/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Recall from Section 14.3 that th... [FREE SOLUTION] | 91Ó°ÊÓ

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Recall from Section 14.3 that the average kinetic energy of an atom in a monatomic ideal gas is given by \(\overline{\mathrm{KE}}=\frac{3}{2} k T,\) where \(k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}\) and \(T\) is the Kelvin temperature of the gas. Determine the de Broglie wavelength of a helium atom (mass \(=6.65 \times 10^{-27} \mathrm{~kg}\) ) that has the average kinetic energy at room temperature \((293 \mathrm{~K})\)

Short Answer

Expert verified
The de Broglie wavelength of the helium atom is approximately \(7.14 \times 10^{-11} \mathrm{m}.\)

Step by step solution

01

Calculate Average Kinetic Energy

First, we need to calculate the average kinetic energy of a helium atom at room temperature \(T = 293\, K\). This is done using the formula \(\overline{\mathrm{KE}} = \frac{3}{2} k T \), where \(k = 1.38 \times 10^{-23}\, \mathrm{J/K}\). Substitute the values: \[ \overline{\mathrm{KE}} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 293 \approx 6.0669 \times 10^{-21} \mathrm{J}. \]
02

Determine Velocity of Helium Atom

The kinetic energy of an atom is also given by \(KE = \frac{1}{2}mv^2 \), where \(m\) is the mass of the atom and \(v\) is its velocity. Using the average kinetic energy calculated earlier, set \( \frac{1}{2} \times 6.65 \times 10^{-27} \times v^2 = 6.0669 \times 10^{-21} \) to solve for \(v\): \[ v^2 = \frac{2 \times 6.0669 \times 10^{-21}}{6.65 \times 10^{-27}} \approx 1.82465 \times 10^{6}\, \mathrm{m^2/s^2}. \] \[ v \approx 1.351 \times 10^{3} \mathrm{m/s}. \]
03

Calculate de Broglie Wavelength

Now that we have the velocity of the helium atom, use the de Broglie wavelength formula \(\lambda = \frac{h}{mv} \), where \(h = 6.626 \times 10^{-34}\, \mathrm{J\cdot s}\) is Planck's constant. Substituting the known values in gives: \[ \lambda = \frac{6.626 \times 10^{-34}}{6.65 \times 10^{-27} \times 1.351 \times 10^3} \approx 7.14 \times 10^{-11} \mathrm{m}. \]

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Key Concepts

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

Average Kinetic Energy
Average kinetic energy is a measure of the energy an atom possesses due to its motion. In the context of a monatomic ideal gas, this is a simple yet powerful concept that reflects the kinetic activity of gas atoms. The formula for average kinetic energy, \[\overline{\mathrm{KE}} = \frac{3}{2} k T\] helps us understand how temperature affects the motion of gas particles. Here, \(k\) is the Boltzmann constant with a value of \(1.38 \times 10^{-23}\, \mathrm{J/K}\), and \(T\) is the absolute temperature in Kelvin. - The higher the temperature, the more kinetic energy and hence more velocity the particles attain. - At room temperature \((293\, K)\), gases have a moderate level of kinetic energy.This formula shows a direct relationship between temperature and kinetic energy, indicating that with increasing temperature, each particle's speed increases.
Monatomic Ideal Gas
A monatomic ideal gas consists of individual atoms that move independently in space, contributing to the overall behavior of the gas. These gases, like helium, are usually noble gases that do not form molecules. - The term "ideal" implies that the gas follows the ideal gas laws, where interactions between atoms are negligible and occupy no volume themselves. - In an ideal gas model, the behavior of gas can be precisely described by macroscopic parameters such as pressure, volume, and temperature, using the ideal gas law. With monatomic gases, the simplicity of dealing with single atoms makes calculations straightforward, such as determining average kinetic energy. Understanding their behavior helps simplify complex thermodynamic systems.
Planck's Constant
Planck's constant is a fundamental constant in physics that represents the smallest possible unit of energy. Its value is approximately \(6.626 \times 10^{-34}\, \mathrm{J \cdot s}\). - It is a key factor in quantum mechanics, relating the energy carried by a photon to its frequency via the equation, \[E = hu\],where \(E\) is energy, \(h\) is Planck's constant, and \(u\) is frequency. - In calculations of the de Broglie wavelength, it provides a bridge between the wave nature and particle nature of matter.Planck's constant is crucial when understanding phenomena at very small scales, defining the scale at which quantum mechanical effects become significant.
Room Temperature
Room temperature is a typical ambient temperature environment, generally considered to be about 25°C or 298 K. However, in scientific calculations, 293 K (20°C) is often used for simplicity. - Room temperature provides a standard reference point for scientific experiments and calculations. - It is a convenient baseline for evaluating the behavior of substances under normal conditions without requiring extreme heating or cooling. In our example, using room temperature allows us to calculate the average kinetic energy of helium atoms effectively. This temperature is close to the temperatures encountered in everyday environments, providing a practical insight into the theoretical calculations.

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Most popular questions from this chapter

Multiple-Concept Example 3 reviews the concepts necessary to solve this problem. Light is incident on the surface of metallic sodium, whose work function is \(2.3 \mathrm{eV}\). The maximum speed of the photoelectrons emitted by the surface is \(1.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What is the wavelength of the light?

A magnesium surface has a work function of \(3.68 \mathrm{eV}\). Electro magnetic waves with a wavelength of \(215 \mathrm{nm}\) strike the surface and eject electrons. Find the maximum kinetic energy of the ejected electrons. Express your answer in electron volts.

(a) How many photons (wavelength \(=620 \mathrm{nm}\) ) must be absorbed to melt a \(2.0-\mathrm{kg}\) block of ice at \(0^{\circ} \mathrm{C}\) into water at \(0^{\circ} \mathrm{C} ?\) (b) On the average, how many \(\mathrm{H}_{2} \mathrm{O}\) molecules does one photon convert from the ice phase to the water phase?

Concept Questions The maximum speed of the electrons emitted from a metal surface is \(v_{\mathrm{A}}\) when the wavelength of the light used is \(\lambda_{\mathrm{A}}\). However, a smaller maximum speed \(v_{\mathrm{B}}\) is observed when a wavelength \(\lambda_{\mathrm{B}}\) is used. (a) In which case is there a greater maximum kinetic energy \(\mathrm{KE}_{\text {max }} ?\) (b) In which case is the frequency of the light greater? (c) Is wavelength \(\lambda_{\text {A }}\) greater or smaller than \(\lambda_{\mathrm{B}}\) ? Explain your answers. Problem The work function of a metal surface is \(4.80 \times 10^{-19} \mathrm{~J}\). The maximum speed of the electrons emitted from the surface is \(v_{A}=7.30 \times 10^{5} \mathrm{~m} / \mathrm{s}\) when the wavelength of the light is \(\lambda_{A}\). However, a maximum speed of \(v_{B}=5.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\) is observed when the wavelength is \(\lambda_{B}\). Find the wavelengths \(\lambda_{A}\) and \(\lambda_{B}\). Be sure your answers are consistent with your answers to the Concept Questions.

An electron, starting from rest, accelerates through a potential difference of \(418 \mathrm{~V}\). What is the final de Broglie wavelength of the electron, assuming that its final speed is much less than the speed of light?

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