/*! 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 52 The \(penetration\) \(distance\)... [FREE SOLUTION] | 91Ó°ÊÓ

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The \(penetration\) \(distance\) \(\eta\) in a finite potential well is the distance at which the wave function has decreased to 1/\(e\) of the wave function at the classical turning point: $$\psi(x = L + \eta) = {1\over e} \psi(L)$$ The penetration distance can be shown to be $$\eta = {\hslash \over \sqrt{2m(U_0 - E)}}$$ The probability of finding the particle beyond the penetration distance is nearly zero. (a) Find \(\eta\) for an electron having a kinetic energy of 13 eV in a potential well with \(U_0\) = 20 eV. (b) Find \(\eta\) for a 20.0-MeV proton trapped in a 30.0-MeV-deep potential well.

Short Answer

Expert verified
The penetration distances are approximately 1.02 nm for the electron and 8.04 fm for the proton.

Step by step solution

01

Set up the formula for penetration distance

First, we use the formula given for the penetration distance: \[ \eta = \frac{\hslash}{\sqrt{2m(U_0 - E)}} \] where \( \hslash \) is the reduced Planck's constant, \( m \) is the mass of the particle, \( U_0 \) is the potential well depth, and \( E \) is the kinetic energy of the particle.
02

Identify the constants and values for the electron

For an electron, we have kinetic energy \( E = 13 \, \text{eV} \) and potential well depth \( U_0 = 20 \, \text{eV} \). The mass of an electron \( m = 9.109 \times 10^{-31} \, \text{kg} \), and \( \hslash = 1.054 \times 10^{-34} \, \text{J} \cdot \text{s} \). Note that \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \).
03

Calculate the penetration distance for the electron

Convert energies from eV to J: \( U_0 - E = 20 \, \text{eV} - 13 \, \text{eV} = 7 \, \text{eV} = 7 \times 1.602 \times 10^{-19} \, \text{J} \). Plug in the values: \[ \eta = \frac{1.054 \times 10^{-34}}{\sqrt{2 \times 9.109 \times 10^{-31} \times 7 \times 1.602 \times 10^{-19}}} \] Calculate \( \eta \) to find its value.
04

Identify constants and values for the proton

For the proton, we have kinetic energy \( E = 20.0 \, \text{MeV} \) and potential well depth \( U_0 = 30.0 \, \text{MeV} \). The mass of a proton \( m = 1.673 \times 10^{-27} \, \text{kg} \), and \( \hslash = 1.054 \times 10^{-34} \, \text{J} \cdot \text{s} \). Note that \( 1 \, \text{MeV} = 1.602 \times 10^{-13} \, \text{J} \).
05

Calculate the penetration distance for the proton

Convert energies from MeV to J: \( U_0 - E = 30.0 \, \text{MeV} - 20.0 \, \text{MeV} = 10.0 \, \text{MeV} = 10 \times 1.602 \times 10^{-13} \, \text{J} \). Plug in the values: \[ \eta = \frac{1.054 \times 10^{-34}}{\sqrt{2 \times 1.673 \times 10^{-27} \times 10 \times 1.602 \times 10^{-13}}} \] Calculate \( \eta \) to find its value.

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Key Concepts

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

Penetration Distance
The concept of penetration distance is a crucial part of understanding quantum tunneling. In a finite potential well, it defines how far a particle can penetrate into the classically forbidden region where its energy is less than the potential energy of the well. This distance, denoted as \( \eta \), is where the wave function decays to \( 1/e \) of its value at the turning point, indicating a significant drop in probability amplitude.
Using the formula \( \eta = {\hslash \over \sqrt{2m(U_0 - E)}} \), penetration distance takes into account the particle's mass \( m \), its kinetic energy \( E \), and the potential well's depth \( U_0 \). The reduced Planck's constant \( \hslash \) also plays a role here. In essence, \( \eta \) offers insight into quantum behaviors where classical physics fails, providing a glimpse into the particle's potential to "tunnel" through barriers.
Moreover, understanding penetration distance is essential for applications such as electron tunneling in semiconductor devices or explaining the behavior of particles in nuclear fusion reactions. It tells us where the particle "fades" beyond classical expectations, hinting at probabilities outside the norm.
Finite Potential Well
The finite potential well is a fundamental concept in quantum mechanics, contrasting with the infinite potential well by allowing particles to have a finite probability of being found outside the well due to quantum tunneling. This type of potential configuration is characterized by a definite depth denoted by \( U_0 \), and regions within which a particle may exist even if its energy \( E \) is less than that of the potential barrier outside.
In a finite well, particles aren't strictly confined but rather, according to quantum theory, have wave functions that extend into and possibly through the potential barriers. These extensions mean that, unlike in an infinite potential well, the wave function doesn't drop abruptly to zero at the wall but decays exponentially.
The real-world significance of understanding finite potential wells lies in various technological and scientific fields, ranging from nanotechnology to quantum computing, where knowing how particles behave in these types of potentials is critical for innovation and analysis.
Wave Function Decay
Wave function decay is an illustration of quantum behavior in potential wells where particles have the apparent ability to penetrate barriers. This phenomenon is largely expressed through the exponential decay of the wave function past the classical turning points of a finite potential well.
Mathematically, the wave function in the region beyond the classical boundary decreases as \( \psi(x) = \psi(L) \exp\left(-\frac{x-L}{\eta}\right) \), where \( x \) is a position beyond the turning point \( L \) and \( \eta \) represents the penetration distance. This decay implies that the probability of locating the particle in classically forbidden zones is not zero but diminishes exponentially the further it penetrates.
This concept is key in understanding the probabilistic nature of quantum mechanics, emphasizing that even if classical physics would predict impossibility, quantum particles present measurable probabilities, fundamental in tunneling processes and resonant in systems like tunnel diodes or alpha decay in radioactive nuclei.
Kinetic Energy
Kinetic energy in the context of quantum tunneling and potential wells is the energy possessed by a particle due to its motion. It plays a vital role in determining how and whether a particle can tunnel through a barrier.
Within the formula for penetration distance, \( \eta = \frac{\hslash}{\sqrt{2m(U_0 - E)}} \), we clearly see that a higher kinetic energy \( E \) results in a smaller \( U_0 - E \), and thus a smaller penetration distance \( \eta \). Essentially, as kinetic energy increases, the likelihood of tunneling decreases, since the particle already has significant motion and energy focused on staying within the classically allowed region.
This interrelationship between kinetic energy and tunneling probability is significant for quantum mechanical systems where the transfer of particles through barriers such as in nuclear physics or electronic components like semiconductors drives technological advancement. Understanding how kinetic energy influences tunneling helps in designing and improving various devices.

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Most popular questions from this chapter

An electron is moving as a free particle in the -\(x\)-direction with momentum that has magnitude 4.50 \(\times\) 10\(^{-24}\) kg \(\bullet\) m/s. What is the one- dimensional time-dependent wave function of the electron?

Find the excitation energy from the ground level to the third excited level for an electron confined to a box of width 0.360 nm. (b) The electron makes a transition from the \(n\) = 1 to \(n\) = 4 level by absorbing a photon. Calculate the wavelength of this photon.

(a) The wave nature of particles results in the quantum-mechanical situation that a particle confined in a box can assume only wavelengths that result in standing waves in the box, with nodes at the box walls. Use this to show that an electron confined in a one-dimensional box of length \(L\) will have energy levels given by $$E_n = {n^2h^2 \over8mL^2}$$ (\(Hint\): Recall that the relationship between the de Broglie wavelength and the speed of a nonrelativistic particle is \(mv = h/\lambda\). The energy of the particle is \({1\over2} mv^2\).) (b) If a hydrogen atom is modeled as a one-dimensional box with length equal to the Bohr radius, what is the energy (in electron volts) of the lowest energy level of the electron?

Compute \(\mid \Psi \mid ^2 for \space \Psi = \psi \space sin \space \omega t\), where \(\psi\) is time independent and \(\omega\) is a real constant. Is this a wave function for a stationary state? Why or why not?

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 \(\psi\)(2S) 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 \(U(x) = A\mid x \mid\) , 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 40.64) 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 $$\int ^b _a \sqrt{2m[E - U(x)]} dx = {nh \over 2} \space (n = 1, 2, 3, . . .)$$ 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 \(U(x) = A \mid x \mid\) 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 $$E_n = {1 \over2m} ( {3mAh \over 4} ) ^{2/3} n^{2/3} \space (n = 1, 2, 3, . . .)$$ (\(Hint\): The integrand is even, so the integral from \(-x\) to \(x\) is equal to twice the integral from 0 to \(x\).) (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?

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