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Chapter 35: Problem 1

Explain qualitatively why a particle confined to a finite region cannot have zero energy.

Short Answer

Expert verified
A particle confined to a finite region cannot have zero energy due to the Heisenberg Uncertainty Principle. Higher the certainty in position (due to the confinement within a finite region), greater is the uncertainty in momentum and hence greater is the kinetic energy. Therefore, the energy of a confined particle is always non-zero.

Step by step solution

01

Understanding the Heisenberg Uncertainty Principle

Heisenberg's Uncertainty Principle is a fundamental concept in Quantum Physics. The principle states that it is impossible to simultaneously determine with full precision the position and momentum of a particle. This principle is formulated mathematically as \( \Delta x . \Delta p \geq \hbar / 2 \), where \( \Delta x \) represents the uncertainty in position, \( \Delta p \) represents the uncertainty in momentum and \( \hbar \) is the Planck constant divided by 2蟺.
02

Understanding the Implication of Confinement

When a particle is confined to a certain region, the certainty of its position increases, because we know it鈥檚 inside that region. Following Heisenberg鈥檚 Uncertainty Principle, if the position is known with a higher level of certainty (small \( \Delta x \)), the uncertainty in momentum would necessarily have to increase (large \( \Delta p \)).
03

Relating the Uncertainty Principle to Energy

In classical mechanics, the kinetic energy of a particle is given by the expression \( K.E = P^2 / 2M \), where P is the momentum and M is the mass of the particle. Higher the momentum, higher would be the kinetic energy. So, when the particle is confined to a certain region, the uncertainty or variability in momentum would increase which subsequently raises the minimum kinetic energy that the particle can have. Hence, the energy would always be non-zero.

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Most popular questions from this chapter

A \(9-\) W laser beam shines on an ensemble of \(10^{24}\) electrons, each in the ground state of a one-dimensional infinite square well \(0.72 \mathrm{nm}\) wide. The photon energy is just high enough to raise an electron to its first excited state. How many electrons can be excited if the beam shines for \(10 \mathrm{ms} ?\)

(a) Using the potential energy \(U=\frac{1}{2} m \omega^{2} x^{2}\) discussed on page \(675,\) develop the Schr枚dinger equation for the harmonic oscillator. (b) Show by substitution that \(\psi_{0}(x)=A_{0} e^{-\alpha^{2} x^{2} / 2}\) satisfies your equation, where \(\alpha^{2}=m \omega / \hbar\) and the energy is given by Equation 35.7 with \(n=0 .\) (c) Find the normalization constant \(A_{0}\) You then have the ground-state wave function for the harmonic oscillator.

You're trying to convince a friend that nuclear energy represents a much more concentrated energy source than fossil fuels, whose combustion involves rearranging atomic electrons. For a rough comparison, you calculate the ground-state energy of a proton confined to 1 -fm-diameter atomic nucleus and that of an electron confined to a 0.1 -nm-diameter atom. Approximating each system as a one-dimensional infinite square well, what's the ratio of their ground-state energies?

An infinite square well extends from \(-L / 2\) to \(L / 2 .\) (a) Find expressions for the normalized wave functions for a particle of mass \(m\) in this well, giving separate expressions for even and odd quantum numbers. (b) Find the corresponding energy levels.

The next three problems solve the Schr枚dinger equation for finite square wells like that shown in Fig. \(35.14 .\) It's convenient to work in dimensionless forms of the particle energy \(E\) and well depth \(U_{0},\) given respectively by \(\epsilon=2 m L^{2} E / \hbar^{2}\) and \(\mu=2 m L^{2} U_{0} / \hbar^{2}\) Assuming that \(E
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