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Ch 20: The Micro/Macro Connection
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 20, Problem 52b

Equation 20.3 is the mean free path of a particle through a gas of identical particles of equal radius. An electron can be thought of as a point particle with zero radius. Electrons travel 3.0 km through the Stanford Linear Accelerator. In order for scattering losses to be negligible, the pressure inside the accelerator tube must be reduced to the point where the mean free path is at least 50 km. What is the maximum possible pressure inside the accelerator tube, assuming T = 20℃? Give your answer in both Pa and atm.

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The mean free path (λ) is given by the formula: λ=1√2πr2n, where r is the radius of the particles, and n is the number density of particles. Since the electron is treated as a point particle with zero radius, the formula simplifies to focus on the relationship between pressure and mean free path.
The number density of particles n is related to the pressure P and temperature T by the ideal gas law: n=PkT, where k is the Boltzmann constant.
Rearrange the mean free path formula to solve for pressure: P=1√2πr2λkT. Substitute the given values: λ=50 km, T=20°C (convert to Kelvin), and r=0 (point particle).
Convert the temperature to Kelvin: T=20+273.15. Also, ensure the mean free path is in meters: λ=50 km=50,000 m.
Finally, calculate the pressure in Pascals using the rearranged formula. To convert the result to atmospheres, use the conversion factor: 1 atm=101,325 Pa. This will give the maximum possible pressure in both units.

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Key Concepts

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

Mean Free Path

The mean free path is the average distance a particle travels between collisions with other particles. It is influenced by the density and size of the particles in a gas. A longer mean free path indicates fewer collisions, which is crucial for applications like particle accelerators where minimizing scattering is essential for maintaining beam quality.
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Mean Free Path

Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas through the equation PV = nRT. This law helps in calculating the pressure of a gas when its volume and temperature are known. In the context of the accelerator, it allows us to determine the maximum pressure that can be achieved while ensuring the mean free path is sufficiently long.
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Scattering Losses

Scattering losses occur when particles, such as electrons, collide with gas molecules, leading to a loss of energy and direction. In particle accelerators, minimizing these losses is critical for maintaining the integrity of the particle beam. By increasing the mean free path through reduced pressure, the likelihood of scattering events is decreased, allowing for more efficient acceleration.
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Related Practice
Textbook Question

On earth, STP is based on the average atmospheric pressure at the surface and on a phase change of water that occurs at an easily produced temperature, being only slightly cooler than the average air temperature. The atmosphere of Venus is almost entirely carbon dioxide (CO2), the pressure at the surface is a staggering 93 atm, and the average temperature is 470℃. Venusian scientists, if they existed, would certainly use the surface pressure as part of their definition of STP. To complete the definition, they would seek a phase change that occurs near the average temperature. Conveniently, the melting point of the element tellurium is 450℃. What are (a) the rms speed and (b) the mean free path of carbon dioxide molecules at Venusian STP based on this phase change in tellurium? The radius of a CO2 molecule is 1.5 x 10-10 m.

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Textbook Question

5.0 x 1023 nitrogen molecules collide with a 10 cm2 wall each second. Assume that the molecules all travel with a speed of 400 m/s and strike the wall head-on. What is the pressure on the wall?

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Textbook Question

Uranium has two naturally occurring isotopes. 238U^{238}\(\text{U}\) has a natural abundance of 99.3%99.3\% and 235U^{235}\(\text{U}\) has an abundance of 0.7%0.7\%. It is the rarer 235U^{235}\(\text{U}\) that is needed for nuclear reactors. The isotopes are separated by forming uranium hexafluoride, UF6\(\text{UF}\)_6, which is a gas, then allowing it to diffuse through a series of porous membranes. 235UF6^{235}UF_6 has a slightly larger rms speed than 238UF6^{238}UF_6 and diffuses slightly faster. Many repetitions of this procedure gradually separate the two isotopes. What is the ratio of the rms speed of 235UF6^{235}UF_6 to that of 238UF6^{238}UF_6?

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Textbook Question

A gas cylinder has a piston at one end that is moving outward at speed vpiston during an isobaric expansion of the gas. Find an expression for the rate at which vrms is changing in terms of vpiston, the instantaneous value of vrms, and the instantaneous value L of the length of the cylinder.

Textbook Question

You are watching a science fiction movie in which the hero shrinks down to the size of an atom and fights villains while jumping from air molecule to air molecule. In one scene, the hero's molecule is about to crash head-on into the molecule on which a villain is riding. The villain's molecule is initially 50 molecular radii away and, in the movie, it takes 3.5 s for the molecules to collide. Estimate the air temperature required for this to be possible. Assume the molecules are nitrogen molecules, each traveling at the rms speed. Is this a plausible temperature for air?

Textbook Question

Photons of light scatter off molecules, and the distance you can see through a gas is proportional to the mean free path of photons through the gas. Photons are not gas molecules, so the mean free path of a photon is not given by Equation 20.3, but its dependence on the number density of the gas and on the molecular radius is the same. Suppose you are in a smoggy city and can barely see buildings 500 m away. How far would you be able to see if the temperature suddenly rose from 20°C to a blazing hot 1500°C with the pressure unchanged?

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