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The density of atoms, mostly hydrogen, in interstellar space is about one per cubic centimeter. Estimate the mean free path of the hydrogen atoms, assuming an atomic diameter of \(10^{-10} \mathrm{~m}\).

Short Answer

Expert verified
The mean free path is approximately \(2.25 \times 10^{13} \text{ m}\).

Step by step solution

01

Convert Atomic Density to SI Units

The given density of hydrogen atoms is one per cubic centimeter. First, convert this density to SI units (cubic meters). Since there are \(1 ext{ m}^3 = 10^6 ext{ cm}^3\), the density in SI units is \(1 ext{ atom/cm}^3 = 10^6 ext{ atoms/m}^3\).
02

Understand the Concept of Mean Free Path

The mean free path \(\lambda\) is the average distance a particle travels before colliding with another particle. For a gas, this can be calculated using the formula \(\lambda = \frac{1}{\sqrt{2} \cdot \pi \cdot d^2 \cdot n}\), where \(d\) is the diameter of the atoms and \(n\) is the number density of the atoms.
03

Substitute the Given Values

From the problem, the atomic diameter \(d = 10^{-10} \text{ m}\), and the number density \(n = 10^6 \text{ atoms/m}^3\). Substitute these into the formula for the mean free path: \(\lambda = \frac{1}{\sqrt{2} \cdot \pi \cdot (10^{-10})^2 \cdot 10^6}\).
04

Calculate the Mean Free Path

Calculate the mean free path by evaluating the expression: \[\lambda = \frac{1}{\sqrt{2} \cdot \pi \cdot 10^{-20} \cdot 10^6} = \frac{1}{\sqrt{2} \cdot \pi \cdot 10^{-14}}\]. Further simplifying, \[\lambda \approx \frac{1}{4.44 \times 10^{-14}} \approx 2.25 \times 10^{13} \text{ m}\].
05

Verify and Interpret the Result

The calculated mean free path is an extraordinarily large value, \(2.25 \times 10^{13} \text{ m}\), due to the very low density of hydrogen in interstellar space. This means that, on average, a hydrogen atom can travel this distance before colliding with another atom.

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Key Concepts

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

Mean Free Path
In physics, the concept of mean free path is crucial when studying the behavior of particles, especially in gases like those found in interstellar space. The mean free path is defined as the average distance a particle, such as an atom or molecule, can travel before it collides with another particle. This measurement helps scientists understand how substances flow and diffuse through space. In the context of our problem, the mean free path \(\lambda\) is calculated using the formula \(\lambda = \frac{1}{\sqrt{2} \cdot \pi \cdot d^2 \cdot n}\), where:
  • d is the diameter of the atom, here given as \(10^{-10} \, \text{m}\),
  • n is the atomic density, or number of atoms per cubic meter.
This equation implies that the mean free path is inversely proportional to both the square of the atomic diameter and the atomic density. In simpler terms, larger atoms or more crowded conditions will reduce the mean free path, causing atoms to collide more frequently.
Given the vast emptiness of interstellar space, hydrogen atoms enjoy a very long mean free path, illustrating the rarity of atomic interactions in this environment.
Atomic Density
Atomic density refers to the number of atoms within a unit volume of space. It's a critical factor when calculating the mean free path of particles, as density influences how often atoms collide. More dense environments lead to shorter mean free paths because atoms are closer together and therefore more likely to collide. In the exercise, the density of hydrogen atoms in interstellar space is given as one per cubic centimeter. To use it in calculations, we convert it into SI units. The metric system provides a consistent framework to measure quantities, and here, it's essential to convert cubic centimeters to cubic meters—a process achieved by multiplying the density by \(10^6\) (since \(1\, \text{m}^3 = 10^6 \, \text{cm}^3\)). Thus, our atomic density becomes \(10^6 \, \text{atoms/m}^3\).
This density in interstellar space is extraordinarily low when compared to Earth-bound environments, resulting in extended mean free paths, illustrating how few and far between hydrogen atoms are amongst the stars.
SI Units Conversion
The International System of Units (SI) is the standard for scientific measurements worldwide. It ensures clarity and uniformity in scientific communication. Conversion of quantities to SI units is often a critical first step in solving physics problems, like the given exercise.In our problem, we began with an atomic density of 1 atom per cubic centimeter. To convert this to the SI unit (cubic meters), we use the conversion:
  • 1 \(\text{cm}^3\) is \(10^{-6} \, \text{m}^3\), so
  • 1 atom/\(\text{cm}^3\) becomes \(10^6 \, \text{atoms/m}^3\).
This conversion allows for ease in computation using the mean free path formula, which requires the atomic density in \(\text{atoms/m}^3\). Converting measurements into SI units is not just helpful but essential for accuracy, especially when dealing with fundamental constants and formulas that rely on these units.

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Most popular questions from this chapter

(III) How well does the ideal gas law describe the pressurized air in a scuba tank? (a) To fill a typical scuba tank, an air compressor intakes about 2300 \(\mathrm{L}\) of air at 1.0 \(\mathrm{atm}\) and compresses this gas into the tank's \(12-\mathrm{L}\) internal volume. If the filling process occurs at \(20^{\circ} \mathrm{C},\) show that a tank holds about 96 \(\mathrm{mol}\) of air. \((b)\) Assume the tank has 96 \(\mathrm{mol}\) of air at \(20^{\circ} \mathrm{C}\) . Use the ideal gas law to predict the air's pressure within the tank. (c) Use the van der Waals equation of state to predict the air's pressure within the tank. For air, the van der Waals constants are \(a=0.1373 \mathrm{N} \cdot \mathrm{m}^{4} / \mathrm{mol}^{2}\) and \(b=3.72 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol} .\) (d) Taking the van der Waals pressure as the true air pressure, show that the ideal gas law predicts a pressure that is in error by only about 3\(\% .\)

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(II) Oxygen diffuses from the surface of insects to the interior through tiny tubes called tracheae. An average trachea is about \(2 \mathrm{~mm}\) long and has cross-sectional area of \(2 \times 10^{-9} \mathrm{~m}^{2}\). Assuming the concentration of oxygen inside is half what it is outside in the atmosphere, \((a)\) show that the concentration of oxygen in the air (assume \(21 \%\) is oxygen) at \(20^{\circ} \mathrm{C}\) is about \(8.7 \mathrm{~mol} / \mathrm{m}^{3},\) then \((b)\) calculate the diffusion rate \(J,\) and \((c)\) estimate the average time for a molecule to diffuse in. Assume the diffusion constant is \(1 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\).

A \(0.50-\mathrm{kg}\) trash-can lid is suspended against gravity by tennis balls thrown vertically upward at it. How many tennis balls per second must rebound from the lid elastically, assuming they have a mass of \(0.060 \mathrm{~kg}\) and are thrown at \(12 \mathrm{~m} / \mathrm{s} ?\)

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