/*! 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 3 An electron has a de Broglie wav... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electron has a de Broglie wavelength of \(2.80 \times 10^{-10} \mathrm{~m}\) Determine (a) the magnitude of its momentum and (b) its kinetic energy (in joules and in electron volts).

Short Answer

Expert verified
The magnitude of the momentum is \(2.37 \times 10^{-24} \mathrm{~ kg \cdot m/s}\). The kinetic energy is \(3.09 \times 10^{-17} \mathrm{~ J}\) and 193 eV.

Step by step solution

01

Calculate the momentum

We rearrange the de Broglie wavelength equation, \(p = h / \lambda\). Substituting the given de Broglie wavelength (\(2.80 \times 10^{-10} \mathrm{~m}\)) and the Planck constant (\(6.626 \times 10^{-34} \mathrm{~ m^2 \cdot kg / s}\)), we find \(p = 6.626 \times 10^{-34} \mathrm{~ m^2 \cdot kg / s} / 2.80 \times 10^{-10} \mathrm{~m} = 2.37 \times 10^{-24} \mathrm{~ kg \cdot m/s}\).
02

Calculate the kinetic energy in Joules

The momentum of a particle is \(p = mv\), where \(m\) is the mass and \(v\) is the velocity. As the particle is an electron, the mass is \(9.11 \times 10^{-31} \mathrm{~ kg}\). We can solve for \(v\) as follows: \(v = p / m = 2.37 \times 10^{-24} \mathrm{~ kg \cdot m/s} / 9.11 \times 10^{-31} \mathrm{~ kg} = 2.6 \times 10^6 \mathrm{~ m/s}\). We then plug \(v\) into the kinetic energy equation, \(KE = 1/2 mv^2\), to find \(KE = 1/2 * 9.11 \times 10^{-31} \mathrm{~ kg} * (2.6 \times 10^6 \mathrm{~ m/s})^2 = 3.09 \times 10^{-17}\) J.
03

Convert kinetic energy to electron volts

We convert Joules to electron volts using the conversion factor 1 eV = \(1.60 \times 10^{-19}\) J. Hence the kinetic energy in electron volts is \(3.09 \times 10^{-17}\) J / \(1.60 \times 10^{-19}\) J/eV = 193 eV.

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 Momentum
When we talk about the momentum of an electron, we refer to the product of its mass and velocity. But in the quantum world, electrons also display wave-like properties. This duality is captured in the de Broglie hypothesis, which states that particles like electrons have wavelengths associated with their momentum. The relationship is given by the equation,
\( p = \frac{h}{\lambda} \)
where \( p \) is the momentum, \( h \) is the Planck constant, and \( \lambda \) is the de Broglie wavelength.
In our textbook example, using the provided wavelength of \( 2.80 \times 10^{-10} \text{ m} \) and Planck's constant, the momentum of an electron is calculated. This momentum reflects the electron's movement and is crucial in determining other characteristics like kinetic energy. Understanding this quantum mechanical concept of momentum can be a stepping stone for students into the fascinating world of quantum physics, where particles are waves, and waves act like particles.
Kinetic Energy Calculation
The kinetic energy of an object is the energy it possesses due to its motion. For an electron, this is particularly significant in various fields, such as chemistry and physics, where the interaction of particles depends greatly on their energies.
The kinetic energy (KE) of an electron is given by
\( KE = \frac{1}{2} mv^2 \)
where \( m \) is the mass of the electron and \( v \) is its velocity. After calculating the electron's momentum, we use it to find its velocity. As electrons have a known mass, once the velocity is determined, we can calculate the kinetic energy.
This process was illustrated in the given solution, where the momentum was used to calculate the velocity, which then allowed us to determine the kinetic energy. The concept of kinetic energy is not just pivotal in solving physics problems, but also in understanding the behavior of particles and the transfer of energy in different systems.
Planck Constant
The Planck constant, denoted as \( h \) and approximately valued at \( 6.626 \times 10^{-34} \text{ m}^2 \cdot \text{kg} / \text{s} \), is a fundamental quantity in quantum mechanics. It relates the energy of a photon to its frequency and plays a central role in the quantum description of matter and radiation.
In terms of the de Broglie wavelength, the Planck constant bridges the concept of particle momentum and wavelength. This constant is vital for not only calculating the momentum of an electron from its de Broglie wavelength, as shown in the exercise, but it also forms the basis of Heisenberg's Uncertainty Principle and is crucial in the quantization of light and energy.
The significance of the Planck constant goes beyond solving equations; it is part of the foundation that explains why the world at a microscopic level operates so differently from our macroscopic experiences. Its very existence requires a rethinking of classical mechanics and introduces students to the probabilistic nature of quantum phenomena.

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 certain atom has an energy state \(3.50 \mathrm{eV}\) above the ground state. When excited to this state, the atom remains for \(2.0 \mu \mathrm{s},\) on average, before it emits a photon and returns to the ground state. (a) What are the energy and wavelength of the photon? (b) What is the smallest possible uncertainty in energy of the photon?

A triply ionized beryllium ion, \(\mathrm{Be}^{3+}\) (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. (a) What is the ground-level energy of \(\mathrm{Be}^{3+}\) ? How does this compare to the groundlevel energy of the hydrogen atom? (b) What is the ionization energy of \(\mathrm{Be}^{3+} ?\) How does this compare to the ionization energy of the hydrogen atom? (c) For the hydrogen atom, the wavelength of the photon emitted in the \(n=2\) to \(n=1\) transition is \(122 \mathrm{nm}\) (see Example 39.6 ). What is the wavelength of the photon emitted when a \(\mathrm{Be}^{3+}\) ion undergoes this transition? (d) For a given value of \(n\), how does the radius of an orbit in \(\mathrm{Be}^{3+}\) compare to that for hydrogen?

Calculate the de Broglie wavelength of (a) a \(50.0 \mathrm{~kg}\) woman jogging leisurely at \(2.0 \mathrm{~m} / \mathrm{s},\) (b) a free electron with kinetic energy \(2.0 \mathrm{MeV}\). and (c) a free electron with kinetic energy \(20 \mathrm{eV}\). Use the proper relativistic expression when necessary.

\(\mathrm{CP}\) (a) A particle with mass \(m\) has kinetic energy equal to three times its rest energy.What is the de Broglie wavelength of this particle? (Hint: You must use the relativistic expressions for momentum and kinetic energy: \(E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}\) and \(\left.K=E-m c^{2} .\right)\) (b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (ii) a proton.

Can the first type of helium-ion microscope, used for surface imaging. produce helium ions with a wavelength of \(0.1 \mathrm{pm} ?\) (a) Yes; the voltage required is \(21 \mathrm{kV}\). (b) Yes; the voltage required is \(42 \mathrm{kV}\). (c) No; a voltage higher than \(50 \mathrm{kV}\) is required. (d) No; a voltage lower than \(10 \mathrm{kV}\) is required.

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.