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The cross section for scattering a certain nuclear particle by a nitrogen nucleus is 0.5 barns. If \(10^{11}\) of these particles are fired through a cloud chamber of length \(10 \mathrm{cm},\) containing nitrogen at STP, how many particles are scattered? (Use the ideal gas law and remember that each nitrogen molecule has two atoms. The scattering by any atomic electrons is completely negligible.)

Short Answer

Expert verified
Approximately 26860 particles are scattered.

Step by step solution

01

Understanding the Problem

We need to calculate how many particles are scattered when a certain number of nuclear particles pass through a cloud chamber containing nitrogen gas. Given the cross-section of 0.5 barns and the number of particles as \(10^{11}\), we'll use properties of nitrogen gas and the ideal gas law for our solution.
02

Convert Cross-Section

The cross-section is given in barns. Recall that 1 barn is equivalent to \(10^{-28} \text{ m}^2\). So, the cross-section \(\sigma\) in \(\text{m}^2\) is:\[\sigma = 0.5 \times 10^{-28} \text{ m}^2 = 5 \times 10^{-29} \text{ m}^2\]
03

Calculate Volume of Chamber

The length of the chamber is 10 cm, equivalent to \(0.1 \text{ m}\). We need the volume to be compatible with the ideal gas calculations. Assuming the chamber's cross-sectional area is maximized by the particle's trajectory, this length per unit area V is \(0.1 \text{ m}\).
04

Use Ideal Gas Law to Find Molecules

The Ideal Gas Law, \(PV = nRT\), gives the number of molecules at STP (Standard Temperature and Pressure). At STP, \(P = 101325 \text{ Pa}\), \(V = 0.1 \text{ m}^3\), \(R = 8.314 \text{ J/mol K}\), and \(T = 273.15 \text{ K}\). Solving for \(n\) gives the number of moles:\[n = \frac{PV}{RT} = \frac{101325 \times 0.1}{8.314 \times 273.15} \approx 0.00446 \text{ mol}\]Converting moles to molecules:\[\text{Number of molecules} = n \times 6.022 \times 10^{23} = 0.00446 \times 6.022 \times 10^{23} \approx 2.686 \times 10^{21}\]
05

Consider Nitrogen Molecules

Each nitrogen molecule consists of 2 nitrogen atoms. Thus, the number of nitrogen atoms is:\[2.686 \times 10^{21} \times 2 = 5.372 \times 10^{21} \text{ atoms}\]
06

Find Scattered Particles Using Cross-Section

The number of particles scattered (\(N_s\)) is given by the formula:\[N_s = \sigma \times \text{Number of target atoms} \times \text{Number of incident particles}\]Substituting the values we calculated:\[N_s = (5 \times 10^{-29}) \times (5.372 \times 10^{21}) \times (10^{11})\]\[N_s \approx 2.686 \times 10^{4}\]
07

Final Calculation

The calculated result yields approximately 26860 particles being scattered as the nuclear particles pass through the cloud chamber.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross-Section
The concept of cross-section in particle physics is essentially the effective area where interactions, like scattering, occur between particles. Imagine it like aiming at a target; the cross-section determines how likely it is for the particle to hit the target when fired.

### Cross-Section in BarnsIn scientific terms, the cross-section is typically measured in "barns", where 1 barn equals \(10^{-28} \text{ m}^2\). This unit is surprisingly named after barns on farms, perhaps to suggest the size is quite small, like missing the broad side of a barn!
  • Cross-sections are crucial for understanding reaction probabilities.
  • A larger cross-section means a higher probability of interaction.
For the problem at hand, we converted the cross-section to \(0.5 \times 10^{-28} = 5 \times 10^{-29} \text{ m}^2\) to use in further calculations.The interaction described here is the scattering of particles by a nitrogen nucleus, a fundamental aspect when studying particle movements in environments like cloud chambers.
Ideal Gas Law
The Ideal Gas Law is an essential principle for understanding the behavior of gases under various conditions. It helps relate the pressure, volume, and temperature of a gas with its quantity of particles. The formula is given by \(PV = nRT\), where:
  • \(P\) represents pressure
  • \(V\) is the volume
  • \(n\) is the number of moles
  • \(R\) is the ideal gas constant (\(8.314 \, \text{J/mol K}\))
  • \(T\) is the temperature in Kelvin
In this exercise, we use the ideal gas law to find the number of nitrogen molecules in the cloud chamber at standard temperature and pressure (STP). At STP, conditions are simplified to 1 atmosphere (101325 Pa) and 273.15 K for calculations. This allows for straightforward conversions from volume to the number of moles and then to individual molecules using Avogadro's number (\(6.022 \times 10^{23}\)).
This principle helps connect microscopic atomic interactions to macroscopic gas properties.
Nitrogen Nucleus
A nitrogen nucleus consists of protons and neutrons and is a key component in understanding the interaction in this particle scattering problem. Each nitrogen atom has seven protons and typically seven neutrons in its nucleus, which means the nucleus is a significant factor in particle scattering events.

### Structure and InteractionWhen particles travel through a medium like nitrogen gas, their interactions are primarily with the nucleus because of its dense mass compared to electrons.
  • The particle scattering event predominantly involves the nucleus due to its relatively large mass and charge concentration.
  • Nitrogen molecules, typically diatomic (\(\text{N}_2\)), consist of two nitrogen atoms, increasing the potential interaction targets.
The high number of target atoms calculated (\(5.372 \times 10^{21}\)) reflect the availability of these nucleonic targets for scattering in our particle physics problem.
Cloud Chamber
A cloud chamber is a fascinating device that is used to visualize the passage of ionizing radiation. It functioned importantly in this problem by being filled with nitrogen gas, creating a medium where these particles can interact.

### How it Works The cloud chamber works on the principle of visible condensation trails, much like your breath making a cloud in cold air. When the charged particles move through the supersaturated vapor inside, they ionize the gas molecules, and droplets form along these paths. This allows scientists to observe paths and interactions.
  • It serves as a visual tool to study particle physics phenomena.
  • Offers qualitative data about particle paths and interaction frequencies.
In this particular exercise, using cloud chambers in conjunction with the cross-section concept provides an essential method for quantifying how many particles like those from nuclear interactions scatter upon interacting with nitrogen nuclei. Thus, it serves as a bridge between theoretical calculations and practical visualization within gas environments.

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Most popular questions from this chapter

In deriving the cross section for scattering by a hard sphere, we used the "law of reflection," that the angles of incidence and reflection of a particle bouncing off a hard sphere are equal, as in Figure 14.10. Use conservation of energy and angular momentum to prove this law. (The definition of "hard-sphere scattering" is that a projectile bounces with its kinetic energy unchanged. That the force is spherically symmetric implies, as usual, that angular momentum about the sphere's center is conserved.)

In their famous experiment, Rutherford's assistants, Geiger and Marsden, detected the scattered alpha particles using a zinc sulphide screen, which produced a tiny flash of light when struck by an alpha particle. If their screen had area \(1 \mathrm{mm}^{2}\) and was \(1 \mathrm{cm}\) from the target, what solid angle did it subtend?

The differential cross section for scattering 6.5-MeV alpha particles at 120^ off a silver nucleus is about 0.5 barns/sr. If a total of \(10^{10}\) alphas impinge on a silver foil of thickness \(1 \mu \mathrm{m}\) and if we detect the scattered particles using a counter of area \(0.1 \mathrm{mm}^{2}\) at \(120^{\circ}\) and \(1 \mathrm{cm}\) from the target, about how many scattered alphas should we expect to count? (Silver has a specific gravity of \(10.5,\) and atomic mass of 108.)

Our definition of the scattering cross section, \(N_{\mathrm{sc}}=N_{\text {inc }} n_{\text {tar }} \sigma,\) applies to an experiment using a narrow beam of projectiles all of which pass through a wide target assembly. Experimenters sometimes use a wide incident beam, which completely engulfs a small target assembly (the beam of photons from a car's headlamp, directed at a small piece of plastic, for example). Show that in this case \(N_{\mathrm{sc}}=n_{\text {inc }} N_{\mathrm{tar}} \sigma\) where \(n_{\text {inc }}\) is the density (number/area) of the incident beam, viewed head-on, and \(N_{\text {tar }}\) is the total number of targets in the target assembly.

(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 \(0.1 \mathrm{nm}\). \(\left(1 \mathrm{nm}=10^{-9} \mathrm{m} .\right)\)

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