Chapter 14: Problem 2
(a) A certain nucleus has radius \(5 \mathrm{fm}\). ( \(1 \mathrm{fm}=10^{-15} \mathrm{m}\).) Find its cross section \(\sigma\) in barns. (1 barn \(=10^{-28} \mathrm{m}^{2} .\) ) (b) Do the same for an atom of radius \(0.1 \mathrm{nm}\). \(\left(1 \mathrm{nm}=10^{-9} \mathrm{m} .\right)\)
Short Answer
Expert verified
(a) \( \approx 0.785 \times 10^{-2} \) barns; (b) \( \approx 3.14 \times 10^{8} \) barns.
Step by step solution
01
Understand the Problem
We need to calculate the cross-sectional area of two spheres with given radii and express these areas in units of barns. One radius is given in femtometers (fm) for a nucleus, and the other in nanometers (nm) for an atom.
02
Calculate the Cross Section of the Nucleus
The cross-sectional area \( \sigma \) of a sphere is given by the formula \( \sigma = \pi \times r^2 \), where \( r \) is the radius.1. Convert the given nucleus's radius from femtometers to meters: \[ r_{nucleus} = 5 \text{ fm} = 5 \times 10^{-15} \text{ m} \]2. Calculate the cross-sectional area: \[ \sigma_{nucleus} = \pi \times (5 \times 10^{-15})^2 = 25\pi \times 10^{-30} \text{ m}^2 \]
03
Convert the Cross Section of the Nucleus to Barns
Convert the cross-sectional area from square meters to barns:1. Use the conversion factor: \( 1 \text{ barn} = 10^{-28} \text{ m}^2 \).2. Convert: \[ \sigma_{nucleus}(\text{in barns}) = \frac{25\pi \times 10^{-30} \text{ m}^2}{10^{-28} \text{ m}^2/\text{barn}} = 25\pi \times 10^{-2} \text{ barns} \]
04
Calculate the Cross Section of the Atom
Repeat the cross-sectional area calculation for the atom:1. Convert the given atom's radius from nanometers to meters: \[ r_{atom} = 0.1 \text{ nm} = 0.1 \times 10^{-9} \text{ m} \]2. Calculate the cross-sectional area: \[ \sigma_{atom} = \pi \times (0.1 \times 10^{-9})^2 = 0.01\pi \times 10^{-18} \text{ m}^2 \]
05
Convert the Cross Section of the Atom to Barns
Convert the cross-sectional area from square meters to barns:1. Using the same conversion factor: \( 1 \text{ barn} = 10^{-28} \text{ m}^2 \).2. Convert: \[ \sigma_{atom}(\text{in barns}) = \frac{0.01\pi \times 10^{-18} \text{ m}^2}{10^{-28} \text{ m}^2/\text{barn}} = 0.01\pi \times 10^{10} \text{ barns} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Cross-Sectional Area
The cross-sectional area is an important concept within physics, particularly in fields like classical mechanics and nuclear physics. When you encounter the term "cross-section," it refers to the size of a slice through an object. Imagine slicing an apple in half – the exposed part is the cross-sectional area. This is a 2D representation of the object's cut surface area.
In equations, the cross-sectional area of a sphere is calculated using the formula: \( \sigma = \pi \times r^2 \). Here, \( \pi \) is a constant value approximately equal to 3.14159, and \( r \) is the radius of the sphere. The radius is essentially the distance from the center of the circle (or sphere) to any point on its edge. Calculating cross-sectional areas becomes vital when determining how much space objects occupy in a given field, such as nuclei and atoms in nuclear physics.
In equations, the cross-sectional area of a sphere is calculated using the formula: \( \sigma = \pi \times r^2 \). Here, \( \pi \) is a constant value approximately equal to 3.14159, and \( r \) is the radius of the sphere. The radius is essentially the distance from the center of the circle (or sphere) to any point on its edge. Calculating cross-sectional areas becomes vital when determining how much space objects occupy in a given field, such as nuclei and atoms in nuclear physics.
Mastering Conversion Factors
Conversion factors are essential tools that help us switch between different units of measurement, which can make problem-solving understandable and efficient. In physics, converting measurements is critical since different units are often used across various fields and scientific disciplines.
To convert between square meters and barns, we need to know the specific conversion factor: \( 1 \text{ barn} = 10^{-28} \text{ m}^2 \). Using conversion involves setting up a ratio that lets us express a value in a different unit. For instance, if you have a value in square meters and want it in barns, you'd set up your conversion to match units, thereby allowing you to calculate the result directly in barns. Conversion factors ensure you can communicate measurements accurately in science.
To convert between square meters and barns, we need to know the specific conversion factor: \( 1 \text{ barn} = 10^{-28} \text{ m}^2 \). Using conversion involves setting up a ratio that lets us express a value in a different unit. For instance, if you have a value in square meters and want it in barns, you'd set up your conversion to match units, thereby allowing you to calculate the result directly in barns. Conversion factors ensure you can communicate measurements accurately in science.
Attributes of Spherical Objects
Spherical objects are shapes where every point on the surface is equidistant from the center, making them perfect circles in three dimensions. This symmetry simplifies calculations such as volume and surface area, as formulas remain consistent across scenarios.
The radius is crucial when examining spherical objects. In problems involving nuclei and atoms – both approximated as spheres for simplicity – you generally start with a given radius. Knowing how to manipulate these dimensions can help in computing essential properties like cross-sectional area. Spheres appear frequently in physics due to their inherent simplicity and are an ideal model for many natural phenomena.
The radius is crucial when examining spherical objects. In problems involving nuclei and atoms – both approximated as spheres for simplicity – you generally start with a given radius. Knowing how to manipulate these dimensions can help in computing essential properties like cross-sectional area. Spheres appear frequently in physics due to their inherent simplicity and are an ideal model for many natural phenomena.
Introduction to Nuclear Physics
Nuclear physics is a branch dealing with atomic nuclei and their constituents and interactions. It covers topics ranging from nuclear power to particle physics, exploring the forces binding neutrons and protons in the nucleus.
In nuclear physics, understanding the size and cross-sectional area of particles, like nuclei, is crucial. These insights help in comprehending reactions such as nuclear fission and fusion. Calculating their cross-sectional areas allows scientists to evaluate the likelihood of certain reactions occurring when nuclei collide, which is pivotal in many applications, including energy production and medical imaging. As such, the interplay between geometry and physics is vital for advancements in technology and science.
In nuclear physics, understanding the size and cross-sectional area of particles, like nuclei, is crucial. These insights help in comprehending reactions such as nuclear fission and fusion. Calculating their cross-sectional areas allows scientists to evaluate the likelihood of certain reactions occurring when nuclei collide, which is pivotal in many applications, including energy production and medical imaging. As such, the interplay between geometry and physics is vital for advancements in technology and science.