Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Two long, straight wires that lie along the same line have a separation at their tips of \(2.00 \mathrm{nm}\). The potential energy of an electron in the gap is about \(1.00 \mathrm{eV}\) higher than it is in the conduction band of the two wires. Conduction-band electrons have enough energy to contribute to the current flowing in the wire. What is the probability that a conduction electron in one wire will be found in the other wire after arriving at the gap?

Short Answer

Expert verified
Answer: The probability that a conduction electron in one wire will be found in the other wire after arriving at the gap is approximately \(7.18 \times 10^{-4}\).

Step by step solution

01

Convert given quantities to SI units

First, we need to convert the given quantities to SI units. The width of the gap \(L = 2.00 \mathrm{nm} = 2.00 \times 10^{-9}\) meters. The potential energy difference between the gap and the conduction band \(U = 1.00 \mathrm{eV} = 1.00 \times 1.602 \times 10^{-19}\) Joules
02

Calculate the constant K

The constant \(K\) depends on the mass of the electron, its potential energy difference and the reduced Planck's constant, \(h\). The formula for \(K\) is given by: \(K = \frac{\sqrt{2mU}}{\hbar}\) Where \(m\) is the mass of the electron, which is \(9.11 \times 10^{-31}\) kg, and \(\hbar\) is the reduced Planck's constant, which is approximately \(1.055 \times 10^{-34} \mathrm{Js}\). Substitute the values for the mass and potential energy difference to get: \(K = \frac{\sqrt{2(9.11 \times 10^{-31}\mathrm{kg})(1.00 \times 1.602 \times 10^{-19}\mathrm{J})}}{1.055 \times 10^{-34} \mathrm{Js}}\)
03

Calculate the tunneling probability

Now that we have the value of \(K\), we can calculate the tunneling probability using the formula: \(P = e^{-2KL}\) Substitute the values for \(K\) and \(L\) into the formula: \(P = e^{-2\left(\frac{\sqrt{2(9.11 \times 10^{-31}\mathrm{kg})(1.00 \times 1.602 \times 10^{-19}\mathrm{J})}}{1.055 \times 10^{-34} \mathrm{Js}}\right)(2.00 \times 10^{-9}\mathrm{m})}\)
04

Evaluate the probability

Evaluate the expression to find the tunneling probability: \(P \approx 7.18 \times 10^{-4}\) Thus, the probability that a conduction electron in one wire will be found in the other wire after arriving at the gap is approximately \(7.18 \times 10^{-4}\).

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 Vaia!

Key Concepts

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

Conduction Band
In the realm of solid-state physics, the conduction band is paramount when discussing electronic properties of materials. It represents an energy range where electrons are free to move within a material, thereby allowing electrical conductivity. Think of it as a track for electrons to race along, with sufficient energy making them move swiftly to contribute to an electric current.

More specifically, in metals, the conduction band is partially filled, meaning some electrons have natural access to these energies even at room temperature. In semiconductors and insulators, the conduction band is typically empty at absolute zero. For them to conduct electricity, electrons must gain enough energy, such as from an external source like heat or light, to jump up from the valence band (where they are usually 'stuck') to the conduction band – this is known as the band gap.

Understanding the conduction band is crucial to solving problems related to electronic conduction and hence is a vital concept in the scenario of electron tunneling probability.
Electron Tunneling Probability
One of the most fascinating phenomena in quantum mechanics is undoubtedly electron tunneling, where particles like electrons can pass through barriers that would be insurmountable according to classical physics. This effect is due to the probabilistic nature of quantum mechanics, where particles are described by wavefunctions that can extend beyond potential barriers.

Calculating the tunneling probability of an electron allows us to predict how likely it is for the electron to appear on the other side of a barrier – like the gap between two wires in this exercise. The process of finding this probability involves knowing the electron's wavefunction and the characteristics of the potential barrier, such as its height (the potential energy) and width. In the case of the exercise, the narrower the gap and the lower the potential energy barrier, the higher the probability of the electron making it to the other side.

The probability is exponentially related to the width of the gap and the square root of the mass and potential energy. This relationship emphasizes the incredibly delicate balance of factors that govern electron tunneling in materials and is crucial for designing nanoscale electronic devices.
Potential Energy in Quantum Mechanics
The concept of potential energy in quantum mechanics plays a pivotal role when we talk about forces acting on particles at microscopic levels. Potential energy is the energy stored by an object due to its position relative to other objects or fields it interacts with. In the context of quantum mechanics, instead of witnessing particles behaving predictably as we would expect in a Newtonian framework, we encounter probabilities and wavefunctions.

In quantum systems, the potential energy is often described by potential wells or barriers. For example, in the exercise, an electron must overcome a potential energy 'hill' to move from one wire to the other. Quantum mechanics allows us to calculate the probability of scenarios that would be impossible in classical mechanics, such as 'tunneling' through this hill. The ability to calculate this potential and predict particle behavior in quantum systems is essential for understanding a wide range of phenomena, from the behavior of atoms and molecules to the workings of transistors in electronics.

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

Calculate the ground state energy for an electron confined to a cube with sides equal to twice the Bohr radius \((R=0.0529 \mathrm{nm}) .\) Determine the spring constant that would give this same ground state energy for a harmonic oscillator.

An experimental measurement of the energy levels of a hydrogen molecule, \(\mathrm{H}_{2}\), shows that the energy levels are evenly spaced and separated by about \(9 \cdot 10^{-20} \mathrm{~J}\). A reasonable model of one of the hydrogen atoms would then seem to be that of a hydrogen atom in a simple harmonic oscillator potential. Assuming that the hydrogen atom is attached by a spring with a spring constant \(k\) to the molecule, what is the spring constant \(k\) ?

The neutrons in a parallel beam, each having kinetic energy \(1 / 40 \mathrm{eV}\) (which is approximately corresponding to “room temperature"), are directed through two slits \(0.50 \mathrm{~mm}\) apart. How far apart will the interference peaks be on a screen \(1.5 \mathrm{~m}\) away?

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 a water vapor molecule in a room \(4.00 \mathrm{~m} \times\) \(10.0 \mathrm{~m} \times 10.0 \mathrm{~m}\). a) What is the ground state energy of this molecule, treating it as a simple particle in a box? b) Compare this energy to the average thermal energy of such a molecule, taking the temperature to be \(300 . \mathrm{K}\). c) What can you conclude from the two numbers you just calculated?

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.

Sign-up for free