/*! 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 33 An electron with initial kinetic... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electron with initial kinetic energy 6.0 \(\mathrm{eV}\) encounters a barrier with height 11.0 \(\mathrm{eV}\) . What is the probability of tunneling if the width of the barrier is (a) 0.80 \(\mathrm{nm}\) and (b) 0.40 \(\mathrm{nm} ?\)

Short Answer

Expert verified
Tunneling probabilities are 1.86 x 10^-8 for 0.80 nm and 4.31 x 10^-4 for 0.40 nm.

Step by step solution

01

Define the Problem

We have a quantum mechanical tunneling problem where an electron with initial kinetic energy of 6.0 eV approaches a barrier with a height of 11.0 eV. We need to find the probability of tunneling for two different barrier widths: 0.80 nm and 0.40 nm.
02

Understand Tunneling Probability Formula

The tunneling probability through a barrier in quantum mechanics can be estimated using the formula: \( P = e^{-2\kappa a} \), where \( \kappa = \sqrt{\frac{2m(U-E)}{\hbar^2}} \), \( m \) is the mass of the electron, \( U \) is the potential energy barrier height, \( E \) is the kinetic energy of the electron, and \( a \) is the width of the barrier.
03

Calculate \( \kappa \)

First, calculate \( \kappa \) using the values: \( U = 11.0 \text{ eV}, E = 6.0 \text{ eV} \), and the known electron mass \( m = 9.11 \times 10^{-31} \) kg, and \( \hbar = 1.054 \times 10^{-34} \) J·s. Convert eV to Joules (1 eV = \( 1.602 \times 10^{-19} \) J). Thus, \( \kappa = \sqrt{\frac{2(9.11 \times 10^{-31})(11 - 6)(1.602 \times 10^{-19})}{(1.054 \times 10^{-34})^2}} \).
04

Calculate \( \kappa \) Value

Substituting the values, \( \kappa = \sqrt{\frac{2(9.11 \times 10^{-31})(5)(1.602 \times 10^{-19})}{(1.054 \times 10^{-34})^2}} \approx 1.145 \times 10^{10} \text{ m}^{-1} \).
05

Calculate Tunneling Probability for Width 0.80 nm

Convert 0.80 nm to meters (0.80 nm = \( 0.80 \times 10^{-9} \) m). Use the tunneling formula: \( P = e^{-2(1.145 \times 10^{10})(0.80 \times 10^{-9})} \).
06

Compute Probability for 0.80 nm

Using the formula, \( P = e^{-2(1.145 \times 10^{10})(0.80 \times 10^{-9})} \approx 1.86 \times 10^{-8} \). Thus, the probability of tunneling for 0.80 nm is very low.
07

Calculate Tunneling Probability for Width 0.40 nm

Convert 0.40 nm to meters (0.40 nm = \( 0.40 \times 10^{-9} \) m). Apply the probability formula: \( P = e^{-2(1.145 \times 10^{10})(0.40 \times 10^{-9})} \).
08

Compute Probability for 0.40 nm

Substitute the values, \( P = e^{-2(1.145 \times 10^{10})(0.40 \times 10^{-9})} \approx 4.31 \times 10^{-4} \). The probability of tunneling for 0.40 nm is significantly higher than for 0.80 nm.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Electron Tunneling Probability
In quantum mechanics, the tunneling probability is a fascinating phenomenon that describes how an electron can cross a potential energy barrier, even when it doesn't have enough energy to do so classically. This probability is determined by how the electron's wave-like nature interacts with the barrier.

Here is a simple method to understand it:
  • The electron behaves as a wave with a certain energy.
  • The potential energy barrier is like a wall that the electron wave approaches.
  • Unlike in classical mechanics, part of the electron wave can "leak" through the wall and continue on the other side, allowing the electron to tunnel through.
The probability of this tunneling happening is exponentially low when the barrier is wider or taller in terms of energy. The tunneling probability, denoted as \( P \), is given by the formula:
\( P = e^{-2\kappa a} \), where \( \kappa \) depends on the electron's characteristics and the barrier's properties.

In our specific problem, by manipulating this formula with given values, we can see how extreme the differences in tunneling probability are for barriers of the same height but different widths.
Potential Energy Barrier
The potential energy barrier is a crucial component in understanding why and how quantum tunneling occurs. Imagine the barrier as an energetic hill that a particle needs to overcome. Classically, only particles with energy equal to or greater than the barrier’s height can cross it. However, in quantum mechanics, things work a bit differently.

The energy barrier is defined by its height (energy) and width (physical thickness).

  • The height corresponds to the energy required to overcome the barrier classically.
  • The width impacts the amount of space over which the tunneling effect occurs.
In the given exercise, the electron has an initial kinetic energy of 6.0 eV but encounters a barrier of 11.0 eV.
Here, the barrier's height prevents the electron from crossing over by simply rolling over. Instead, it must tunnel, briefly existing in states of higher and lower energy than the barrier allows, thanks to its wave-like probability cloud that quantum mechanics predicts.
Quantum Mechanics
Quantum mechanics revolutionizes our understanding of the microscopic world. It provides a framework to explain phenomena like tunneling, which do not agree with classical physics.

In quantum mechanics:

  • Particles such as electrons are described by wave functions which give the probability of finding them in a certain place.
  • The principle of superposition allows these particles to exist in multiple states at once until measured.
  • Heisenberg's uncertainty principle indicates that we can't precisely know both the position and momentum of a particle simultaneously.
These principles allow electrons to tunnel through barriers, existing momentarily in places they classically shouldn't. It's this non-intuitive leap that quantum mechanics can predict, showing that the world of the very small follows different rules than the macro-scale world we experience daily.
Electron Kinetic Energy
The kinetic energy of an electron in quantum mechanics provides insights into its ability to approach and interact with potential barriers. Kinetic energy (\( E \)) is usually expressed in electron volts (eV) in quantum physics, which makes it easier to calculate other parameters involved in tunneling.

  • In our problem, the electron starts with a kinetic energy of 6.0 eV.
  • This energy is what drives it forward towards the barrier.
  • However, compared to the barrier's height of 11.0 eV, the electron doesn't have enough energy to cross over in classical terms.
Yet, due to its kinetic energy and the quantum tunneling effect, the electron can traverse this barrier with certain probabilities that depend on the width of the barrier as calculated in the steps. Essentially, while the electron would stop at the barrier classically, the quantum world offers it a unique path through – though in practice, it might emerge with diminished or altered energy.

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 is described by a wave function \(\psi(x)=A e^{-\alpha x^{2}}\) where \(A\) and \(\alpha\) are real, positive constants. If the value of \(\alpha\) is increased, what effect does this have on (a) the particle's uncertainty in position and (b) the particle's uncertainty in momentum? Explain your answers.

Normalization of the Wave Function. Consider a particle moving in one dimension, which we shall call the \(x\) -axis. (a) What does it mean for the wave function of this particle to be normalized? (b) Is the wave function \(\psi(x)=e^{a x},\) where \(a\) is a positive real number, normalized? Could this be a valid wave function? (c) If the particle described by the wave function \(\psi(x)=A e^{b x},\) where \(A\) and \(b\) are positive real numbers, is confined to the range \(x \geq 0\) , determine \(A\) (including its units) so that the wave function is normalized.

Photon in a Dye Laser. An electron in a long, organic molecule used in a dye laser behaves approximately like a particle in a box with width 4.18 nm. What is the wavelength of the photon emitted when the electron undergoes a transition (a) from the first excited level to the ground level and (b) from the second excited level to the first excited level?

The ground-state energy of a harmonic oscillator is 5.60 \(\mathrm{eV} .\) If the oscillator undergoes a transition from its \(n=3\) to \(n=2\) level by emitting a photon, what is the wavelength of the photon?

Protons, neutrons, and many other particles are made of more fundamental particles called quarks and antiquarks (the antimatter equivalent of quarks). A quark and an antiquark can form a bound state with a variety of different energy levels, each of which corresponds to a different particle observed in the laboratory. As an example, the \(\psi\) particle is a low-energy bound state of a so-called charm quark and its antiquark, with a rest energy of 3097 MeV; the \(\psi(2 S)\) particle is an excited state of this same quark-antiquark combination, with a rest energy of 3686 MeV. A simplified representation of the potential energy of interaction between a quark and an antiquark is \(U(x)=A|x|,\) where \(A\) is a positive constant and \(x\) represents the distance between the quark and the antiquark. You can use the WKB approximation (see Challenge Problem 40.72 ) to determine the bound-state energy levels for this potential-energy function. In the WKB approximation, the energy levels are the solutions to the equation $$\int_{a}^{b} \sqrt{2 m[E-U(x)]} d x=\frac{n h}{2} \quad(n=1,2,3, \ldots)$$ Here \(E\) is the energy, \(U(x)\) is the potential-energy function, and \(x=a\) and \(x=b\) are the classical turning points (the points at which \(E\) is equal to the potential energy, so the Newtonian kinetic energy would be zero). (a) Determine the classical turning points for the potential \(U(x)=A|x|\) and for an energy \(E\) . (b) Carry out the above integral and show that the allowed energy levels in the WKB approximation are given by $$E_{n}=\frac{1}{2 m}\left(\frac{3 m A h}{4}\right)^{2 / 3} n^{2 / 3} \quad(n=1,2,3, \ldots)$$ (Hint: The integrand is even, so the integral from \(-x\) to \(x\) is equal to twice the integral from 0 to \(x\) . \((\mathrm{c})\) Does the difference in energy between successive levels increase, decrease, or remain the same as \(n\) increases? How does this compare to the behavior of the energy levels for the harmonic oscillator? For the particle in a box? Can you suggest a simple rule that relates the difference in energy between successive levels to the shape of the potential-energy function?

See all solutions

Recommended explanations on Physics Textbooks

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.