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Chapter 38: Problem 58

What are (a) the energy of a photon corresponding to wavelength \(1.00 \mathrm{nm},\) (b) the kinetic energy of an electron with de Broglie wavelength \(1.00 \mathrm{nm}\), (c) the energy of a photon corresponding to wavelength \(1.00 \mathrm{fm},\) and \((\mathrm{d})\) the kinetic energy of an electron with de Broglie wavelength \(1.00 \mathrm{fm} ?\)

Short Answer

Expert verified
(a) 1.99x10^-16 J, (b) 2.42x10^-19 J, (c) 1.99x10^-13 J, (d) 2.42x10^-13 J

Step by step solution

01

Calculate Energy of a Photon (1.00 nm)

The energy of a photon is calculated using the formula: \( E = \frac{hc}{\lambda} \). Here, \( h = 6.626 \times 10^{-34} \text{ Js} \) is Planck's constant and \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light. For a wavelength (\( \lambda \)) of 1.00 nm (which is 1.00 \( \times 10^{-9} \) m), we plug in the values:\[ E = \frac{(6.626 \times 10^{-34} \text{ Js})(3.00 \times 10^8 \text{ m/s})}{1.00 \times 10^{-9} \text{ m}} \approx 1.99 \times 10^{-16} \text{ J} \]
02

Calculate Kinetic Energy of Electron (1.00 nm)

The de Broglie wavelength is given by \( \lambda = \frac{h}{p} \), where \( p \) is momentum. For an electron, \( p = mv \) (mass \( m = 9.109 \times 10^{-31} \) kg). The kinetic energy \( K = \frac{p^2}{2m} \). Rearranging for \( p \), we have \( p = \frac{h}{\lambda} \). Substitute \( \lambda = 1.00 \times 10^{-9} \) m:\[ p = \frac{6.626 \times 10^{-34}}{1.00 \times 10^{-9}} \approx 6.626 \times 10^{-25} \text{ kg m/s} \]\[ K = \frac{(6.626 \times 10^{-25})^2}{2 \times 9.109 \times 10^{-31}} \approx 2.42 \times 10^{-19} \text{ J} \]
03

Calculate Energy of a Photon (1.00 fm)

Using the same formula for photon energy, but with \( \lambda = 1.00 \times 10^{-15} \) m (converting 1.00 fm to meters):\[ E = \frac{(6.626 \times 10^{-34} \text{ Js})(3.00 \times 10^8 \text{ m/s})}{1.00 \times 10^{-15} \text{ m}} \approx 1.99 \times 10^{-13} \text{ J} \]
04

Calculate Kinetic Energy of Electron (1.00 fm)

Using the de Broglie wavelength formula again with \( \lambda = 1.00 \times 10^{-15} \) m:\[ p = \frac{6.626 \times 10^{-34}}{1.00 \times 10^{-15}} \approx 6.626 \times 10^{-19} \text{ kg m/s} \]\[ K = \frac{(6.626 \times 10^{-19})^2}{2 \times 9.109 \times 10^{-31}} \approx 2.42 \times 10^{-13} \text{ J} \]

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Key Concepts

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

Planck's Constant
Planck's constant, denoted as \( h \), is a fundamental constant in physics that plays a crucial role in quantum mechanics. Its value is \( 6.626 \times 10^{-34} \text{ Js} \). This constant is instrumental in relating the energy of a photon to its frequency. The energy \( E \) of a photon is calculated through the equation:
  • \( E = hf \)
where \( f \) represents the frequency of the photon. Planck's constant also appears in other key relations, such as the de Broglie wavelength equation, showcasing its importance in describing particle-wave duality. Understanding this constant helps bridge the gap between classical and quantum physics.
de Broglie Wavelength
The concept of de Broglie wavelength arises from the idea that particles can exhibit wave-like behavior. Named after the French physicist Louis de Broglie, this wavelength is defined by the equation:
  • \( \lambda = \frac{h}{p} \)
where:
  • \( \lambda \) is the wavelength
  • \( h \) is Planck's constant
  • \( p \) is the momentum of the particle
This equation illustrates that the wavelength is inversely proportional to the particle's momentum. Thus, heavier particles moving at higher speeds exhibit shorter wavelengths. This concept is vital in understanding electron behavior at a quantum level, such as when calculating kinetic energy of particles.
Kinetic Energy
Kinetic energy represents the energy an object possesses due to its motion. For particles like electrons, this can be calculated using the formula:
  • \( K = \frac{p^2}{2m} \)
where:
  • \( p \) is the momentum
  • \( m \) is the mass of the particle
When expressed in terms of de Broglie wavelength as \( \lambda \), we can substitute \( p = \frac{h}{\lambda} \) to find kinetic energy:
  • \( K = \frac{h^2}{2m\lambda^2} \)
This equation helps in determining the kinetic energy of particles like electrons, using their wave-like properties, contributing to a better understanding of their behavior in quantum systems.
Electron
Electrons are subatomic particles with a negative charge that orbit the nucleus of an atom. They are fundamental constituents of matter and play a crucial role in electricity, magnetism, and thermal conductivity. The mass of an electron is approximately \( 9.109 \times 10^{-31} \text{ kg} \). Electrons can exhibit both wave-like and particle-like properties, which is where the concept of de Broglie wavelength becomes important. By calculating the kinetic energy of electrons through their de Broglie wavelength, we gain insight into their behavior at the nanoscale. This is essential in fields like quantum physics and nanotechnology.
Wavelength Calculation
Wavelength calculation involves determining the distance between consecutive peaks of a wave, which can be applied to both photons and particles. For photons, the energy \( E \) can be computed using:
  • \( E = \frac{hc}{\lambda} \)
where \( c \) is the speed of light. The same wavelength calculation concept is pivotal for particles through the de Broglie relation, revealing their dual wave-particle nature. Converting units is often necessary, such as translating nanometers to meters, in order to properly use these equations, ensuring accurate and meaningful results. Understanding wavelength allows us to explore and calculate various properties of electromagnetic waves and particles.

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Most popular questions from this chapter

(a) If the work function for a certain metal is \(1.8 \mathrm{eV}\), what is the stopping potential for electrons ejected from the metal when light of wavelength \(400 \mathrm{nm}\) shines on the metal? (b) What is the maximum speed of the ejected electrons?

A light detector (your eye) has an area of \(2.00 \times 10^{-6} \mathrm{~m}^{2}\) and absorbs \(80 \%\) of the incident light, which is at wavelength \(500 \mathrm{nm}\). The detector faces an isotropic source, \(3.00 \mathrm{~m}\) from the source. If the detector absorbs photons at the rate of exactly \(4.000 \mathrm{~s}^{-1}\), at what power does the emitter emit light?

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.)

A \(100 \mathrm{~W}\) sodium lamp \((\lambda=589 \mathrm{nm})\) radiates energy uniformly in all directions. (a) At what rate are photons emitted by the lamp? (b) At what distance from the lamp will a totally absorbing screen absorb photons at the rate of 1.00 photon \(/ \mathrm{cm}^{2} \cdot \mathrm{s} ?\) (c) What is the photon flux (photons per unit area per unit time) on a small screen \(2.00 \mathrm{~m}\) from the lamp?

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.
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