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Chapter 4: Problem 42

An apparatus is used to prepare an atomic beam by heating a collection of atoms to a temperature \(T\) and allowing the beam to emerge through a hole of diameter \(d\) in one side of the oven. The beam then travels through a straight path of length \(L\). Show that the uncertainty principle causes the diameter of the beam at the end of the path to be larger than \(d\) by an amount of order \(L \hbar / d \sqrt{3 m k T}\), where \(m\) is the mass of an atom. Make a numerical estimate for typical values of \(T=1500 \mathrm{K}, m=7 \mathrm{u}\) (lithium atoms), \(d=3 \mathrm{mm},\) and \(L=2 \mathrm{m}\)

Short Answer

Expert verified
The diameter of the beam at the end of the path is increased by approximately \(10^{-7}\) meters due to the uncertainty principle.

Step by step solution

01

- Understand the problem setup

The problem involves an atomic beam emerging through a hole and traveling a certain distance. The goal is to find how much the diameter of the beam spreads due to the uncertainty principle.
02

- Recall the uncertainty principle

The Heisenberg uncertainty principle states that \(\triangle x \triangle p_x \geq \hbar/2\), where \(\triangle x\) is the position uncertainty and \(\triangle p_x\) is the momentum uncertainty.
03

- Relate temperature to momentum uncertainty

At temperature \(T\), the average kinetic energy of atoms is given by \( \frac{3}{2}k_B T \). The momentum uncertainty can be expressed as \(\triangle p_x \approx \sqrt{3m k_B T}\).
04

- Apply the uncertainty principle

\(\triangle x \geq \frac{\hbar}{2 \triangle p_x } \approx \frac{\hbar}{2 \sqrt{3m k_B T}} \). The uncertainty in position will cause a spread in the beam diameter.
05

- Calculate beam spread over distance L

The additional spread in beam diameter can be estimated by multiplying the position uncertainty by the distance \(L\): \(\triangle x \approx L \frac{\hbar}{d \sqrt{3m k_B T}}\).
06

- Plug in numerical values

Given \(T = 1500K\), \(m = 7 \times 1.66 \times 10^{-27} kg\) (mass of lithium atom), \( d = 3 \times 10^{-3} m\), and \(L = 2 m\): \( \triangle x_L \approx 2m \frac{1.054 \times 10^{-34} Js}{3 \times 10^{-3} m \sqrt{3 \times 7 \times 1.66 \times 10^{-27} kg \times 1.38 \times 10^{-23} J/K \times 1500K }} \approx 10^{-7} m\).

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Key Concepts

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

Atomic Beam Dynamics
Atomic beam dynamics focuses on how atoms move and spread as they travel in a beam. In our scenario, an atomic beam is created by heating atoms so they can escape through a small hole and then move in a straight line. The key point here is how the atoms in the beam spread out over a distance due to their momentum and initial velocity directions. We end up with a beam that is wider at a distance than at the source hole. Understanding this spread helps us comprehend the motion and distribution of atoms in a beam.
Temperature Effects on Atomic Motion
When the temperature of the atoms in the oven increases, their kinetic energy also increases. This relationship is foundational in physics. For our problem, temperature plays a crucial role in the beam spread. The kinetic energy, \[ \frac{3}{2}k_B T \], translates into the atoms' momentum, where higher temperatures cause higher uncertainties in momentum. This higher momentum uncertainty means the atoms will have a greater range of velocities, leading to a wider spread as they travel through space.
Quantum Mechanics in Physical Systems
The Heisenberg Uncertainty Principle tells us that we cannot precisely measure both the position and momentum of a particle. This principle is crucial in explaining the spread of our atomic beam. If the position of an atom passing through the hole is somewhat known, its momentum (or speed and direction) becomes less certain. This uncertainty then results in the beam's diameter growing as the atoms travel along the path. The longer the path, the more noticeable this spread becomes.
Calculating Uncertainties in Physics
Calculating uncertainties involves applying the Heisenberg Uncertainty Principle: \[ \Delta x \Delta p_x \geq \frac{\hbar}{2} \]. For our beam鈥檚 spread, we relate the momentum uncertainty due to temperature, \[ \Delta p_x \approx \sqrt{3m k_B T} \]. The position uncertainty or beam spread becomes \[ \Delta x \approx \frac{\hbar}{2 \sqrt{3m k_B T}} \]. Multiplying this by the travel distance L gives us the additional diameter spread. By substituting specific values (T=1500K, m=7u, d=3mm, L=2m), we quantify the spread, which is shown to be on the order of 10^-7 meters.

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Most popular questions from this chapter

A certain crystal is cut so that the rows of atoms on its surface are separated by a distance of \(0.352 \mathrm{nm} .\) A beam of electrons is accelerated through a potential difference of \(175 \mathrm{V}\) and is incident normally on the surface. If all possible diffraction orders could be observed, at what angles (relative to the incident beam) would the diffracted beams be found?

By doing a nuclear diffraction experiment, you measure the de Broglie wavelength of a proton to be \(8.29 \mathrm{fm}\) (a) What is the speed of the proton? (b) Through what potential difference must it be accelerated to achieve that speed?

Certain surface waves in a fluid travel with phase velocity \(\sqrt{b / \lambda}\), where \(b\) is a constant. Find the group velocity of a packet of surface waves, in terms of the phase velocity.

It is possible to construct oscillatory wave packets without using trigonometric functions. Consider the function \(y(x)=\left(64 x^{6}-240 x^{4}+180 x^{2}-15\right) e^{-x^{2}}\). Wave packets using polynomials occur in quantum mechanics as solutions to the simple harmonic oscillator and the hydrogen atom, as we discuss later in this text. \((a)\) Sketch this function in the region where it has reasonably large amplitude. (b) What is the width of this wave packet? Make a rough estimate from your sketch. (c) Estimate the average wavelength. (d) Estimate the uncertainty in the Wavelength.

In designing an experiment, you want a beam of photons and a beam of electrons with the same wavelength of \(0.281 \mathrm{nm},\) equal to the separation of the \(\mathrm{Na}\) and \(\mathrm{Cl}\) ions in a crystal of NaCl. Find the energy of the photons and the kinetic energy of the electrons.
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