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Chapter 38: Problem 71

For the arrangement of Figs. \(38-14\) and \(38-15\), electrons in the incident beam in region 1 have energy \(E=800 \mathrm{eV}\) and the potential step has a height of \(U_{1}=600 \mathrm{eV}\). What is the angular wave number in (a) region 1 and (b) region \(2 ?\) (c) What is the reflection coefficient? (d) If the incident beam sends \(5.00 \times 10^{5}\) electrons against the potential step, approximately how many will be reflected?

Short Answer

Expert verified
(a) ~1.44 脳 10鹿鈦� m鈦宦�, (b) ~7.20 脳 10鈦� m鈦宦�, (c) ~0.111, (d) ~5.55 脳 10鈦� electrons.

Step by step solution

01

Understand the Problem

We are given two regions for an electron beam. In region 1, electrons have an energy of 800 eV. There is a potential step of 600 eV between region 1 and region 2. We need to find the angular wave numbers in both regions and the reflection coefficient, as well as calculate how many electrons are reflected.
02

Calculate Angular Wave Number in Region 1

The angular wave number, \( k_1 \), in region 1 is given by \( k_1 = \frac{\sqrt{2mE}}{\hbar} \). Substituting \( E = 800 \text{ eV} \), and using \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \) and \( \hbar = 1.054 \times 10^{-34} \text{ Js} \), we find \( k_1 = \frac{\sqrt{2 \cdot 9.11 \times 10^{-31} \cdot 800 \times 1.602 \times 10^{-19}}}{1.054 \times 10^{-34}} \). Solving this gives \( k_1 \approx 1.44 \times 10^{10} \text{ m}^{-1} \).
03

Calculate Angular Wave Number in Region 2

In region 2, the potential energy \( U_1 = 600 \text{ eV} \) must be subtracted from the total energy \( E = 800 \text{ eV} \). The kinetic energy becomes \( E - U_1 = 200 \text{ eV} \). Thus, \( k_2 = \frac{\sqrt{2m(E - U_1)}}{\hbar} \). Solving \( k_2 = \frac{\sqrt{2 \cdot 9.11 \times 10^{-31} \cdot 200 \times 1.602 \times 10^{-19}}}{1.054 \times 10^{-34}} \), we find \( k_2 \approx 7.20 \times 10^{9} \text{ m}^{-1} \).
04

Calculate Reflection Coefficient

The reflection coefficient \( R \) is given by \( R = \left| \frac{k_1 - k_2}{k_1 + k_2} \right|^2 \). Substituting \( k_1 = 1.44 \times 10^{10} \text{ m}^{-1} \) and \( k_2 = 7.20 \times 10^{9} \text{ m}^{-1} \), we find \( R = \left| \frac{1.44 \times 10^{10} - 7.20 \times 10^{9}}{1.44 \times 10^{10} + 7.20 \times 10^{9}} \right|^2 \approx 0.111 \).
05

Calculate Number of Reflected Electrons

To find the number of reflected electrons, multiply the rate of incident electrons by the reflection coefficient. Thus, \( N_{\text{reflected}} = 5.00 \times 10^5 \times 0.111 \approx 5.55 \times 10^4 \).

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Key Concepts

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

Angular Wave Number
In quantum mechanics, the angular wave number is a fundamental concept used to describe the behavior of wave-like particles, such as electrons, in different regions of space. It is intricately linked to the wavelength of the particle. To understand the angular wave number, we define it as:
  • Angular wave number, denoted as \( k \), is given by \( k = \frac{2\pi}{\lambda} \), where \( \lambda \) is the wavelength.
  • It can also be related to the energy of the particle through the expression \( k = \frac{\sqrt{2mE}}{\hbar} \), where \( E \) is the particle's energy, \( m \) is its mass, and \( \hbar \) is the reduced Planck's constant.
In the context of the problem, the angular wave number helps determine how electrons behave in different regions when subjected to potential energy changes, such as a potential step.
Potential Step
A potential step in the realm of quantum mechanics represents a sudden change in potential energy in a specific region of space. This change can be seen as a barrier that incoming particles, like electrons, must overcome.
  • In this problem, we are considering a step from 0 eV to 600 eV, meaning electrons moving from a lower energy state to a higher energy state.
  • This step creates different conditions in regions 1 and 2, necessitating the calculation of different angular wave numbers and affecting the motion of the electrons.
  • The potential step determines how much kinetic energy the electrons will have once they move past the step, and is a critical factor in calculating the reflection coefficient.
The potential step is essential for understanding phenomena such as reflection and transmission of wave-packets in quantum mechanics.
Reflection Coefficient
In quantum mechanics, the reflection coefficient quantifies the fraction of a wave-like particle, such as an electron, that gets reflected when encountering a potential barrier. It tells us how much of the wave is sent back towards the point of origin.
  • The reflection coefficient \( R \) is calculated through the formula \( R = \left| \frac{k_1 - k_2}{k_1 + k_2} \right|^2 \), where \( k_1 \) and \( k_2 \) are the angular wave numbers in the respective regions.
  • For this problem, using \( k_1 = 1.44 \times 10^{10} \, \text{m}^{-1} \) and \( k_2 = 7.20 \times 10^{9} \, \text{m}^{-1} \), we computed \( R \) to be approximately 0.111.
  • An \( R \) value less than 1 indicates partial reflection, where only a fraction of the electrons are reflected back.
Understanding the reflection coefficient is crucial for predicting the behavior of wave-particles when encountering changes in potential energy.
Electron Beam Energy
Electron beam energy in quantum mechanics is crucial in determining how electrons will interact with potential barriers and steps. It defines the total energy that an electron possesses as it moves through a given medium.
  • Here, the energy of the electron beam is given as 800 eV, which means each electron carries this energy into the first region before encountering the potential step.
  • Upon reaching the potential step of 600 eV, the electrons face a shift in energy dynamics, as the potential energy affects their kinetic energy.
  • This energy difference between the regions influences how electrons transmit past the potential step, and contributes to calculations like angular wave numbers and reflection coefficients.
A fundamental grasp of electron beam energy allows students to frame how momentum and energy conservation principles apply to quantum scenarios and observe real-world quantum mechanical applications.

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Most popular questions from this chapter

An electron and a photon each have a wavelength of \(0.20 \mathrm{~nm}\). What is the momentum (in \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) ) of the (a) electron and (b) photon? What is the energy (in \(\mathrm{eV}\) ) of the (c) electron and (d) photon?

An electron with total energy \(E=5.1 \mathrm{eV}\) approaches a barrier of height \(U_{b}=6.8 \mathrm{eV}\) and thickness \(L=\) \(750 \mathrm{pm}\). What percentage change in the transmission coefficient \(T\) occurs for a \(1.0 \%\) change in (a) the barrier height, (b) the barrier thickness, and (c) the kinetic energy of the incident electron?

An orbiting satellite can become charged by the photoelectric effect when sunlight ejects electrons from its outer surface. Satellites must be designed to minimize such charging because it can ruin the sensitive microelectronics. Suppose a satellite is coated with platinum, a metal with a very large work function \((\Phi=5.32 \mathrm{eV})\). Find the longest wavelength of incident sunlight that can eject an electron from the platinum.

Just after detonation, the fireball in a nuclear blast is approximately an ideal blackbody radiator with a surface temperature of about \(1.0 \times 10^{7} \mathrm{~K} .\) (a) Find the wavelength at which the thermal radiation is maximum and (b) identify the type of electromagnetic wave corresponding to that wavelength. (See Fig. 33-1.) This radiation is almost immediately absorbed by the surrounding air molecules, which produces another ideal blackbody radiator with a surface temperature of about \(1.0 \times 10^{5} \mathrm{~K}\). (c) Find the wavelength at which the thermal radiation is maximum and (d) identify the type of electromagnetic wave corresponding to that wavelength.

You will find in Chapter 39 that electrons cannot move in definite orbits within atoms, like the planets in our solar system. To see why, let us try to "observe" such an orbiting electron by using a light microscope to measure the electron's presumed orbital position with a precision of, say, \(10 \mathrm{pm}\) (a typical atom has a radius of about \(100 \mathrm{pm}\) ). The wavelength of the light used in the microscope must then be about \(10 \mathrm{pm}\). (a) What would be the photon energy of this light? (b) How much energy would such a photon impart to an electron in a head-on collision? (c) What do these results tell you about the possibility of "viewing" an atomic electron at two or more points along its presumed orbital path? (Hint: The outer electrons of atoms are bound to the atom by energies of only a few electron-volts.)
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