A proton, which is the nucleus of a hydrogen atom, can be modeled as a sphere with a diameter of 2.4 fm and a mass of 1.67 × 10^-27 kg.
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Reference
Serway & Jewett's "Physics For Scientists And Engineers With Modern Physics" (9th edition) is a comprehensive textbook covering various physics topics. It delves into mechanics, thermodynamics, electromagnetism, optics, and modern physics. The book is known for its clear explanations, detailed examples, and practical applications, making it a valuable resource for students and professionals in scientific and engineering fields.
Question
A proton, which is the nucleus of a hydrogen atom, can be modeled as a sphere with a diameter of 2.4 fm and a mass of 1.67 × 10^-27 kg. (a) Determine the density of the proton. (b) State how your answer to part (a) compares with the density of osmium, given in Table 14.1 in Chapter 14
Explain Question
This question focuses on calculating the density of a proton and comparing it to the density of osmium. To solve this problem, we need to: 1. Calculate the volume of the proton using the given diameter. 2. Use the given mass and calculated volume to determine the proton's density. 3. Compare the calculated proton density with the density of osmium provided in the reference table. The question tests understanding of density calculations, unit conversions, and the ability to interpret and compare scientific data.
Explain Solve
Let's solve this problem step by step: 1. Calculate the volume of the proton: - The proton is modeled as a sphere with diameter d = 2.4 fm (femtometers). - Volume of a sphere: V = (4/3)πr³, where r is the radius (d/2). - V = (4/3)π(d/2)³ = (π/6)d³ - V = (π/6) × (2.4 × 10^-15 m)³ = 7.24 × 10^-45 m³ 2. Calculate the density of the proton: - Density ρ = mass / volume - ρ = (1.67 × 10^-27 kg) / (7.24 × 10^-45 m³) - ρ = 2.31 × 10^17 kg/m³ 3. Compare with osmium density: - Osmium density (from Table 14.1): 22.6 × 10³ kg/m³ - Ratio = (Proton density) / (Osmium density) - Ratio = (2.31 × 10^17) / (22.6 × 10³) = 1.02 × 10^13 The proton's density is approximately 1.02 × 10^13 times greater than osmium's density.
Summary
This problem illustrates the incredibly high density of subatomic particles like protons compared to even the densest known elements. By calculating the proton's volume using its diameter and then dividing its mass by this volume, we determined its density. The comparison with osmium, one of the densest elements, highlights the extreme nature of nuclear densities, showcasing the vast differences in scale between atomic and subatomic realms.
Solve
(a) ρ = m/V and V = (4/3)πr³ = (4/3)π(d/2)³ = πd³/6, where d is the diameter. Then ρ = 6m/πd³ = 6(1.67 × 10^-27 kg) / (π(2.4 × 10^-15 m)³) = 2.3 × 10^17 kg/m³
(b) 2.3 × 10^17 kg/m³ / 22.6 × 10³ kg/m³ = 1.0 × 10^13 times the density of osmium
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