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The kinetic energy of a particle is equal to the energy of a photon. The particle moves at 5.0% of the speed of light. Find the ratio of the photon wavelength to the de Broglie wavelength of the particle.

Short Answer

Expert verified
The ratio of the photon's wavelength to the de Broglie wavelength of the particle is 0.1.

Step by step solution

01

Understand the Given Parameters

We are given that the kinetic energy of a particle is equal to the energy of a photon. The speed of the particle is 5.0% of the speed of light (denoted as \( c \)), which means \( v = 0.05c \). We need to find the ratio of the photon's wavelength to the de Broglie wavelength of the particle.
02

Write the Kinetic Energy of the Particle

The kinetic energy (KE) of a particle moving at speed \( v \) is given by:\[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass of the particle and \( v = 0.05c \).
03

Write the Energy of the Photon

The energy of a photon is given by the formula:\[ E_{photon} = \frac{hc}{\lambda_{photon}} \]where \( h \) is Planck's constant and \( \lambda_{photon} \) is the wavelength of the photon.
04

Set the Kinetic Energy of the Particle Equal to the Energy of the Photon

According to the problem, the kinetic energy of the particle equals the energy of the photon:\[ \frac{1}{2} mv^2 = \frac{hc}{\lambda_{photon}} \]Solving for \( \lambda_{photon} \), we get:\[ \lambda_{photon} = \frac{2hc}{mv^2} \]
05

Determine de Broglie Wavelength of the Particle

The de Broglie wavelength \( \lambda_{dB} \) for a particle is given by:\[ \lambda_{dB} = \frac{h}{mv} \]
06

Calculate the Ratio of Wavelengths

The required ratio is \( \frac{\lambda_{photon}}{\lambda_{dB}} \):\[ \frac{\lambda_{photon}}{\lambda_{dB}} = \frac{\frac{2hc}{mv^2}}{\frac{h}{mv}} \]Simplifying, we have:\[ \frac{\lambda_{photon}}{\lambda_{dB}} = 2v \]Substituting \( v = 0.05c \), we get:\[ \frac{\lambda_{photon}}{\lambda_{dB}} = 2(0.05c) = 0.1c/c = 0.1 \]
07

Final Calculation

Since \( c/c = 1 \), the ratio becomes:\[ \frac{\lambda_{photon}}{\lambda_{dB}} = 0.1 \]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It depends on both the mass of an object and the square of its velocity.
The formula for kinetic energy, specifically for objects moving at non-relativistic speeds (speeds much less than light speed), is given by: \[ KE = \frac{1}{2} mv^2 \] where:
  • \( KE \) is the kinetic energy,
  • \( m \) stand for the mass of the object,
  • \( v \) is the velocity (in this scenario, \( v = 0.05c \) where \( c \) is the speed of light).
When a particle moves, it uses some of its internal energy to move, which we can measure using the above kinetic energy formula. For the particle in the exercise, the kinetic energy is equated to the photon energy to find the desired wavelength ratios.
Photon Energy
Photon energy pertains to the energy carried by a single photon, the basic unit of light and all other forms of electromagnetic radiation. It is an essential concept in understanding quantum physics. The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength. The formula is given by: \[ E_{photon} = \frac{hc}{\lambda_{photon}} \] where:
  • \( E_{photon} \) stands for photon energy,
  • \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) Js),
  • \( c \) is the speed of light in a vacuum (approximately \( 3.00 \times 10^8 \) m/s),
  • \( \lambda_{photon} \) is the wavelength of the photon.
This formula helps bridge the classical and quantum worlds by linking the wavelength of light to its energy. Understanding photon energy is crucial in technologies like solar power, lasers, and various quantum mechanisms.
Wavelength Ratio
The wavelength ratio is a comparative measure between two different types of wavelengths. In our scenario, this involves comparing the photon's wavelength with the de Broglie wavelength of the particle. Each of these wavelengths is determined by separate principles, but their comparison is instrumental for unraveling quantum mechanical phenomena. The photon's wavelength, based on its energy, is determined using the photon energy formula: \[ \lambda_{photon} = \frac{2hc}{mv^2} \] The de Broglie wavelength for a particle is derived from its momentum: \[ \lambda_{dB} = \frac{h}{mv} \] To find their ratio, the expression becomes: \[ \frac{\lambda_{photon}}{\lambda_{dB}} = \frac{2v}{v} = 0.1 \] This result shows that the photon's wavelength is a tenth of the de Broglie wavelength, given the conditions of the movement speed of the particle. Wavelength ratios like these are critical in understanding how particles behave like waves, fundamental in the world of quantum mechanics.

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Most popular questions from this chapter

In a Young's double-slit experiment that uses electrons, the angle that locates the first-order bright fringes is \(\theta_{\mathrm{A}}=1.6 \times 10^{-4}\) degrees when the magnitude of the electron momentum is \(p_{\mathrm{A}}=1.2 \times 10^{-22} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) . With the same double slit, what momentum magnitude \(p_{\mathrm{B}}\) is necessary so that an angle of \(\theta_{\mathrm{B}}=4.0 \times 10^{-4}\) degrees locates the first-order bright fringes?

What is the maximum amount by which the wavelength of an incident photon could change when it undergoes Compton scattering from a nitrogen molecule \(\left(\mathrm{N}_{2}\right) ?\)

A glass plate has a mass of 0.50 \(\mathrm{kg}\) and a specific heat capacity of 840 \(\mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) The wavelength of infrared light is \(6.0 \times 10^{-5} m\) , while the wavelength of blue light is \(4.7 \times 10^{-7} \mathrm{m}\) . Find the number of infrared photons and the number of blue photons needed to raise the temperature of the glass plate by \(2.0 \mathrm{C}^{\prime},\) assuming that all the photons are absorbed by the glass.

Particles pass through a single slit of width 0.200 mm (see Figure 29.14). The de Broglie wavelength of each particle is 633 nm. After the particles pass through the slit, they spread out over a range of angles. Assume that the uncertainty in the position of the particles is one- half the width of the slit, and use the Heisenberg uncertainty principle to determine the minimum range of angles.

A particle has a de Broglie wavelength of \(2.7 \times 10^{-10} \mathrm{m} .\) Then its kinetic energy doubles. What is the particle's new de Broglie wavelength, assuming that relativistic effects can be ignored?

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