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A gas of helium atoms at \(273 \mathrm{K}\) is in a cubical container with \(25.0 \mathrm{cm}\) on a side. (a) What is the minimum uncertainty in momentum components of helium atoms? (b) What is the minimum uncertainty in velocity components? (c) Find the ratio of the uncertainties in (b) to the mean speed of an atom in each direction.

Short Answer

Expert verified
The minimum uncertainty in momentum components of helium atoms can be calculated using Heisenberg's Uncertainty principle. The minimum uncertainty in velocity components can be determined from this value, knowing the relationship between momentum and velocity. Finally, the ratio of these uncertainties to the mean speed of an atom in each direction can be calculated using the derived values and the formula for the mean speed of an atom.

Step by step solution

01

Calculate the minimum uncertainty in momentum components

Heisenberg's Uncertainty Principle states that the minimum uncertainty in momentum is determined by the equation \(\Delta p \Delta x \geq \hbar/2\) where \(\hbar\) is Planck's constant divided by \(2\pi\) and \(\Delta x\) is the uncertainty in position which is equivalent to the width of the box in our case. So, we can rearrange the equation and solve for \(\Delta p\) which is the minimum uncertainty in momentum components. In this case, \(\Delta x\) is \(25.0 cm = 0.25 m\), and \(\hbar = h/(2\pi) = 1.055 * 10^{-34} Js\). So, \(\Delta p = \hbar/(2\Delta x)\)
02

Calculate the minimum uncertainty in velocity components

Uncertainty in velocity (\(\Delta v\)) can be calculated using the relationship between momentum and velocity which is \(p = m*v\) where `m` is the mass. So, we can rearrange the equation and solve for \(\Delta v\) that gives us \(\Delta v = \Delta p/m\). The mass of a helium atom is approximately \(6.64*10^{-27} kg\). Substituting the values we get \(\Delta v = \(\Delta p / m)\
03

Calculate the ratio of the uncertainties to the mean speed in each direction

The mean speed (\(v\)) of an atom in each direction can be calculated using the equation \(v = (8kT/(\pi m))^0.5\) where \(k\) is the Boltzmann constant, \(T\) is the Kelvin temperature and \(m\) is the atom's mass. Substituting \(k = 1.38*10^{-23}\), \(T = 273K\), and \(m = 6.64*10^{-27} kg\), we get the mean speed. Finally, the ratio of the uncertainties to the mean speed in each direction (\(\Delta v/v\)) can be calculated.

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Key Concepts

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

Minimum Uncertainty in Momentum
When we delve into the fascinating world of quantum physics, we encounter the enigmatic Heisenberg's Uncertainty Principle. This principle introduces a fundamental limit to our ability to measure certain properties of particles simultaneously. One important aspect of this principle affects the momentum of particles.

According to Heisenberg's formula, \(\Delta p \Delta x \geq \hbar/2\), the product of the uncertainties in position (\(\Delta x\)) and momentum (\(\Delta p\)) of a particle must be greater than or equal to one-half of the reduced Planck's constant (\(\hbar\)).

For practical purposes, in an exercise scenario like the one with helium atoms in a cubical container, the minimum uncertainty in momentum components can be thought of as the 'fuzziness' or 'spread' in momentum that is inherently present due to quantum constraints. Helium atoms confined within a definitive space, such as a cubical container, have their position somewhat known, forcing a corresponding uncertainty in their momentum.

The exercise illustrates that by knowing the width of the cubical box (which defines \(\Delta x\)) and employing Planck's constant, we can determine the minimum uncertainty in the momentum of helium atoms. Interestingly, this isn't just theoretical; it's a quantifiable limit that informs us about the intrinsic properties of matter at the microscopic level.
Minimum Uncertainty in Velocity
The uncertainty in velocity is inextricably linked to the uncertainty in momentum. However, where momentum incorporates mass into the equation, velocity stands alone as a measure of speed without mass. For atoms or particles of a known mass, once we have a grasp of the uncertainty in momentum (\(\Delta p\)), we can extrapolate the minimum uncertainty in velocity (\(\Delta v\)) with a relatively simple conversion: \(\Delta v = \Delta p / m\), where \(m\) is the mass of the particle.

Our exercise focuses on helium atoms—tiny entities that abide by the rules of quantum mechanics. Given the mass of a single helium atom, we can calculate the minimum uncertainty in its velocity by dividing the previously computed \(\Delta p\) by this mass. This result isn't just a theoretical concept but has real implications, for instance, in understanding the behavior of gases and in fields like spectroscopy, where the movement of atoms and their interactions with light are pivotal.

The notion of minimum uncertainty in velocity allows us to peak into the nuances of a particle's ability to 'decide' its speed, thus demystifying a key quantum behavior that classical physics would struggle to explain.
Mean Speed of an Atom
Moving on to a macroscopic perspective, the mean speed of an atom in a gas offers us a bridge between quantum uncertainty and the observable world of thermodynamics. At a given temperature, atoms in a gas move chaotically, colliding with one another and the walls of their container. This random motion has a certain average speed, known statistically as the mean speed.

On a mathematical level, the mean speed \(v\) of an atom can be derived from the formula \(v = (8kT/(\pi m))^{0.5}\), involving the temperature (\(T\)) of the gas, Boltzmann's constant (\(k\)), and the atom's mass (\(m\)). For our helium gas at 273K, the calculation is straightforward and reveals the kinetic thermal motion experienced by the atoms.

Understanding the mean speed ties into not just equations and numbers; it reflects the energy within a system and plays a key role in phenomena like diffusion, sound propagation, and even the simple act of a balloon deflating. The mean speed creates a context for us to better appreciate the preceding calculations of uncertainty, by comparing them to a tangible property of the gas—revealing a full picture of the microscopic and macroscopic dance of particles.

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Most popular questions from this chapter

A simple model of a radioactive nuclear decay assumes that \(\alpha\) -particles are trapped inside a well of nuclear potential that walls are the barriers of a finite width \(2.0 \mathrm{fm}\) and height \(30.0 \mathrm{MeV}\). Find the tunneling probability across the potential barrier of the wall for \(\alpha\) -particles having kinetic energy (a) \(29.0 \mathrm{MeV}\) and \((\mathrm{b}) 20.0 \mathrm{MeV} .\) The mass of the \(\alpha\) -particle is \(m=6.64 \times 10^{-27} \mathrm{kg}\)

A velocity measurement of an \(\alpha\) -particle has been performed with a precision of \(0.02 \mathrm{mm} / \mathrm{s}\). What is the minimum uncertainty in its position?

A quantum particle with initial kinetic energy 32.0 eV encounters a square barrier with height 41.0 eV and width \(0.25 \mathrm{nm} .\) Find probability that the particle tunnels through this barrier if the particle is (a) an electron and, (b) a proton.

Suppose an electron is confined to a region of length \(0.1 \mathrm{nm}\) (of the order of the size of a hydrogen atom) and its kinetic energy is equal to the ground state energy of the hydrogen atom in Bohr's model (13.6 eV). (a) What is the minimum uncertainty of its momentum? What fraction of its momentum is it? (b) What would the uncertainty in kinetic energy of this electron be if its momentum were equal to your answer in part (a)? What fraction of its kinetic energy is it?

Which one of the following functions, and why, qualifies to be a wave function of a particle that can move along the entire real axis? (a) \(\psi(x)=A e^{-x^{2}}\) (b) \(\psi(x)=A e^{-x} ;\) (c) \(\psi(x)=A \tan x\) (d) \(\psi(x)=A(\sin x) / x ;\) (e) \(\psi(x)=A e^{-|x|}\)

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