/*! 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 60 The penetration distance \(\eta\... [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: $$\psi(x=L+\eta)=\frac{1}{e} \psi(L)$$ The penetration distance can be shown to be $$\eta=\frac{\hbar}{\sqrt{2 m\left(U_{0}-E\right)}}$$ 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 \(U_{0}=20 \mathrm{eV} .\) (b) Find \(\eta\) for a 20.0 -MeV proton trapped in a 30.0 -Me \(\mathrm{V}\) -deep potential well.

Short Answer

Expert verified
For the electron, \(\eta\) is approximately 1.04 nm. For the proton, \(\eta\) is approximately 1.41 fm.

Step by step solution

01

Identify Constants for the Electron

For this part of the question, we are dealing with an electron in a potential well. We know that: - The electron's kinetic energy, \(E = 13 \text{ eV}\).- The depth of the potential well, \(U_0 = 20 \text{ eV}\).- The rest mass of an electron \( m = 9.11 \times 10^{-31} \text{ kg} \).- Planck's constant \( \hbar = 1.055 \times 10^{-34} \text{ J s} \).
02

Calculate Energy Difference for Electron

Compute the energy difference \( U_0 - E \):\[ U_0 - E = 20 \text{ eV} - 13 \text{ eV} = 7 \text{ eV} \]Convert electron volts to joules (1 eV = \( 1.602 \times 10^{-19} \text{ J}\)):\[ U_0 - E = 7 \times 1.602 \times 10^{-19} \text{ J} \]
03

Calculate Penetration Distance (\( \eta \)) for Electron

Substitute the values into the formula for \( \eta \):\[ \eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}} \]Substitute known values:\[ \eta = \frac{1.055 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times (7 \times 1.602 \times 10^{-19})}} \]Perform the calculations to find \( \eta \).
04

Identify Constants for the Proton

For the second part, we are dealing with a proton. We know:- The proton's kinetic energy, \(E = 20 \text{ MeV}\).- The depth of the potential well, \(U_0 = 30 \text{ MeV}\).- The mass of a proton \( m = 1.67 \times 10^{-27} \text{ kg} \).- Convert MeV to joules (1 MeV = \( 1.602 \times 10^{-13} \text{ J}\)).
05

Calculate Energy Difference for Proton

Compute the energy difference \( U_0 - E \):\[ U_0 - E = 30 \text{ MeV} - 20 \text{ MeV} = 10 \text{ MeV} \]Convert to joules:\[ U_0 - E = 10 \times 1.602 \times 10^{-13} \text{ J} \]
06

Calculate Penetration Distance (\( \eta \)) for Proton

Substitute the values into the formula:\[ \eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}} \]Substitute:\[ \eta = \frac{1.055 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times (10 \times 1.602 \times 10^{-13})}} \]Compute \( \eta \) for the proton.

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Key Concepts

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

Potential Well
A potential well is a region in space where a particle's potential energy is comparatively lower than in the surrounding areas. Think of it like a dip or a valley in an energy landscape. This concept helps to describe how particles, such as electrons or protons, are confined to certain regions unless they have enough energy to escape. In quantum mechanics, potential wells highlight interesting phenomena such as quantum tunneling. In the exercise above, both the electron and the proton are trapped in potential wells with finite depths, which affects their behavior and the energy calculations necessary for determining the penetration distance.
Wave Function Decay
The wave function, denoted as \(\psi\), is a fundamental concept in quantum mechanics that describes the probability amplitude of a particle's position and state. When we discuss wave function decay, we mean how this probability decreases as we move away from a certain point, particularly in a potential well.
For example, after reaching a certain distance from the classical turning point in the well, the wave function decreases significantly. This decay is so substantial that the probability of finding the particle decreases to 1/e of its value at the turning point. It's this decay that gives rise to the concept of the penetration distance, \(\eta\), beyond which a particle is unlikely to be found.
Kinetic Energy Calculations
Kinetic energy is the energy that a particle possesses due to its motion, calculated using the expression \(E_{k} = \frac{1}{2}mv^2\). In the context of finite potential wells, kinetic energy helps to determine how deeply a particle can penetrate into or escape from these wells.
The given exercise requires the calculation of penetration distance, \(\eta\), for both an electron and a proton based on their kinetic energies and the respective depths of the potential wells. Calculations involve comparing the potential well's depth with the particle's kinetic energy to determine how much further beyond the classical limits the particle can tunnel.
Planck's Constant
Planck's constant, denoted \(\hbar\), is a crucial element in quantum mechanics. It is approximately \(1.055 \times 10^{-34} \text{ J s} \) and serves as the foundation for the quantum of action. In the context of this exercise, Planck's constant is used in calculating the penetration distance, \(\eta\).
The penetration distance formula \(\eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}}\) directly involves \(\hbar\), illustrating its role in connecting energy and temporal measurements. By incorporating Planck's constant into the calculations, we accurately capture the quantum mechanical nature of particles in potential wells.
Electron Proton Mass
In quantum mechanics, the mass of a particle is a crucial factor in determining its behavior, especially when considering its penetration into a potential well. The mass of an electron is \(9.11 \times 10^{-31} \text{ kg} \), whereas the mass of a proton is \(1.67 \times 10^{-27} \text{ kg}\).
These vastly different masses significantly affect their respective penetration distances because the kinetic energy relationship through the mass term \(m\) in the formula \(\eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}}\) is nonlinear. Heavier particles, like protons, tend to have smaller penetration distances in the same potential well settings compared to lighter electrons, highlighting the mass's impact.

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Most popular questions from this chapter

A fellow student proposes that a possible wave function for a free particle with mass \(m\) (one for which the potential-energy function \(U(x)\) is zero) is $$\psi(x)=\left\\{\begin{array}{ll}{e^{+\kappa x},} & {x<0} \\ {e^{-\kappa x},} & {x \geq 0}\end{array}\right.$$ where \(\kappa\) is a positive constant. (a) Graph this proposed wave function. (b) Show that the proposed wave function satisfies the Schrodinger equation for \(x < 0\) if the energy is \(E=-\hbar^{2} \kappa^{2} / 2 m-\) that is, if the energy of the particle is negative. (c) Show that the proposed wave function also satisfies the Schrodinger equation for \(x \geq 0\) with the same energy as in part (b). (d) Explain why the proposed wave function is nonetheless not an acceptable solution of the Schrodinger equation for a free particle. (Hint: What is the behavior of the function at \(x=0 ?\) ) It is in fact impossible for a free particle (one for which \(U(x)=0 )\) to have an energy less than zero.

Chemists use infrared absorption spectra to identify chemicals in a sample. In one sample, a chemist finds that light of wavelength 5.8\(\mu \mathrm{m}\) is absorbed when a molecule makes a transition from its ground harmonic oscillator level to its first excited level. (a) Find the energy of this transition. (b) If the molecule can be treated as a harmonic oscillator with mass \(5.6 \times 10^{-26} \mathrm{kg},\) find the force constant.

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.

(a) An electron with initial kinetic energy 32 eV encounters a square barrier with height 41 \(\mathrm{eV}\) and width 0.25 \(\mathrm{nm}\) . What is the probability that the electron will tunnel through the barrier? (b) A proton with the same kinetic energy encounters the same barrier. What is the probability that the proton will tunnel through the barrier?

Wave functions like the one in Problem 40.50 can represent free particles moving with velocity \(v=p / m\) in the \(x\) -direction. Consider a beam of such particles incident on a potential-energy step \(U(x)=0,\) for \(x < 0,\) and \(U(x)=U_{0} < E,\) for \(x > 0 .\) The wave function for \(x < 0\) is \(\psi(x)=A e^{i k_{1} x}+B e^{-i k_{1} x}\) representing incident and reflected particles, and for \(x > 0\) is \(\psi(x)=C e^{i k_{2} x},\) representing transmitted particles. Use the conditions that both \(\psi\) and its first derivative must be continuous at \(x=0\) to find the constants \(B\) and \(C\) in terms of \(k_{1}, k_{2},\) and \(A .\)

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