Calculation

Atomic spacing refers to the distance between adjacent atoms in a crystal lattice. To calculate this spacing in KBr, one must consider the total number of atoms in a unit cell and the volume of that cell derived from the density. This calculation is essential for understanding the physical properties of the material, such as its conductivity and reactivity.

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