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Chapter 29: Problem 44

Two sources produce electromagnetic waves. Source \(\mathrm{B}\) produces a wavelength that is three times the wavelength produced by source A. Each photon from source A has an energy of \(2.1 \times 10^{-18} \mathrm{J} .\) What is the energy of a photon from source \(\mathrm{B} ?\)

Short Answer

Expert verified
The energy of a photon from source B is \(0.7 \times 10^{-18} \text{ J}\."

Step by step solution

01

Understand the Relationship Between Wavelength and Energy

The energy of a photon is given by the formula: \( E = \frac{hc}{\lambda} \), where \( E \) is photon energy, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ m}^2 \text{ kg/s}) \), \( c \) is the speed of light \( (3 \times 10^8 \text{ m/s}) \), and \( \lambda \) is the wavelength.
02

Identify the Wavelengths of Sources A and B

Let \( \lambda_A \) be the wavelength of source A and \( \lambda_B \) be the wavelength of source B. It is given that \( \lambda_B = 3\lambda_A \).
03

Determine the Energy Equation for Source B

Given that the energy of a photon from source A is \( E_A = 2.1 \times 10^{-18} \text{ J} \), and \( E_A = \frac{hc}{\lambda_A} \), the energy of a photon from source B, \( E_B \), is \( E_B = \frac{hc}{3\lambda_A} \).
04

Relate the Energies of the Photons

Since \( E_B = \frac{hc}{3\lambda_A} \), we can write \( E_B = \frac{E_A}{3} \). This implies \( E_B = \frac{2.1 \times 10^{-18}}{3} \).
05

Calculate the Energy of a Photon from Source B

Compute \( E_B = \frac{2.1 \times 10^{-18}}{3} = 0.7 \times 10^{-18} \text{ J} \).
06

Verify the Solution

Each calculation is consistent with the relationship between wavelength and energy, confirming the energy of a photon from source B.

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Key Concepts

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

Wavelength
Wavelength is a fundamental concept when it comes to understanding electromagnetic waves. It refers to the distance between two consecutive peaks (or troughs) of a wave. It's commonly denoted by the Greek letter lambda (\( \lambda \)). In our exercise, the wavelength is crucial to determining the energy of photons from different sources.
  • Wavelength is inversely proportional to energy: as wavelength increases, photon energy decreases.
  • Unit of measurement for wavelength is meters (m).
To understand how wavelength affects photon energy, imagine stretching a wave. As the wave gets longer (wavelength increases), the energy becomes spread out more, leading to lower energy within each photon.
Planck's constant
Planck's constant (\( h \)) is a fundamental constant in quantum mechanics, crucial for calculating photon energies. It represents the proportionality factor between the energy of a photon and the frequency of the electromagnetic wave.
  • Value: \( 6.626 \times 10^{-34} \, \text{m}^2 \, \text{kg/s} \).
  • Relates energy (E) to frequency (\( u \)): \( E = hu \).
In the formula \( E = \frac{hc}{\lambda} \), Planck's constant helps link the speed of light and wavelength to determine the energy of a photon. Even though it鈥檚 a tiny number, Planck鈥檚 constant plays a huge role in the world of photons and waves.
Speed of Light
The speed of light (denoted as \( c \)) is a constant that represents the speed at which electromagnetic waves travel in a vacuum. Its value is approximately \( 3 \times 10^8 \, \text{m/s} \). This speed is not only incredibly fast, but it鈥檚 also essential for calculating photon energy.
  • Key value in the equation \( E = \frac{hc}{\lambda} \).
  • Makes the link between wavelength and the energy of a photon possible.
In photon energy calculations, the speed of light allows us to convert between the frequency, wavelength, and energy of light waves. Its constant nature ensures that the relationship among these quantities is stable and predictable.
Electromagnetic Waves
Electromagnetic waves are waves propagated through the electromagnetic field, encompassing a broad spectrum that includes visible light, radio waves, X-rays, and more. They carry energy and travel through space.
  • Comprised of oscillating electric and magnetic fields.
  • Do not require a medium to travel; can move through a vacuum.
The exercise focuses on calculating the energy of photons from sources emitting electromagnetic waves of different wavelengths. Understanding that these waves behave consistently according to principles like wavelength and frequency helps in unraveling how they carry energy and influence its quantification according to their properties.

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Most popular questions from this chapter

The minimum uncertainty \(\Delta y\) in the position \(y\) of a particle is equal to its de Broglie wavelength. Determine the minimum uncertainty in the speed of the particle, where this minimum uncertainty \(\Delta v_{y}\) is expressed as a percentage of the particle's speed \(v_{y}\left(\right.\) Percentage \(\left.=\frac{\Delta v_{y}}{v_{y}} \times 100 \%\right)\) Assume that relativistic effects can be ignored.

The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is \(485 \mathrm{nm} .\) What is the work function \(W_{0}\) of this metal? Express your answer in electron volts.

ssm An electron and a proton have the same kinetic energy and are moving at speeds much less than the speed of light. Concepts: (i) How is the de Broglie wavelength \(\lambda\) related to the magnitude \(p\) of the momentum? (ii) How is the magnitude of the momentum related to the kinetic energy of a particle of mass \(m\) that is moving at a speed that is much less than the speed of light? (iii) Which has the greater de Broglie wavelength, the electron or the proton? Calculations: Determine the ratio of the de Broglie wavelength of the electron to that of the proton.

A proton is confined to a nucleus that has a diameter of \(5.5 \times 10^{-15} \mathrm{m} .\) If this distance is considered to be the uncertainty in the position of the proton, what is the minimum uncertainty in its momentum?

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?
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