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Chapter 3: Problem 51

The average intensity of an electromagnetic wave is \(\frac{1}{2} s_{0} c E_{0}^{2}\). where \(E_{0}\) is the amplitude of the electric-field portion of the wave. Find a general expression for the photon flux \(j\) (measured in photons/s \(-m^{2}\) ) in terms of \(E_{0}\) and wavelength \(\lambda .\)

Short Answer

Expert verified
The general expression for the photon flux \(j\) in terms of \(E_{0}\) and wavelength \(\lambda\) is \(j = \frac{E_{0}^{2}}{2 h \lambda }\).

Step by step solution

01

Understand the Units and Variables

Firstly, it is important to understand the variables in the equation and their units. The symbol \(s_{0}\) represents the average energy per photon, \(c\) is the speed of light, and \(E_{0}^{2}\) is the square of the amplitude of the electric-field portion of the wave. Photon flux \(j\) is the quantity to be found, which is the number of photons per unit time per unit area (photons/s-\(m^{2}\)).
02

Write the Given Formula

The given formula for the average intensity of an electromagnetic wave is \(\frac{1}{2} s_{0} c E_{0}^{2}\) .
03

Write the Energy of a Single Photon

The energy of a single photon can be given as \(s_{0} = h c / \lambda\), where \(h\) is Plank's constant and \(\lambda\) is the wavelength of the photon.
04

Replace s0 in the intensity equation

Replace \(s_{0}\) with its expression in the intensity equation, which gives intensity \(I = \frac{1}{2} (h c / \lambda) E_{0}^{2}\) c . Simplifying the equation gives \(I = \frac{1}{2} (h E_{0}^{2}) / \lambda \) .
05

Write the expression of photon flux

The photon flux can be understood as the intensity divided by the energy per photon. Therefore, \(j = I/s_{0}\). Replace \(I\) and \(s_{0}\) with their respective expression we derived, \(j = \frac{\frac{1}{2} (h E_{0}^{2})}{h c / \lambda} / \lambda\). After simplifying this equation, it gives the photon flux \(j = \frac{E_{0}^{2}}{2 h \lambda }\).

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Key Concepts

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

Electromagnetic Wave Intensity
Understanding electromagnetic wave intensity is essential for grasping the principles of photon flux. Simply put, the intensity of an electromagnetic wave reflects the amount of energy that the wave carries per unit area per second. It's an expression of power distributed over a surface area, and it relates directly to how strong or weak an electromagnetic wave is.

When the intensity equation is provided as \( I = \frac{1}{2} \rho_{0} c E_{0}^{2} \), it incorporates the amplitude of the electric-field \( E_{0} \), the speed of light \( c \), and a constant \( \rho_{0} \) that links all these values together. The square of the electric-field amplitude \( E_{0}^{2} \) is especially important, as this term determines how changes in the electric field's strength affect the overall intensity. In essence, if you increase the electric-field amplitude, the intensity of the wave grows, and this is crucial when it comes to calculating the photon flux.

Please note, in context with the textbook, \( s_{0} \) represents the average energy per photon which is a typographic error, the correct symbol should be \( \rho_{0} \).
Electric-Field Amplitude
At the heart of an electromagnetic wave, electric-field amplitude \( E_{0} \) plays a pivotal role. It's the maximum strength of the electric field at a point in space, determining how powerful the influence of the electric field is at that spot.

Why is \( E_{0} \) significant for understanding photon flux? The term \( E_{0}^{2} \) in the intensity formula is directly proportional to the energy carried by the wave. A higher electric-field amplitude means a stronger electric field, and thus more energy is transmitted through the wave. For students, picture the amplitude like the height of a wave in the ocean - the taller the wave, the more energy it has.

Therefore, when calculating the photon flux, which involves the number of photons passing through a certain area, it's pivotal to note the electric-field amplitude. A higher amplitude results in more energetic waves, and thus, a greater number of photons passing through a unit area in a second.
Wavelength
Wavelength \( \lambda \) is a fundamental property of both visible light and all other electromagnetic waves. It refers to the distance between consecutive crests or troughs in a wave. It is inversely related to the frequency and energy of the photons that make up the wave.

In the context of photon flux, wavelength is a critical factor that helps you understand the nature of the photons being studied. Shorter wavelengths correspond to more energetic photons, since the energy of a photon is given by \( E = h c / \lambda \), where \( h \) is Planck's constant, and \( c \) is the speed of light. When you're working through photon flux problems, keep in mind how wavelength affects the calculation: a shorter wavelength indicates higher photon energy and, consequently, changes the number of photons that make up the wave's intensity.

This relationship is utilized when deducing formulas for photon flux, as seen when we replace the energy per photon \( s_{0} \) in terms of wavelength - providing a direct way to link the concept of wavelength with the measurable quantity of photon flux.

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Most popular questions from this chapter

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