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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
Momentum: \(2.366 \times 10^{-24}\text{ kg m/s}\); Kinetic Energy: \(3.07 \times 10^{-18} \text{ J} \) or \(19.15 \text{ eV} \).

Step by step solution

01

Use the de Broglie Wavelength Formula to find Momentum

The de Broglie wavelength formula is given by \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ Js}) \), and \( p \) is the momentum. We can rearrange this formula to solve for momentum: \( p = \frac{h}{\lambda} \). Substituting the given wavelength, \( \lambda = 2.80 \times 10^{-10} \text{ m} \), gives us the momentum \( p = \frac{6.626 \times 10^{-34} \text{ Js}}{2.80 \times 10^{-10} \text{ m}} = 2.366 \times 10^{-24} \text{ kg m/s} \).
02

Use the Momentum to Calculate Kinetic Energy in Joules

The kinetic energy \( K \) of an electron is given by the formula \( K = \frac{p^2}{2m} \), where \( p \) is the momentum and \( m \) is the electron's mass \( (9.109 \times 10^{-31} \text{ kg}) \). Using the momentum calculated in Step 1, we find the kinetic energy in joules: \( K = \frac{{(2.366 \times 10^{-24})}^2}{2 \times 9.109 \times 10^{-31}} = 3.07 \times 10^{-18} \text{ J} \).
03

Convert Kinetic Energy to Electron Volts

To convert the kinetic energy from joules to electron volts, we use the conversion factor \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). Thus, \( K = \frac{3.07 \times 10^{-18} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} = 19.15 \text{ eV} \).

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics. It represents the quantity of motion an object has and is determined by both its mass and velocity.
For particles at the atomic scale, like electrons, momentum can be calculated with the de Broglie wavelength formula: \[ p = \frac{h}{\lambda} \] where:
  • \( p \) is the momentum,
  • \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ Js}) \), and
  • \( \lambda \) is the de Broglie wavelength.
This formula highlights how smaller wavelengths, typical for particles with greater momentum, relate to the wave-particle duality of matter. For example, by substituting an electron's de Broglie wavelength into this formula, we calculate its momentum. Such calculations are essential for understanding behaviors of microscopic particles and their interactions, which are quintessential for fields like quantum mechanics.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In classical mechanics, it is given by the expression: \[ K = \frac{1}{2} m v^2 \] where:
  • \( K \) is the kinetic energy,
  • \( m \) is the mass, and
  • \( v \) is the velocity of that object.
For electrons, we often work in quantum realms where momentum can be more useful than velocity. Thus, another formula expresses kinetic energy in terms of momentum: \[ K = \frac{p^2}{2m} \] where
  • \( p \) is momentum, and
  • \( m \) is mass.
By first determining momentum with the de Broglie wavelength, we can subsequently find kinetic energy. This approach is particularly helpful for calculating energies of subatomic particles, which is crucial for contexts like particle accelerators and quantum phenomena exploration.
Electron Volts
An electron volt is a unit of energy commonly used in atomic and particle physics. It simplifies the process of working with the tiny energies involved in these fields.
One electron volt (eV) is defined as the energy gained by an electron when it moves through an electric potential difference of one volt. It is equivalent to \( 1 ext{ eV} = 1.602 \times 10^{-19} ext{ joules} \).
To convert energy from joules to electron volts, you use the relationship: \[ K_e = \frac{K_j}{1.602 \times 10^{-19} \] where:
  • \( K_e \) is the energy in electron volts, and
  • \( K_j \) is the energy in joules.
This conversion allows scientists and engineers to express measurements in more digestible forms, which are easier to interpret and compare, especially when dealing with energies of light particles or low energy systems.

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Most popular questions from this chapter

A beam of \(40-\mathrm{eV}\) electrons traveling in the \(+x-\) direction passes through a slit that is parallel to the \(y\) -axis and 5.0\(\mu \mathrm{m}\) wide. The diffraction pattern is recorded on a screen 2.5 \(\mathrm{m}\) from the slit. (a) What is the de Broglie wavelength of the electrons? (b) How much time does it take the electrons to travel from the slit to the screen? (c) Use the width of the central diffraction pattern to calculate the uncertainty in the \(y\) -component of momentum of an electron just after it has passed through the slit. (d) Use the result of part (c) and the Heisenberg uncertainty principle (Eq. 39.29 for \(y\) ) to estimate the minimum uncertainty in the \(y\) -coordinate of an electron just after it has passed through the slit. Compare your result to the width of the slit.

Find the wavelengths of a photon and an electron that have the same energy of 25 eV. (Note: The energy of the electron is its kinetic energy.)

Light from an ideal spherical blackbody 15.0 \(\mathrm{cm}\) in diameter is analyzed using a diffraction grating having 3850 lines/cm. When you shine this light through the grating, you observe that the peak-intensity wavelength forms a first-order bright fringe at \(\pm 11.6^{\circ}\) from the central bright fringe. (a) What is the temperature of the blackbody? (b) How long will it take this sphere to radiate 12.0 \(\mathrm{MJ}\) of energy?

A sample of hydrogen atoms is irradiated with light with wavelength \(85.5 \mathrm{nm},\) and electrons are observed leaving the gas. (a) If each hydrogen atom were initially in its ground level, what would be the maximum kinetic energy in electron volts of these photoelectrons? (b) A few electrons are detected with energies as much as 10.2 eV greater than the maximum kinetic energy calculated in part (a). How can this be?

(a) An atom initially in an energy level with \(E=\) \(-6.52 \mathrm{eV}\) absorbs a photon that has wavelength 860 \(\mathrm{nm} .\) What is the internal energy of the atom after it absorbs the photon? (b) An atom initially in an energy level with \(E=-2.68\) eV emits a photon that has wavelength 420 \(\mathrm{nm} .\) What is the internal energy of the atom after it emits the photon?

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