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Chapter 29: Problem 36

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 is 40.

Step by step solution

01

Define Kinetic Energy of the Particle

The kinetic energy (KE) of a particle with mass \(m\) moving at velocity \(v\) is given by \( \text{KE} = \frac{1}{2} m v^2 \). The particle moves at 5% the speed of light \(c\), so \(v = 0.05c\). Thus, \( \text{KE} = \frac{1}{2} m (0.05c)^2 = 0.00125 m c^2 \).
02

Define Energy of the Photon

The energy of a photon is given by \( E = \frac{hc}{\lambda_{\text{photon}}} \), where \(h\) is Planck's constant and \(\lambda_{\text{photon}}\) is the wavelength of the photon. Since the kinetic energy of the particle equals the energy of the photon, \(0.00125 m c^2 = \frac{hc}{\lambda_{\text{photon}}} \).
03

Find De Broglie Wavelength of the Particle

The de Broglie wavelength \(\lambda_{\text{particle}}\) of the particle is given by \( \lambda_{\text{particle}} = \frac{h}{p} \), where \(p\) is the momentum of the particle. The momentum \(p = mv\), so \(\lambda_{\text{particle}} = \frac{h}{m(0.05c)} = \frac{h}{0.05mc} \).
04

Relate Photon Energy to De Broglie Wavelength

Substitute the expression for \(m\) from Step 2 into Step 3. From \(0.00125 m c^2 = \frac{hc}{\lambda_{\text{photon}}}\), solve for \(m\): \(m = \frac{hc}{0.00125 c^2 \lambda_{\text{photon}}}\). Substitute this into \(\lambda_{\text{particle}} \): \(\lambda_{\text{particle}} = \frac{h \times 0.00125 c \lambda_{\text{photon}}}{0.05 hc^2}\).
05

Calculate Ratio of Wavelengths

Simplifying the expression from Step 4 gives \(\lambda_{\text{particle}} = 0.025 \lambda_{\text{photon}}\). The ratio of the photon wavelength to the de Broglie wavelength of the particle is \(\frac{\lambda_{\text{photon}}}{\lambda_{\text{particle}}} = \frac{1}{0.025} = 40\).

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It's a crucial concept in physics, especially when dealing with moving particles.
For a particle with a given mass \(m\) and velocity \(v\), the kinetic energy (KE) is calculated using the formula:
  • \( KE = \frac{1}{2} m v^2 \)
In the exercise, the particle is moving at 5% of the speed of light, which means \(v = 0.05c\). Substituting this into the equation gives:
  • \( KE = \frac{1}{2} m (0.05c)^2 \)
  • \( KE = 0.00125 m c^2 \)
This expression represents the kinetic energy of the particle related to its mass and speed, showing how even small speeds, relative to the speed of light, can significantly impact kinetic energy.
Exploring Photon Energy
Photon energy is the energy carried by a photon, the basic unit of light and all other forms of electromagnetic radiation.
Photon energy can be calculated using Planck's equation, which states:
  • \( E = \frac{hc}{\lambda_{\text{photon}}} \)
Here, \(h\) is Planck's constant and \(\lambda_{\text{photon}}\) is the wavelength of the photon.
In the exercise, the kinetic energy of the particle equals the energy of the photon, leading to the equation:
  • \( 0.00125 m c^2 = \frac{hc}{\lambda_{\text{photon}}} \)
This relationship allows us to connect the energy concepts of particles and photon wavelengths, ultimately setting the stage for comparing their respective wavelengths.
Analyzing the Wavelength Ratio
The wavelength ratio involves comparing the photon's wavelength to the de Broglie wavelength of the particle.
The de Broglie wavelength \(\lambda_{\text{particle}}\) is given by:
  • \( \lambda_{\text{particle}} = \frac{h}{p} \)
where \(p\) is the momentum of the particle (\(p = mv\)).
Substituting \(v = 0.05c\), we get:
  • \( \lambda_{\text{particle}} = \frac{h}{0.05mc} \)
To find the ratio of the wavelengths, we solve for \(m\) using the expression from the photon energy equation and substitute it back:
  • \( m = \frac{hc}{0.00125 c^2 \lambda_{\text{photon}}} \)
  • \( \lambda_{\text{particle}} = \frac{h \times 0.00125 c \lambda_{\text{photon}}}{0.05 hc^2} \)
After simplifying, we find:
  • \( \lambda_{\text{particle}} = 0.025 \lambda_{\text{photon}} \)
Therefore, the ratio of the photon's wavelength to the de Broglie wavelength is \(\frac{1}{0.025} = 40\).
This ratio highlights the comparative difference in the nature of particle waves versus photon waves.

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Most popular questions from this chapter

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 \(6.0 \times 10^{-5} \mathrm{m},\) 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 \(2.0 \mathrm{C}^{\circ},\) assuming that all the photons are absorbed by the glass.

Light is shining perpendicularly on the surface of the earth with an intensity of \(680 \mathrm{W} / \mathrm{m}^{2}\). Assuming that all the photons in the light have the same wavelength (in vacuum) of \(730 \mathrm{nm},\) determine the number of photons per second per square meter that reach the earth.

ssm A laser emits \(1.30 \times 10^{18}\) photons per second in a beam of light that has a diameter of \(2.00 \mathrm{mm}\) and a wavelength of \(514.5 \mathrm{nm}\). Determine (a) the average electric field strength and (b) the average magnetic field strength for the electromagnetic wave that constitutes the beam.

Particle A is at rest, and particle \(\mathrm{B}\) collides head-on with it. The collision is completely inelastic, so the two particles stick together after the collision and move off with a common velocity. The masses of the particles are different, and no external forces act on them. The de Broglie wavelength of particle \(\mathrm{B}\) before the collision is \(2.0 \times 10^{-34} \mathrm{m} .\) What is the de Broglie wavelength of the object that moves off after the collision?

The width of the central bright fringe in a diffraction pattern on a screen is identical when either electrons or red light (vacuum wavelength \(=661 \mathrm{nm}\) ) pass through a single slit. The distance between the screen and the slit is the same in each case and is large compared to the slit width. How fast are the electrons moving?
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