Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 37: Problem 11

True or False: The larger the amplitude of a Schrödinger wave function, the larger its kinetic energy. Explain your answer.

Short Answer

Expert verified
Explain your reasoning. Answer: No, a larger amplitude of a Schrödinger wave function does not necessarily mean a larger kinetic energy. The kinetic energy is influenced by the second derivative of the wave function rather than the amplitude.

Step by step solution

01

Review the Schrödinger equation

The time-independent Schrödinger equation is given by: \[ \hat{H} \psi(x) = E \psi(x) \] Where \(\hat{H}\) is the Hamiltonian operator, which describes the total energy of the system, \(E\) is the energy eigenvalue, and \(\psi(x)\) is the wave function. The Hamiltonian operator can be broken down into two components - the kinetic energy operator, \(\hat{T}\), and the potential energy operator, \(\hat{V}\): \[ \hat{H} = \hat{T} + \hat{V} \]
02

Review the kinetic energy operator

The kinetic energy operator for a particle in one dimension is given by: \[ \hat{T} = -\dfrac{\hbar^2}{2m} \dfrac{d^2}{dx^2} \] Where \(m\) is the particle's mass, and \(\hbar\) is the reduced Planck constant.
03

Relate the amplitude of the wave function to kinetic energy

The amplitude of the wave function, which is the square root of the probability density, is related to the modulus of the wave function: \[ \lvert \psi(x) \rvert ^2 \] However, the kinetic energy of a particle is not directly related to the wave function's amplitude (or the square root of the probability density). Instead, it is connected to the spatial derivatives of the wave function. We can see from the kinetic energy operator that the second derivative of the wave function influences the kinetic energy.
04

Determine the validity of the given statement

So, when we consider the Schrödinger equation and its components, we find that the amplitude of the wave function doesn't have a direct impact on the kinetic energy of the particle. The second derivative of the wave function is the primary factor influencing the kinetic energy. Therefore, the given statement is: False. The larger the amplitude of a Schrödinger wave function does not necessarily mean a larger kinetic energy. The kinetic energy is influenced by the second derivative of the wave function rather than the amplitude.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle of energy \(E=5 \mathrm{eV}\) approaches an energy barrier of height \(U=8 \mathrm{eV}\). Quantum mechanically there is a finite probability that the particle tunnels through the barrier. If the barrier height is slowly decreased, the probability that the particle will reflect from the barrier will a) decrease. b) increase. c) not change.

Consider an electron in a three-dimensional box-with infinite potential walls- of dimensions \(1.00 \mathrm{nm} \times 2.00 \mathrm{nm} \times 3.00 \mathrm{nm}\). Find the quantum numbers \(n_{x}, n_{y}, n_{z}\) and energies in \(\mathrm{eV}\) of the six lowest energy levels. Are any of these levels degenerate, that is, do any distinct quantum states have identical energies?

An oxygen molecule has a vibrational mode that behaves approximately like a simple harmonic oscillator with frequency \(2.99 \cdot 10^{14} \mathrm{rad} / \mathrm{s} .\) Calculate the energy of the ground state and the first two excited states.

A beam of electrons moving in the positive \(x\) -direction encounters a potential barrier that is \(2.51 \mathrm{eV}\) high and \(1.00 \mathrm{nm}\) wide. Each electron has a kinetic energy of \(2.50 \mathrm{eV},\) and the electrons arrive at the barrier at a rate of 1000 electrons/s (1000. electrons every second). What is the rate \(\mathrm{I}_{\mathrm{T}}\) in electrons/s at which electrons pass through the barrier, on average? What is the rate \(\mathrm{I}_{\mathrm{R}}\) in electrons/s at which electrons reflect back from the barrier, on average? Determine and compare the wavelengths of the electrons before and after they pass through the barrier.

An electron is confined in a one-dimensional infinite potential well of \(1.0 \mathrm{nm}\). Calculate the energy difference between a) the second excited state and the ground state, and b) the wavelength of light emitted by this radiative transition.
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Scientific Method Physics

Read Explanation

Electrostatics

Read Explanation

Electricity

Read Explanation

Particle Model of Matter

Read Explanation

Circular Motion and Gravitation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free